Business Mathematics Code 1429

2. Contents UNIT- 1 PROBABILITY THEORY 1 1.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 OBJECTIVES . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 BASIC PROBABILTY THEORY . . . . . . . . . . . . . . . 3 1.4 DEFINITIONS . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.5 INDEPENDENT AND DEPENDENT EVENTS . . . . . . 5 1.5.1 Events . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.5.2 Mutually Exclusive Events . . . . . . . . . . . . . . . 5 1.5.3 Collectively Exhaustive Events . . . . . . . . . . . . 6 1.5.4 Relative Frequency . . . . . . . . . . . . . . . . . . . 9 1.6 RULES OF PROBABILITY . . . . . . . . . . . . . . . . . . 11 1.7 INDEPENDENT EVENTS . . . . . . . . . . . . . . . . . . 17 1.7.1 Marginal Probability . . . . . . . . . . . . . . . . . . 17 1.7.2 Joint Probability . . . . . . . . . . . . . . . . . . . . 17 1.8 LAWS OF PROBABILITY . . . . . . . . . . . . . . . . . . 20 1.8.1 Conditional Probability . . . . . . . . . . . . . . . . 20 1.8.2 Dependent Events . . . . . . . . . . . . . . . . . . . . 21 1.9 SELF ASSESMENT QUESTIONS . . . . . . . . . . . . . . 25 UNIT- 2 RANDOM VARIABLES 27 2.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . 28 2.2 OBJECTIVES . . . . . . . . . . . . . . . . . . . . . . . . . 28 ix

3. 2.3 DISCRETE AND CONTINUOUS RANDOM VARIABLES 31 2.3.1 Discrete Random Variables . . . . . . . . . . . . . . . 31 2.3.2 Continuous Random Variables . . . . . . . . . . . . . 31 2.4 PROBABILITY DISTRIBUTIONS . . . . . . . . . . . . . . 34 2.4.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . 34 2.4.2 Discrete Probability Distribution . . . . . . . . . . . 34 2.5 PROBABILITY DISTRIBUTION PROPERTIES . . . . . . 36 2.6 APPLICATIONS . . . . . . . . . . . . . . . . . . . . . . . . 41 2.7 SELF ASSESMENT QUESTIONS . . . . . . . . . . . . . . 42 UNIT- 3 EQUATIONS 44 3.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . 45 3.2 OBJECTIVES . . . . . . . . . . . . . . . . . . . . . . . . . 45 3.3 SOLVING FIRST-DEGREE EQUATIONS IN ONE VARI- ABLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 3.3.1 Equations and their Properties . . . . . . . . . . . . 46 3.3.2 Types of Equations . . . . . . . . . . . . . . . . . . . 46 3.4 SOLVING SECOND-DEGREE EQUATIONS IN ONE VARI- ABLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 3.4.1 Methods to Solve Quadratic Equations . . . . . . . . 48 3.5 INEQUALITIES AND THEIR SOLUTIONS . . . . . . . . . 52 3.5.1 Inequalities . . . . . . . . . . . . . . . . . . . . . . . 52 3.5.2 Types of Inequalities . . . . . . . . . . . . . . . . . . 52 3.6 INTERVAL NOTATION . . . . . . . . . . . . . . . . . . . . 53 3.6.1 Solving Inequalities . . . . . . . . . . . . . . . . . . . 53 3.6.2 Second-Degree Inequalities . . . . . . . . . . . . . . . 54 3.7 ABSOLUTE VALUE . . . . . . . . . . . . . . . . . . . . . 56 3.7.1 Some Properties of Absolute Values . . . . . . . . . . 57 x

4. 3.7.2 Solving equations and inequalities involving absolute values . . . . . . . . . . . . . . . . . . . . . . . . . . 57 3.8 RECTANGULAR COORDINATE SYSTEM . . . . . . . . . 60 3.8.1 Applications . . . . . . . . . . . . . . . . . . . . . . . 60 3.8.2 Rectangular Coordinates . . . . . . . . . . . . . . . . 60 3.8.3 The Cartesian Coordinates . . . . . . . . . . . . . . . 61 3.8.4 The Midpoint Formula . . . . . . . . . . . . . . . . . 62 3.8.5 The Distance Formula . . . . . . . . . . . . . . . . . 62 3.9 SELF ASSESMENT QUESTIONS . . . . . . . . . . . . . . 64 UNIT- 4 LINEAR EQUATIONS 66 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 4.2 OBJECTIVES . . . . . . . . . . . . . . . . . . . . . . . . . 67 4.3 CHARACTERISTICS OF LINEAR EQUATIONS . . . . . . 69 4.4 REPRESENTATION USING LINEAR EQUATIONS . . . . 69 4.5 LINEAR EQUATIONS WITH n VARIABLES . . . . . . . . 74 4.6 GRAPHICAL CHARACTERISTICS . . . . . . . . . . . . . 74 4.6.1 Graphing Two-Variable Equations . . . . . . . . . . . 74 4.7 INTERCEPTS . . . . . . . . . . . . . . . . . . . . . . . . . 76 4.7.1 x−Intercept . . . . . . . . . . . . . . . . . . . . . . . 76 4.7.2 y−Intercept . . . . . . . . . . . . . . . . . . . . . . . 77 4.8 SLOPE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 4.9 TWO POINT FORMULA . . . . . . . . . . . . . . . . . . . 80 4.10 SLOPE INTERCEPT FORM . . . . . . . . . . . . . . . . . 82 4.11 DETERMINING THE EQUATION OF A STRAIGHT LINE 83 4.11.1 Slope and Intercept Form . . . . . . . . . . . . . . . 83 4.11.2 Point and Slope Form . . . . . . . . . . . . . . . . . 84 4.11.3 Point-Slope Formula . . . . . . . . . . . . . . . . . . 84 4.11.4 Two Points Form . . . . . . . . . . . . . . . . . . . . 85 xi

5. 4.11.5 Parallel Lines and Perpendicular Lines . . . . . . . . 87 4.11.6 Supply and Demand Analysis . . . . . . . . . . . . . 88 4.12 LINEAR EQUATIONS INVOLVING MORE THAN TWO VARIABLES . . . . . . . . . . . . . . . . . . . . . . . . . . 90 4.12.1 Three-Dimensional Coordinate Systems . . . . . . . . 90 4.12.2 Equations Involving Three Variables . . . . . . . . . 91 4.12.3 Equations Involving more than three Variables . . . . 91 4.13 SELF ASSESMENT QUESTIONS . . . . . . . . . . . . . . 93 UNIT- 5 MATRICES 95 5.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . 96 5.2 OBJECTIVES . . . . . . . . . . . . . . . . . . . . . . . . . 96 5.3 INTRODUCTION TO MATRICES . . . . . . . . . . . . . . 97 5.3.1 Generalized Form of a Matrix . . . . . . . . . . . . . 97 5.3.2 Purpose of Studying Matrix Algebra . . . . . . . . . 97 5.4 SPECIAL TYPES OF MATRICES . . . . . . . . . . . . . . 97 5.4.1 Vector . . . . . . . . . . . . . . . . . . . . . . . . . . 97 5.4.2 Row Vector . . . . . . . . . . . . . . . . . . . . . . . 97 5.4.3 Column Vector . . . . . . . . . . . . . . . . . . . . . 98 5.4.4 Square Matrices . . . . . . . . . . . . . . . . . . . . . 98 5.4.5 Identity Matrix . . . . . . . . . . . . . . . . . . . . . 98 5.4.6 Transpose . . . . . . . . . . . . . . . . . . . . . . . . 98 5.5 MATRICES OPERATIONS . . . . . . . . . . . . . . . . . . 99 5.5.1 Matrix Addition and Subtraction . . . . . . . . . . . 99 5.5.2 Scalar Multiplication . . . . . . . . . . . . . . . . . . 100 5.5.3 Inner Product . . . . . . . . . . . . . . . . . . . . . . 100 5.5.4 Matrix Multiplication . . . . . . . . . . . . . . . . . . 101 5.6 REPRESENTATION OF AN EQUATION . . . . . . . . . 103 5.7 REPRESENTATION OF A SYSTEM OF EQUATIONS . . 103 xii

6. 5.8 SELF ASSESMENT QUESTIONS . . . . . . . . . . . . . . 107 UNIT- 6 DETERMINANTS AND INVERSES OF MATRICES 109 6.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . 110 6.2 OBJECTIVES . . . . . . . . . . . . . . . . . . . . . . . . . 110 6.3 THE DETERMINANT . . . . . . . . . . . . . . . . . . . . 111 6.3.1 Determinant of a (1 × 1) Matrix . . . . . . . . . . . 111 6.3.2 Determinant of a (2 × 2) Matrix . . . . . . . . . . . . 111 6.3.3 Determinant of (3 × 3) Matrix . . . . . . . . . . . . 111 6.4 METHOD OF COFACTORS . . . . . . . . . . . . . . . . . 113 6.4.1 Method of Cofactors Expansion . . . . . . . . . . . . 114 6.4.2 Adjoint of a Matrix of Order n ≥ 3 . . . . . . . . . . 117 6.5 PROPERTIES OF DETERMINANTS . . . . . . . . . . . . 117 6.6 CRAMER’s RULE . . . . . . . . . . . . . . . . . . . . . . . 118 6.7 INVERSE OF A MATRIX . . . . . . . . . . . . . . . . . . . 121 6.7.1 Some Facts Regarding the Inverse . . . . . . . . . . . 122 6.7.2 Determining the Inverse . . . . . . . . . . . . . . . . 122 6.7.3 Gaussian Reduction Procedure . . . . . . . . . . . . 122 6.7.4 Finding the Inverse Using the Cofactors . . . . . . . 123 6.7.5 The Inverse and System of Equations . . . . . . . . . 125 6.8 SELF ASSESMENT QUESTIONS . . . . . . . . . . . . . . 131 UNIT- 7 DIFFERENTIATION 133 7.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . 134 7.2 OBECTIVES . . . . . . . . . . . . . . . . . . . . . . . . . . 134 7.3 LIMITS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 7.3.1 Limits of Function . . . . . . . . . . . . . . . . . . . 135 7.3.2 Test for Existence of a Limit . . . . . . . . . . . . . . 135 7.4 PROPERTIES OF LIMIT . . . . . . . . . . . . . . . . . . . 139 7.4.1 Some Properties of Limit . . . . . . . . . . . . . . . . 139 xiii

7. 7.4.2 Limit at Infinity . . . . . . . . . . . . . . . . . . . . . 140 7.4.3 Continuity Over an Interval . . . . . . . . . . . . . . 140 7.5 HORIZONTAL AND VERTICAL ASYMPOTOTES . . . . 142 7.5.1 Horizontal Asymptote . . . . . . . . . . . . . . . . . 142 7.5.2 Vertical Asymptote . . . . . . . . . . . . . . . . . . 142 7.6 AVERAGE RATE OF CHANGE . . . . . . . . . . . . . . . 143 7.6.1 Average Rate of Change and Slope . . . . . . . . . . 143 7.7 DERIVATIVES . . . . . . . . . . . . . . . . . . . . . . . . . 144 7.7.1 Instantaneous Rate of Change . . . . . . . . . . . . 144 7.7.2 Limit Approach for the Derivatives . . . . . . . . . . 144 7.8 RULES OF DIFFERENTIATION . . . . . . . . . . . . . . . 147 7.8.1 Additional Rules of Differentiation . . . . . . . . . . 150 7.8.2 Average Cost . . . . . . . . . . . . . . . . . . . . . . 152 7.8.3 Consumption Function . . . . . . . . . . . . . . . . . 154 7.8.4 Marginal Revenue Product . . . . . . . . . . . . . . . 155 7.9 INSTANTANEOUS RATE OF CHANGE AND THE IN- TERPRETATION . . . . . . . . . . . . . . . . . . . . . . . 158 7.10 HIGHER ORDER DERIVATIVES . . . . . . . . . . . . . . 159 7.10.1 The Second Derivative . . . . . . . . . . . . . . . . . 159 7.10.2 Third and Higher Order Derivatives . . . . . . . . . . 160 7.11 DIFFERENTIATION OF EXPLICIT AND IMPLICIT FUNC- TIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 7.11.1 Explicit Function . . . . . . . . . . . . . . . . . . . . 160 7.11.2 Implicit Function . . . . . . . . . . . . . . . . . . . . 161 7.12 MAXIMA AND MINIMA WITH APPLICATIONS . . . . . 163 7.12.1 Absolute Maximum . . . . . . . . . . . . . . . . . . . 163 7.12.2 Absolute Minimum . . . . . . . . . . . . . . . . . . . 163 7.13 APPLICATIONS OF DERIVATIVES . . . . . . . . . . . . 165 7.14 SELF ASSESMENT QUESTIONS . . . . . . . . . . . . . . 167 xiv

8. UNIT- 8 PARTIAL DERIVATIVES 170 8.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . 171 8.2 OBJECTIVES . . . . . . . . . . . . . . . . . . . . . . . . . 171 8.3 GEOMETRIC INTERPRETATION OF PARTIAL DERIVA- TIVES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 8.4 PARTIAL DERIVATIVES WITH RESPECT TO x . . . . . 172 8.5 PARTIAL DERIVATIVES WITH RESPECT TO y . . . . . 173 8.6 MAXIMA AND MINIMA OF FUNCTIONS OF MULTI- VARIABLES . . . . . . . . . . . . . . . . . . . . . . . . . . 174 8.6.1 The Second Partial Derivative Test . . . . . . . . . . 174 8.7 CRITICAL AND SADDLE POINTS . . . . . . . . . . . . . 175 8.7.1 Critical Point . . . . . . . . . . . . . . . . . . . . . . 175 8.7.2 Saddle Point . . . . . . . . . . . . . . . . . . . . . . . 175 8.8 APPLICATIONS OF PARTIAL DERIVATIVES . . . . . . 179 8.9 SELF ASSESMENT QUESTIONS . . . . . . . . . . . . . . 183 UNIT- 9 OPTIMIZATION 185 9.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . 186 9.2 OBJECTIVES . . . . . . . . . . . . . . . . . . . . . . . . . 186 9.3 INCREASING AND DECREASING FUNCTIONS . . . . . 187 9.4 CONCAVITY . . . . . . . . . . . . . . . . . . . . . . . . . . 188 9.5 FIRST DERIVATIVE TEST . . . . . . . . . . . . . . . . . . 192 9.6 SECOND DERIVATIVE TEST . . . . . . . . . . . . . . . . 192 9.7 CURVE SKETCHING . . . . . . . . . . . . . . . . . . . . . 192 9.7.1 Key Data Points . . . . . . . . . . . . . . . . . . . . 193 9.8 REVENUE, COST AND PROFIT APPLICATIONS IN BUSI- NESS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 9.8.1 Revenue . . . . . . . . . . . . . . . . . . . . . . . . . 197 9.8.2 Cost . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 xv

9. 9.8.3 Variable Cost . . . . . . . . . . . . . . . . . . . . . . 199 9.8.4 Fixed Cost . . . . . . . . . . . . . . . . . . . . . . . . 199 9.8.5 Profit Function . . . . . . . . . . . . . . . . . . . . . 200 9.8.6 Marginal Approach to Profit Maximization . . . . . . 201 9.9 SELF ASSESMENT QUESTIONS . . . . . . . . . . . . . . 205 10 BIBLIOGRAPHY 207 xvi

10. List of Tables 1.1 Number of Males and Females with Different Degrees (Cour- tesy F.S. Budnick) . . . . . . . . . . . . . . . . . . . . . . . 7 1.2 Number of Males and Females with Different Degrees . . . . 9 1.3 Probabilities of the Selected Applicants with Different Char- acteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1.4 Probability of Number of Alarms Pulled . . . . . . . . . . . 13 2.1 Sample Space for Events E . . . . . . . . . . . . . . . . . . . 31 2.2 Discrete Probability Distribution . . . . . . . . . . . . . . . 34 2.3 Random Variable and Probability Distribution of Event S . 35 2.4 Probability Distribution of Number of Radios in a Household 36 2.5 Probability Distribution of White Balls . . . . . . . . . . . . 36 2.6 Probability Distribution Function of White Balls . . . . . . . 37 2.7 Frequency Distribution for the Major Snowfalls of the Last 60 Years . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 2.8 Probability Distribution for Snowfalls . . . . . . . . . . . . . 38 2.9 Probability Distribution Function for Head . . . . . . . . . . 39 2.10 Probability Distribution of 500 Customers . . . . . . . . . . 39 2.11 Frequency Distribution of Number of False Alarms . . . . . 42 4.1 Solution Set of Linear Equation y = 2x + 1 . . . . . . . . . . 70 4.2 Total Cost at Varying Level of Output . . . . . . . . . . . . 70 4.3 Solution Set of the Linear Equation 2x − 3y = 12 . . . . . . 75 xvii

11. 9.1 Behaviour of the Function f(x) on Different Intervals . . . . 188 9.2 Behaviour of the Function f(x) on Different Intervals . . . . 190 9.3 Behaviour of the function f(x) on Different Intervals . . . . 191 9.4 Second Derivative Test of f . . . . . . . . . . . . . . . . . . 196 9.5 First Derivative Test of f . . . . . . . . . . . . . . . . . . . . 196 9.6 Second Derivative Test of c . . . . . . . . . . . . . . . . . . . 204 xviii

12. List of Figures 1.1 Venn Diagram of Events . . . . . . . . . . . . . . . . . . . . 5 1.2 Venn Diagram of Mutually Exclusive Events . . . . . . . . . 6 1.3 Venn Diagram of not Mutually Exclusive Events . . . . . . . 6 1.4 Venn Diagram for Rule 3 . . . . . . . . . . . . . . . . . . . . 12 1.5 Venn Diagram for Rule 5 . . . . . . . . . . . . . . . . . . . . 12 1.6 Probability Tree Diagram for Coin Toss (Courtesy F.S. Bud- nick) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.1 Probability Tree Diagram(Courtesy F.S. Budnick) . . . . . 30 3.1 Representation on the Number Line . . . . . . . . . . . . . . 55 3.2 Representation on the Number Line . . . . . . . . . . . . . . 55 3.3 Absolute Value Representation . . . . . . . . . . . . . . . . . 56 3.4 Representation on the Number Line . . . . . . . . . . . . . . 58 3.5 Representation on the Number Line . . . . . . . . . . . . . . 58 3.6 Representation on the Number Line . . . . . . . . . . . . . . 59 3.7 Representation on the Number Line . . . . . . . . . . . . . . 59 3.8 Lorenz Curve for Cumulative Income and Household . . . . 60 3.9 Cartesian Plane or Cartesian Coordinate System . . . . . . . 61 3.10 Four Quadrants of the Cartesian Plane . . . . . . . . . . . . 62 3.11 Distance Formula Using Pythagoras Theorem . . . . . . . . 63 4.1 Graph of the Linear Equation y = 2x + 1 . . . . . . . . . . . 70 4.2 Graphical Representation of Linear Equation . . . . . . . . . 71 xix

13. 4.3 Graphical Representation of Linear Equation . . . . . . . . . 72 4.4 Graphical Representation of Linear Equation . . . . . . . . . 76 4.5 x− Intercept at x = 3 . . . . . . . . . . . . . . . . . . . . . 77 4.6 y− Intercept at y = 3 . . . . . . . . . . . . . . . . . . . . . . 78 4.7 Types of Slope . . . . . . . . . . . . . . . . . . . . . . . . . 79 4.8 Mathematical Representation of Slope . . . . . . . . . . . . 80 4.9 Representation of Slope . . . . . . . . . . . . . . . . . . . . . 81 4.10 Graphical Representation of Linear Equation . . . . . . . . . 83 4.11 Graphical Representation of Supply and Demand . . . . . . 89 4.12 Three-Dimensional Coordinate System . . . . . . . . . . . . 90 4.13 Graphical Representation of Given Equation . . . . . . . . . 92 6.1 Cars Manufacturing Industry (Courtesy Budnick) . . . . . . 125 7.1 Graphical Representation of Limit . . . . . . . . . . . . . . . 136 7.2 Graphical Representation of Limit . . . . . . . . . . . . . . . 137 7.3 Graphical Representation of f(x) . . . . . . . . . . . . . . . 141 7.4 Horizontal and Vertical Asymptotes . . . . . . . . . . . . . . 142 7.5 Graphical Representation of Maximum Revenue . . . . . . . 153 8.1 Geometric Interpretation of Partial Derivatives . . . . . . . . 172 9.1 Graphical Representation of Increasing and Decreasing Func- tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 9.2 Graphical Representation of the Given Function . . . . . . . 189 9.3 Concavity and Point of Inflection . . . . . . . . . . . . . . . 189 9.4 Graphical Representation of the Increasing, Decreasing, Con- cave up, Concave down and Point of Inflection . . . . . . . . 191 9.5 Graphical Representation of Maxima, Minima and Point of Inflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 9.6 Graphical Representation of Relative Extrema . . . . . . . . 197 xx

14. 9.7 Graphical Representation of Quadratic Revenue Function . . 198 9.8 Graphical Representation of Quadratic Revenue Function . . 201 xxi

15. Unit-1 PROBABILITY THEORY WRITTEN BY: DR NASIR REHMAN REVIEWED BY: DR BABAR AHMED 1

16. 1.1 INTRODUCTION Probability is the branch of mathematics that deals with the random events, collection, analysis, interpretation and display of numerical data. Proba- bility has its origin in the study of chancing and insurance in the 17th century, and it is now an essential tool of both social and natural sciences. It is a part of mathematics that enhances the subject as a whole by its interactions with other uses of mathematics. In this unit students will be able to learn about. 1. The concept of basic probability theory 2. Important definitions related to probability 3. Differentiate between the independent and dependent events 4. Rules of probability 5. Marginal probability and joint probability 6. Laws of probability 1.2 OBJECTIVES After studying this unit, students will be able to 1. explain basic probability theory ideas 2. apply these concepts to solve practical problems 3. understand the laws of probability from the applications point of view 2

17. The word of probability is associated with the random processes and random experiments. In other words, we can say it is a sequence of different processes, statements, experiments, trials which results in one of the number of different possible outcomes. 1.3 BASIC PROBABILTY THEORY Basically, we call the triplet (Ω, F, P) as the probability space. In this triplet, every member has its own importance and explanations. Here we have the definitions and details of the triplet. 1.4 DEFINITIONS 1. Ω : It is read as omega and called the Sample Space. The sample space for an experiment is the set of outcomes S such that any experiment or trial results in one and only one element of the set S. Where each element in the set S is referred as an outcome of the experiment. Example 1 Suppose we flip the coin twice and note down the outcome whether it is head or tail than the sample space is given by Ω = {HH, HT, TH, TT} Example 2 Suppose you make a phone call to your friend and note down the duration of call. Maximum time allowed is 5 minutes. Then what can be the duration of the call. Here the sample space will be any time t [0, 5] i.e. duration can be any time from 0 minute to 5 minutes so we say that sample space is infinite. Example 3 Find the sample space for selecting a prime number less than 15 at random. Solution 3

18. Ω = {2, 3, 5, 7, 11, 13} Example 4 Find the sample space for selecting one letter from the word MATHEMATICS. Solution Ω = {M, A, T, H, E, I, C, S} 2. F : It is called the σ-field and read as Sigma-field it has the following properties. • φ, Ω F • If A is an event which belongs to F then Ac F • If Ai are events that belong to F then ∪∞ i=1Ai F i.e. their infinite union also belongs to the σ− field F So, we can say that we arrange the sample space in a special order to get the σ− field F. Example 5 Let an event A Ω then we have the following σ− field F i.e. F = {φ, Ω, A, Ac } Hence is this case we have four members in the σ− field F. Example 6 If S = {1, 2, 3} then F = φ, {1} , {2} , {3} , {1, 2} , {1, 3} , {2, 3} , {1, 2, 3} 3. P : It is called the probability measure and it is a function f : Ω → [0, 1] i.e. it is a real valued function with range of [0, 1] taking any value in the interval [0, 1]. If the probability of some events E is 0 it means it is impossible event and if the probability of some event E in the sample 4

19. space Ω is 1, it means that occurrence of this event E is quite sure. And other probabilities lie between these values 0 and 1 i.e. 0 ≤ P(E) ≤ 1 ∀ E Ω 1.5 INDEPENDENT AND DEPENDENT EVENTS Two events are called to be independent if the probability of occurrence of one event is not affected by the occurrence or nonoccurence of the other event. Whereas if the probability of occurrence of one event is affected by the occurrence or nonoccurence of the other event this is known as dependent event. 1.5.1 Events A subset of a sample space Ω is called an event. We can denote it by E and explain by using the Venn Diagram. Figure 1.1: Venn Diagram of Events 1.5.2 Mutually Exclusive Events A set of events is said to be mutually exclusive if the occurrence of any one event precludes the occurrence of another event. i.e. the events E1, E2, ..., En are called mutually exclusive if Ei ∩ Ej = φ if i 6= j ∀ i, j = 1, 2, ....n Example 7 Consider the example of flipping a coin, the possible outcomes 5

20. are head and tail. Since the occurrence of head precludes the occurrence of tail and similarly the occurrence of tail precludes the occurrence of head. So, these two events are mutually exclusive. Figure 1.2: Venn Diagram of Mutually Exclusive Events But if we consider the example of rolling the dice and noting the top measurement e.g. let us suppose that E1 is the event that top shows 1 and E2 is the event top number is less than or equal to three. Then the events E1 and E2 are not mutually exclusive because the occurrence of one does not necessarily precludes the occurrence of other. That is if top side shows 1 then it means that both events E1 and E2 have occurred. Figure 1.3: Venn Diagram of not Mutually Exclusive Events 1.5.3 Collectively Exhaustive Events A set of events E1, E2, ...En are called collectively exhaustive events if their union makes the whole sample space i.e. n [ i=1 Ei = Ω 6

21. Example 8 Consider the previous example of flipping the coin then these two events are collectively exhaustive since the union of head and tail ac- counts for all possible outcomes. Consider now the flipping of coin twice then the events HT, HH and TT are not collectively exhaustive since their union does not make the whole sample space Ω. We know that there is the possible event TH. Hence the set of events HT, HH, TT and TH will be collectively exhaustive events. Example 9 Consider the following table Sex Highest Degree Total College (C) High School (H) No Degree (N) Male (M) 350 100 40 490 Female (F) 275 210 25 510 Total 625 310 65 1000 Table 1.1: Number of Males and Females with Different Degrees (Courtesy F.S. Budnick) Suppose that one applicant is to be selected at random in an experiment then the Sample space Ω consists of the following outcomes. Ω = {MC, MH, MN, FC, FH, FN} Here e.g. MC: Means male applicant having college degree. Now we want to check that whether following statements are mutually exclusive or and collectively exhaustive. i. {M, F} ii. {M, F, H} iii. {C, H, N, M, F} iv. {MC, MH, MN, MF} 7

22. v. {MC, FC, C, H, N} vi. {M, FC, FH} Solution i. {M, F}: First, we check these events for mutually exclusive property the two events M and F are mutually exclusive since the occurrence of a male applicant precludes the occurrence of a female and similarly for the opposite case. These two events are also collectively exhaustive because the two events include all the possible outcomes of the sample space Ω. ii. {M, F, H}: The three events M, F and H are not mutually exclusive because the occurrence of a male applicant M does not preclude the occurrence of a High school degree applicant H. Similar reasoning can be given for F and H. These three events are collectively exhaustive since all the possible outcomes are included if we take union of these three events. iii. {C, H, N, M, F}: These five events are not mutually exclusive. Since if we take C and F then there can be a college applicant which is female. But these five events are collectively exhaustive because C, H and N includes all the applicants. iv. {MC, MH, MN, MF}: These four events are mutually exclusive because the occurrence of any one precludes the occurrence of other. Also, these four events are collectively exhaustive because the union of MC, MH and MN contains all the male and F contains all the female. v. {MC, FC, C, H, N}: These five events are not mutually exclusive because MC and C contain common members. Similarly, FC and C 8

23. contain common entries. But these five events are collectively exhaustive since the union C, H and N contains all applicants. vi. {M, FC, FH}: These three events are mutually exclusive since FC and FH contain females with college degree and High school degree so nothing common. Similarly, the third one is M male applicants. So, nothing common. But it is not collectively exhaustive because females with no degree FN are not present here. 1.5.4 Relative Frequency Suppose that an experiment is performed and the possible outcomes of the random experiment are n. Now if m of these outcomes are our favorable then the probability of the favorable event can be thought of as a relative frequency and is given by m n . For example: We flip a coin and note down the result whether it is head or tail. If we are interested in head then its relative frequency will be 1 2 . It means that there are 50% chances when we flip the coin we get the head. Example 10 We consider the previous example once again where we had the following data: Sex Highest Degree Total College (C) High School (H) No Degree (N) Male (M) 350 100 40 490 Female (F) 275 210 25 510 Total 625 310 65 1000 Table 1.2: Number of Males and Females with Different Degrees Assume that an applicant is selected at random from the pool of 1000 applicants and also that each applicant has an equal chances of being selected. Solution Now using the definition of relative frequency we can estimate the proba- 9

24. bility that a selected applicant has certain characteristics. For example, the probability that the selected applicant is a male is P(M) = Total number of males Total number of applicants P(M) = 490 1000 = 0.49 i.e. there are 49% chances that the selected applicant is male. Similarly the probability that the selected applicant will have a high school diploma as the highest degree is P(H) = No.of applicants having High School Degree(highest) Total number of applicants P(H) = 310 1000 = 0.310 Next the probability that the selected applicant will be a male with no degree is given by P(MN) = No. of male applicants having no degree Total number of applicants P(MN) = 40 1000 = 0.040 The probability that the selected applicant will be female with a college degree is given as P(FC) = No. of female applicants with College Degree Total number of applicants P(FC) = 275 1000 = 0.275 Moving in the same way we summarize the table which gives information about the probabilities of the selected applicant with different characteristics. Some important points about the table: • The sum of probabilities of Male and Female applicants is equal to 1. 10

25. Sex Highest Degree Total College (C) High School (H) No Degree (N) Male (M) 0.350 0.100 0.040 0.490 Female (F) 0.275 0.210 0.025 0.510 Total 0.625 0.310 0.065 1.000 Table 1.3: Probabilities of the Selected Applicants with Different Charac- teristics • The sum of probabilities of applicants having College Degree, High School Degree or No degree is equal to 1. The reason behind is that the events in both the cases are mutually exclusive and collectively exhaustive events. 1.6 RULES OF PROBABILITY Rule 1: The probability of an event E denoted by P(E) is a real number between 0 and 1 both inclusive i.e. 0 ≤ P(E) ≤ 1 Rule 2: If P(E) represents that an event E will occur, then the probability that the event E will not occur is denoted by P(Ec ) and is given by P(Ec ) = 1 − P(E) Rule 3: If the events E1 and E2 are mutually exclusive then the probability that either event E1 will occur or event E2 will occur is given by P(E1 ∪ E2) = P(E1) + P(E2) Rule 4: More generally if the events E1, E2, , En are mutually exclusive then the probability that event E1 or E2 or ... or En will occur is given by P(E1 ∪ E2 ∪ ... ∪ En) = P(E1) + P(E2) + ... + P(En) 11

26. Rule 5: If E1 and E2 are any two events (not necessarily mutually exclusive) then the probability that either event E1 will occur or event E2 will occur or both E1 and E2 will occur is given by P(E1 ∪ E2) = P(E1) + P(E2) − P(E1 ∩ E2) Now we explain the difference between rules 3 and 5 with the help of Venn Diagram. Figure 1.4: Venn Diagram for Rule 3 Here we assume that shaded areas in Venn Diagram represent probability. Area of the whole sample space Ω is taken as 1. Figure 1.5: Venn Diagram for Rule 5 12

27. Example 11 The given table shows the number of fire alarms pulled in one hour on a given day in Fire Bridge office Islamabad. The analysts have estimated the corresponding probabilities of the different number of alarms pulled per hour which is shown in the second column of the table. What is the probability that a More than 8 alarms will be pulled b Between 8 and 10 alarms will be pulled including both 8 and 10 c No more than 8 alarms will be pulled No. of alarms pulled (n) Probability P(n) Fewer than 8 0.16 8 0.20 9 0.24 10 0.28 More than 10 0.12 Table 1.4: Probability of Number of Alarms Pulled Solution a The events corresponding to more than 8 alarms include 9 alarms, 10 alarms, and more than 10 alarms. These three events are mutually exclusive. Therefore by applying Rule 4 we get that. P(More than 8 alarms) = P(9 alarms)+P(10 alarms)+P(more than 10) P(More than 8 alarms) = 0.24 + 0.28 + 0.12 = 0.64 b Here the required events include 8 alarms, 9 alarms, and 10 alarms exactly. These three events are mutually exclusive so again apply Rule 4 we get P(between 8 and 10 alarms both inclusive) = P(8 alarms)+P(9)+P(10) P(between 8 and 10 alarms both inclusive) = 0.20+0.24+0.28 = 0.72 13

28. c If the event corresponding to No more than 8 alarms pulled is denoted by E then the event more than 8 alarms pulled which we calculated in part (a) will be Ec . So we can use Rule 2. P(E) = 1 − P(Ec ) P(No more than 8 alarms) = 1 − P(more than 8 alarms) P(No more than 8 alarms) = 1 − 0.64 = 0.36 Example 12 A card is drawn at random from a well-shuffled deck of cards. What is the probability that the card will be a A king or jack b A face card c A 7 or a spade d A face card or a card from a red suit Solution a Since the selection of king and the selection of jack are two mutually exclusive events so P(King ∪ Jack) = P(King) + P(Jack) P(King ∪ Jack) = 4 52 + 4 52 = 8 52 = 2 13 b Face cards mean king, queen, or jack of any suit. Therefore P(Face cards) = 12 52 = 3 13 14

29. c The events of selecting a 7 and selecting a spade card are not mutually exclusive so we apply Rule 5 to get P(7 ∪ spade card) = P(7) + P(spade card) − P(7 ∩ spade card) P(7 ∪ spade card) = 4 52 + 13 52 − P(7of spades) P(7 ∪ spade card) = 4 52 + 13 52 − 1 52 P(7 ∪ spade card) = 16 52 = 4 13 d The events of selecting a face card and selecting a card from the red suit are not mutually exclusive so we have to apply Rule 5 to get P(face ∪ red suit) = P(face) + P(red suit) − P(face ∩ red suit) P(face ∪ red suit) = P(face) + P(red suit) − P(face card of red suit) P(face ∪ red suit) = 12 52 + 25 52 − 6 52 P(face ∪ red suit) = 36 52 = 9 13 Example 13 An experiment consisting of selecting one card at random from a deck of 52 cards, the events king and spade are not mutually exclusive. Determine the probability of selecting a king, a spade, or both a king and a spade. Solution Apply rule 5 to solve the given problem. P(King ∪ Spade) = P(King) + P(Spade) − P(King ∩ Spade) = P(King) + P(Spade) − P(King of Spade) = 4 52 + 13 52 − 1 52 = 16 52 = 4 13 Example 14 A group of 2,000 people was surveyed regarding policies which might be enacted to conserve oil. Of the 2,000 people 1,000 people said that gas rationing would be acceptable to them, 500 people said that a federal surtax of $0.25 per gallon would be acceptable, and 275 indicate that both 15

30. rationing and the surtax would be acceptable. If a person is selected at random from this group, what is the probability that the person would: a. Find the surtax acceptable? b. Find the surtax acceptable but not gas rationing? c. Find one or both of the alternative acceptable? d. Find neither alternative acceptable? Solution a. Since 500 persons approved of the surtax, the probability that a person would approve is P(T) = Number of persons approving surtax Number of persons surveyed P(T) = 500 2000 = 0.25 b. 225 people indicated approval of the surtax but not rationing. Thus, the probability of selecting such a person is P(Surtax but not gas rationing ) = 225 2000 = 0.1125 c. One or both alternative acceptable P(R ∪ T) = 725 + 275 + 225 2000 P(R ∪ T) = 1225 2000 = 0.6125 d. Neither alternative acceptable P((R ∪ T)0 ) =1 − P(R ∪ T) P((R ∪ T)0 ) =1 − 0.6125 = 0.3875 16

31. 1.7 INDEPENDENT EVENTS Two events are called independent events if the occurrence or nonoccurrence of one event does not effect the occurrence or nonoccurrence of the other event. Example Flipping a fair coin is an example of independent events. Be- cause the occurrence of head does not effect the occurrence of tail in next trial. Similarly if we draw a card from a deck of cards then the probability of selecting card remains the same if we replace the selected card back in the deck. That is probability of selecting a red card stays 26/52 = 1/2 as long as we replace the drawn card back in the deck. 1.7.1 Marginal Probability The simple probability of an event is called marginal probability. But sometimes we are interested in the combination of events. Example We toss a coin three times and we are interested to get three heads in succession. 1.7.2 Joint Probability The probability of the joint occurrence of two or more events is called joint probability. Rule 6:The joint probability of two independent events occurring in succession is equal to the product of their marginal probabilities i.e. we can write P(E1 ∩ E2) = P(E1).P(E2) Rule 7:The joint probability of n independent events occurring in succession is equal to the product of their marginal probabilities. i.e. we can 17

32. Figure 1.6: Probability Tree Diagram for Coin Toss (Courtesy F.S. Bud- nick) write P(E1 ∩ E2 ∩ ... ∩ En) = P(E1).P(E2)...P(En) Example 15 A coin is tossed such that P(H) = 0.45 and P(T) = 0.55. Construct a probability tree showing all possible outcomes if the coin is tossed 3 times. Also what is the probability of getting a Two tails in three tosses b At least one head in three tosses Solution 18

33. a To find the probability of two tails in three tosses if we look into the last column in the tree diagram then we see that 4th , 6th and 7th circle which consist of H1T2T3, T1H2T3 and T1T2H3 contain exactly 2 trails. So our required probability will be the sum of these three probabilities and it is given by 0.1361 + 0.1361 + 0.1361 = 0.4083 b Similarly if we look into the last column we can see that except the last circle which consist of T1T2T3 all others contain at least one head. So our required probability will be given by 1 − 0.1664 = 0.8336 Example 16 A group of university student has started using direct mail- ing campaign as a major way of soliciting donation from alumni. The vice president for development estimates that the probability an alumnus, contacted by mail, will contribute is 0.30. Given two successive contacts, what is the probability that: a The first contacted alumnus will contribute and the second not? b The first will not contribute but the second will? c Both will contribute? d Neither will contribute? Solution If the event A represents the occurrence of a contribution, the event A 0 represents the nonoccurrence of a contribution. a P(A1 ∩ A 0 2) =P(A1).P(A 0 2) =(0.30)(1 − 0.30) =(0.30)(0.7) = 0.21 19

34. b P(A 0 1 ∩ A2) =P(A 0 1).P(A2) =(1 − 0.30)(0.30) =(0.7)(0.30) = 0.21 c P(A1 ∩ A2) =P(A1).P(A2) =(0.30)(0.30) =0.09 d P(A 0 1 ∩ A 0 2) =P(A 0 1).P(A 0 2) =(1 − 0.30)(1 − 0.30) =(0.7)(0.7) = 0.49 1.8 LAWS OF PROBABILITY 1.8.1 Conditional Probability It is represented as P E1 E2 . It is called the conditional probability of event E1 given that event E2 has occurred. Now we discuss conditional probability for independent and dependent events. Rule 8: Suppose we have two independent events E1 and E2 then the conditional probability of event E1 given that event E2 has occurred is the marginal probability of E1 and it is represented as P E1 E2 = P(E1) Example: The conditional probability of getting 6 on the roll of a die given that no 6 has occurred in the last 50 rolls is equal to 1 6 . Which is same as getting 6 on the first roll of a die. 20

35. 1.8.2 Dependent Events Two events are called dependent if the probability of occurrence or nonoccurrence of one event is affected by the occurrence or nonoccurrence of the other event. Example: Suppose we have an urn containing balls of different colors. We want to find the probability of getting a red ball at random from the urn. For the first time we take a ball, whether it is red ball or not. If we dont replace it then in the next time the probability of getting a red ball is different from the previous time probability. Because the count has been changed now. Rule 9: The conditional probability of an event E1 is given that the event E2 has occurred is given by P E1 E2 = P(E1 ∩ E2) P(E2) Example 17 A large jar contains 8 red balls, 6 yellow balls and 6 blue balls. Two balls are to be selected at random from the jar. Assume that each ball in the jar has an equal chance of being selected and that the first ball selected is not replaced back into the jar. a What is the probability that first ball will be red and second yellow? b What is the probability that both balls will be blue? c What is the probability that neither will be red? Solution a Selection of balls from the jar without replacement are the events which are dependent. So we use Rule 9 in the following form P (R1 ∩ Y2) = P Y2 R1 .P(R1) 21

36. Where R1 is the event that first ball is red Y2 is the event that 2nd ball is yellow. P (R1 ∩ Y2) = 6 19 × 8 20 = 48 380 P (R1 ∩ Y2) = 12 95 b Let B1 = The event that first ball is blue. B2 = The event that 2nd ball is blue. Again using Rule 9, we get P (B1 ∩ B2) = P B2 B1 .P(B1) P (B1 ∩ B2) = 5 19 × 6 20 = 30 380 = 3 38 c Let Rc be the event that first ball is not red. Rc be the event that 2nd ball is not red. Here again we use Rule 9 and obtain P (Rc 1 ∩ Rc 2) = P Rc 2 Rc 1 .P(Rc 1) P (Rc 1 ∩ Rc 2) = 11 19 × 12 20 = 132 380 = 33 95 Example 18 Suppose that cards are selected at random, without replacement from a standard 52 cards deck. Determine the probability that a The first two cards are hearts b The first is a spade, second a club, third a heart and fourth a diamond c 3 aces are selected in a row d No aces are included in first four cards. Solution 22

37. a Let H1 be the event that first card is heart H2 be the event that 2nd card is heart Then P (H1 ∩ H2) = P H2 H1 .P(H1) P (H1 ∩ H2) = 12 51 × 13 52 = 3 51 P (H1 ∩ H2) = 1 17 b Here we have to find P(S1 ∩ C2 ∩ H3 ∩ D4) First we will calculate P (S1 ∩ C2) = P C2 S1 .P(S1) P (S1 ∩ C2) = 13 51 × 13 52 = 13 204 Now we will calculate P (S1 ∩ C2 ∩ H3) = P H3 S1 ∩ C2 .P (S1 ∩ C2) P (S1 ∩ C2 ∩ H3) = 13 50 × 13 204 = 169 10200 Therefore P(S1 ∩ C2 ∩ H3 ∩ D4) = P D4 S1 ∩ C2 ∩ H3 .P (S1 ∩ C2 ∩ H3) P(S1 ∩ C2 ∩ H3 ∩ D4) = 13 49 × 169 10200 = 2197 499800 c Here we have to find P(A1 ∩ A2 ∩ A3) First we will calculate P(A1 ∩ A2) = P A2 A1 .P (A1) P(A1 ∩ A2) = 3 51 × 4 52 = 1 221 Now we will calculate P(A1 ∩ A2 ∩ A3) = P A3 A1 ∩ A2 .P (A1 ∩ A2) P(A1 ∩ A2 ∩ A3) = 2 50 × 39 52 = 1 5525 23

38. d Let Ac 1 be the event that 1st card is not ace Ac 2 be the event that 2nd card is not ace Ac 3 be the event that 3rd card is not ace First we will calculate P(Ac 1 ∩ Ac 2) = P Ac 2 Ac 1 .P (Ac 1) P(Ac 1 ∩ Ac 2) = 38 51 × 39 52 = 57 102 P(Ac 1 ∩ Ac 2) = 19 34 Now we will calculate P(Ac 1 ∩ Ac 2 ∩ Ac 3) = P Ac 3 Ac 1 ∩ Ac 2 .P (Ac 1 ∩ Ac 2) P(Ac 1 ∩ Ac 2 ∩ Ac 3) = 37 50 × 19 34 = 703 1700 Example 19 Assume that a person selects a card at random from a deck of 52 cards and tells us that the selected card is red. Find the probability that the card is king of hearts given that is red. Solution The probability that the card is king of hearts given that is red can be determined by using rule 9. P King of hearts red = P(King of hearts ∩ red) P(red) P King of hearts red = 1 52 26 52 P King of hearts red = 1 26 Example 20 Find the joint probability of selecting two aces in a row from a deck without replacement of the first card. Solution P(A1 ∩ A2) =P(ace on first draw).P(ace on second draw an ace on first draw) = 4 52 . 3 51 = 12 2652 24

39. 1.9 SELF ASSESMENT QUESTIONS 1. Find the sample space for choosing an odd number from 1 to 15 at random. 2. What is the difference between mutually exclusive events and collectively exhaustive event? 3. The probability that an applicant for pilot school will be admitted is 0.5. If three applicants are selected at random, what is the probability that a. All three will be admitted b. None will be admitted c. Only one will be admitted 4. A student estimates the probability of attaining an A in math course at 0.4 and the probability of attaining a B at 0.3. What is the probability that he/she a. Will not receive an A b. Will not receive a B c. Will receive neither an A nor a B 5. An urn contains 8 green dotted balls, 10 green striped balls, 12 blue dotted balls and 10 blue striped balls. If a ball is selected at random from the urn, what is the probability that the ball will be a. Green or striped b. Dotted c. Blue or dotted 25

40. 6. A single die is rolled and each side has an equal chance of occurring. what is the probability of rolling four consecutive 6s? 7. A ball is selected at random from an urn containing three red striped balls, eight solid balls, six yellow striped balls, four solid yellow balls and four blue striped balls. a. what is the probability that the ball is yellow, given that is striped? b. What is the probability that the ball is striped, given that is red? c. What is the probability that the ball is blue, given that it is solid colored? 8. Suppose that E and F are events and P(E) = 0.2, P(F) = 0.5 and P(E ∪ F) = 0.6 determine the following a. P(E ∩ F) b. P E F c. P F 0 E 9. What is the probability of drawing three cards, without replacement, from a deck of cards and getting three kings? 10. An urn contains 18 red balls, 14 red striped balls, 16 yellow balls and 12 yellow stripped balls, a. Given that a ball selected from the urn is striped, what is the probability it is yellow? b. Given that a ball selected from the urn is not striped, what is the probability it is red? 26

41. Unit-2 RANDOM VARIABLES WRITTEN BY: DR NASIR REHMAN REVIEWED BY: DR BABAR AHMED 27

42. 2.1 INTRODUCTION In every day life, we base many of our decisions on random outcomes that is chance occurrence. It is important in studying and analyzing the chance events and it is defined as to take care of all the possible outcomes of that event. In this chapter students will study 1. Random variable 2. The concept of discrete and continuous random and what are the discrete and continuous random variables? 3. Random numbers and their generation 4. Probability distribution and discrete probability distribution 5. Differentiate between conditional probability and marginal probability 6. The use or applications of random variable in other fields and daily life 2.2 OBJECTIVES After reading this unit you will be able to 1. differentiate between different types of randomness we come across in our daily life problems and situations 2. formulate and solve these problems using different types of distributions 28

43. A random variable is a function which associates a numerical value to each event in the sample space. Since it is associated to a random experiment outcome therefore its values fluctuate in a unpredictable manner. Example 1 Suppose that a coin is tossed twice so that the sample space is S = {HH, HT, TH, TT}. Let X represent the number of heads which can come up. With each sample point we can associate a number for X as shown in the table below. For example, in the case of HH (two heads) X = 2 while for TH X = 1 (one head). It follows that X is a random variable. Example 2 The Internal Revenue Service (IRS) estimates the probability of an error on personal income tax returns to be 0.4. Suppose that an experiment is conducted in which three returns are selected at random to check the error for the purpose of the audit. Let Ei represents the event that outcome is error for the ith trial and Ni represents the event that outcome is without error for the ith trial then it means that P(Ei) = 0.4 ∀ i P(Ni) = 0.6 ∀ i 29

44. Figure 2.1: Probability Tree Diagram(Courtesy F.S. Budnick) Now the sample space S for this experiment consists of the following events. S = {EEE, EEN, ENE, ENN, NEE, NEN, NNE, NNN} We can associate a random variable with each event in this sample space. Since in this experiment we are more interested in finding the errors in the personal income tax returns. So, we define a random variable X which represents the number of returns found to contain errors. Since the number of errors in a sample containing three returns may be 0, 1, 2, or 3 so our random variable X assigns a number randomly from the set {0, 1, 2, 3} to all events in S. These assignments are shown in the following table. 30

45. Sample events in S Random variable X EEE 3 EEN 2 ENE 2 ENN 1 NEE 2 NEN 1 NNE 1 NNN 0 Table 2.1: Sample Space for Events E 2.3 DISCRETE AND CONTINUOUS RAN- DOM VARIABLES The outcomes of an experiment which measures the number of units of a product demanded each day can be represented by a discrete random variables. Whereas in an experiment which selects people at random and records some attribute such as height or weight, in such cases the outcomes can be represented by continuous random variable. 2.3.1 Discrete Random Variables If a random variable assumes values which consists of a set of discrete values (numbers) then it is called a discrete random variable. For example: In the previous case X is a discrete random variable since it takes values from the set {0, 1, 2, 3}. 2.3.2 Continuous Random Variables If a random variable can assume any value from an interval of real numbers, then it is called continuous random variable. For example: Let X counts the height of a person in Business Math class. Then if a person is selected randomly from this class and his/her height is measured then it can be any real number in the interval for example [0, 300] 31

46. cm. Now to manage the data which we get from the continuous random variables or discrete random variables, we use the tool of probability distribution or frequency distribution. Which summarizes each possible value taken by a random variable and it tells us the no. of occurrence of that special value which is called frequency. Example 3 Suppose that a coin is tossed. Let 0 denote the occurrence of head and 1 denote the occurrence of tail. Thus the random variable X assumes the values 0 and 1. Since the coin is fair, the probability of head is 1 2 and that of tail is also 1 2 . The probability function of X is thus given in the following table. X 0 1 P(X) 1/2 1/2 Example 4 Suppose that a fair die is rolled. Here the random variable X assumes the value 1, 2, 3, 4, 5 or 6 (number of spots on the face of the die) each with probability 1 6 . Thus the probability distribution of X is given in the following table. X 1 2 3 4 5 6 P(X) 1/6 1/6 1/6 1/6 1/6 1/6 Example 5 Find the probability function corresponding to the random variable X of the data given in the following table. Sample Point HH HT TH TT X 2 1 1 0 32

47. Solution Suppose that the coin is fair, we have P(HH) = 1 4 , P(HT) = 1 4 , P(TH) = 1 4 and P(TT) = 1 4 . Then P(X = 0) =P(TT) = 1 4 P(X = 1) =P(HT ∪ TH) = P(HT) + P(TH) = 1 4 + 1 4 = 1 2 P(X = 2) =P(HH) = 1 4 The probability function of X is given in the following table X 0 1 2 P(X) 1/4 1/2 1/4 Example 6 Obtain a probability distribution for the number of heads when three coins are tossed. Solution If H represents head and T represents tail, then the sample space for the experiment of tossing three coin is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} If X represents the number of heads when three coins are tossed, then the elements of the sample space and the values of th random variable X are shown in the following table Sample Point HHH HHT HTH THH HTT THT TTH TTT X 3 2 2 2 1 1 1 0 The random variable X takes the value 0, 1, 2 or 3. The sample space consists of 8 sample points. There is only one sample point which corresponds to X = 0. Thus P(X = 0) = 1 8 . There are three sample points which corresponds to X = 1. Thus P(X = 1) = 3 8 . Similarly, P(X = 2) = 3 8 and 33

48. P(X = 3) = 1 8 . Thus the probability distribution of the random variable X is given in the following table. X 0 1 2 3 P(X) 1/8 3/8 3/8 1/8 2.4 PROBABILITY DISTRIBUTIONS 2.4.1 Definition A probability distribution is a function which gives us a complete listing of all possible values of a random variable along with the probability of each value. So, depending upon the random variable X, probability distributions can be classified into two main types i.e. 1. Discrete Probability Distributions 2. Continuous Probability Distributions 2.4.2 Discrete Probability Distribution Let the discrete random variable X can take following n values x1, x2, ..., xn with probabilities p1, p2, ..., pn respectively then the corresponding discrete probability distribution function is represented in the following table. We Value of random variables X = xi Probability p(xi) x1 p1 x2 p2 . . . . . . xn pn Table 2.2: Discrete Probability Distribution 34

49. include all the possible values of a random variable here in the probability distribution function shown in the table therefore the sum of probabilities is always 1.i.e. p1 + p2 + ... + pn = 1 Example 7 Considering the IRS example again we can see that the probabilities with each event given by Sample events in S Random variable X Probability EEE 3 0.064 EEN 2 0.096 ENE 2 0.096 ENN 1 0.144 NEE 2 0.096 NEN 1 0.144 NNE 1 0.144 NNN 0 0.216 Table 2.3: Random Variable and Probability Distribution of Event S In IRS case, the probability that X ≤ 0 includes P(X = 0) which is 0.216 similarly the probability that X ≤ 1 includes P(X = 0) and P(X = 1) so its value is 0.216 + 0.432 = 0.648. Going in a similar way we get that P((X) ≤ 2 = 0.936) and P((X) ≤ 3 = 1) and it remains 1 for all real numbers greater than 3. The reason is that since our random variable X takes the value from the set {0, 1, 2, 3} so the maximum value of the function F(X) is 1. Example 8 Consider the given data which shows the number of radio in a household estimate the probability mass function. Solution 35

50. No. of Radio No. of households x P(x) 0 1218 0 1218/101501 = 0.012 1 32379 1 32379/101501 = 0.319 2 37961 2 37961/101501 = 0.374 3 19387 3 19387/101501 = 0.191 4 7714 4 7714/101501 = 0.076 5 2842 5 2842/101501 = 0.028 Total 101501 1.000 Table 2.4: Probability Distribution of Number of Radios in a Household 2.5 PROBABILITY DISTRIBUTION PROP- ERTIES Some other properties of the probability distribution function are given below. For a discrete random variable X taking values {x1, x2, ...xn} we have that 1. 0 ≤ P(X = xi) ≤ 1 ∀i = 1, 2, ..., n 2. P(X = x1) + P(X = x2) + ... + P(X = xn) = 1 Example 9 A bag contains two white and three black balls. The probability distribution of the number of white balls if two balls are selected is given in the following table. Find the distribution function for the given probability distribution. x 0 1 2 P(x) 3/10 6/10 1/10 Table 2.5: Probability Distribution of White Balls Solution The distribution function for the given probability distribution is given in the table below. 36

51. x F(x) x 0 0 0 ≤ x 1 3/10 1 ≤ x 2 9/10 x ≥ 2 1 Table 2.6: Probability Distribution Function of White Balls Example 10 The Director of Murree Development Authority has checked the city records to determine the number of major snowfalls which have occurred in each of the last 60 years. Which is shown in the table below a- Construct the probability distribution for this study b- What is the probability that there will be more than two snowfalls in a given year? c- What is the probability that there will be three or less snowfalls in a given year? No. of major snowfalls Frequency 0 3 1 5 2 10 3 13 4 8 5 16 6 5 Total 60 Table 2.7: Frequency Distribution for the Major Snowfalls of the Last 60 Years Solution 37

52. a- First, we complete the given table by adding the column of probability distribution. Which can be constructed using for example since there are 3 years when there is no snowfall so its probability is given by 3 60 = 1 20 = 0.05 No. of major snowfalls Frequency Probability distribution 0 3 0.05 1 5 0.08 2 10 0.17 3 13 0.22 4 8 0.13 5 16 0.27 6 5 0.08 Total 60 1 Table 2.8: Probability Distribution for Snowfalls b- P(more than two snowfalls) = P 2 P 2 = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) P 2 = 0.22 + 0.13 + 0.27 + 0.08 = 0.70 So, there are 70% chances that in given year there will be more than two major snowfalls. c- P(three or less snowfalls) = P ≤ 2 P 2 = P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0) P 2 = 0.22 + 0.17 + 0.08 + 0.05 = 0.50 So, there are 50% chances that in each year there will be three or less major snowfalls. 38

53. Example 11 Find the distribution function for the probability distribution of example 6. Solution x F(x) x 0 0 0 ≤ x 1 1/8 1 ≤ x 2 4/8 2 ≤ x 3 7/8 x ≥ 3 1 Table 2.9: Probability Distribution Function for Head Example 12 A bank has been concerned about the length of time its customers must wait before serviced by a teller. A study of 500 customers has resulted in the probability distribution given in the table below. Waiting time in minutes per customer is the random variable X. a. What is the probability a customer will wait for a teller? b. What is the probability that a customer will wait less than 2 minutes? More than 3 minutes? X P(X) 0 0.32 1 0.24 2 0.18 3 0.12 4 0.09 5 0.05 Total 1.00 Table 2.10: Probability Distribution of 500 Customers Solution 39

54. a. P(wait) =P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) P(wait) =0.24 + 0.18 + 0.12 + 0.09 + 0.05 P(wait) =0.68 b. For less than 2 minutes P(wait 2 minutes) =P(X = 0) + P(X = 1) P(wait 2 minutes) =0.32 + 0.24 P(wait 2 minutes) =0.56 For more than 3 minutes P(wait 3 minutes) =P(X = 4) + P(X = 5) P(wait 3 minutes) =0.09 + 0.05 P(wait 2 minutes) =0.14 Example 13 Determine the constant k in the probability function: P(x) = k(x − 2), x = 3, 4, 5, 6 Solution Firstly we will substitute the value of x = 3, 4, 5, 6 in the function so we can get the values of P(x) as shown in the following table. x 3 4 5 6 P(x) k 2k 3k 4k P P(x) = 10k we know that sum of probability of an event in 1 so P P(x) = 1 or 10k = 1 or k = 1 10 . Thus the probability distribution of X is given in the following table. x 3 4 5 6 P(x) 1/10 2/10 3/10 4/10 40

55. 2.6 APPLICATIONS 1. Random variable is use in any business firm there is a communication system with certain number of lines to communicate data and voice communication. 2. If we need to know the probability of how many lines are working at one time we use discrete variable. 3. We can see application of discrete random variable in airport as well. For example number of airplanes taking off and landing during a given time in an airport. Let say there are two Etihad flights landing from Karachi to Dubai and two Etihad flights from Dubai to Karachi. So 2 is a discrete number and it can be denoted as X discrete random variable. 4. There are numerous applications of continuous random variable in chemical engineering. For example error in the reaction temperature may be defined by continuous random variable with many probability density functions. 5. It can also help to estimate the time at which chemical reaction com- pletes. 41

56. 2.7 SELF ASSESMENT QUESTIONS 1. Define random variable. What is the difference between discrete random variable and continuous random variable? 2. Write down the application of discrete random variable in business mathematics. 3. The fire chief for a small volunteer fire department has compiled data on the number of false alarms called in each day for the past 360 days. the data in the table shows a frequency distribution summarizing the findings. Construct the probability distribution for this study. Number of false alarm Frequency 0 75 1 80 2 77 3 40 4 28 5 24 6 20 7 16 Total 360 Table 2.11: Frequency Distribution of Number of False Alarms 4. Construct the discrete probability distribution which corresponds to the experiment of tossing a fair coin three times. Suppose the random variable X equal the number of heads occurring in three tosses. What is the probability of two or more heads? 5. A bag contains 4 red and 6 black balls. A sample of 4 balls is selected from the bag without replacement. Let X be the number of red balls. Find the probability distribution of X. 42

57. 6. Three balls are drawn from a bag containing 5 white and 3 black balls. If X represents the number of white balls drawn from the bag, then find the probability distribution of X. 7. A fair die is rolled 5 times. Let X represents the number of times the face 3 turns up. Obtain a probability distribution of X. 8. The probability distribution of a random variable X is given in the table below. X 1 2 3 4 5 P(X = x) 0.1 0.3 y 0.2 0.1 Find the value of y 9. Find the missing value of k such that the given distribution is a probability distribution of X. X 2 3 4 5 6 P(X) 0.01 0.25 0.40 k 0.04 10. The probability distribution of a random variable X is given. X 0 2 4 6 8 P(X) K 2K 4K 2k K Rewrite the probability distribution after finding out the values of K. 43

58. Unit-3 EQUATIONS WRITTEN BY: DR NASIR REHMAN REVIEWED BY: DR BABAR AHMED 44

59. 3.1 INTRODUCTION In our business when we make some calculations, when we deliver different types of goods, when we pay our workers etc., we have been using equations. Equations are actually all the mathematical statements where we see the symbol =. Equations are of many types but in this chapter we will explore only linear and quadratic equations. These equations involve one or more variables. And solution of an equation or equations means that we have to find the exact values of these variables. Equations play an important role to solve the different problems of Business, Economics, Banking, Stock Market and Agriculture etc. In this unit students will be able to learn about 1. Linear equation and non linear equation 2. Differentiate between linear and quadratic equations 3. Solve linear and quadratic equations using different methods 4. Inequalities and their solution 5. Interval notation and absolute value 6. Rectangular coordinate system 7. Model mathematical equations in different situations 3.2 OBJECTIVES After studying this unit students will be able to to model different situations in terms of equations and inequalities- differentiate equations of different types understand absolute termi- nology for different practical situations Solve linear and quadratic equations using different methods 45

60. 3.3 SOLVING FIRST-DEGREE EQUATIONS IN ONE VARIABLE Before moving towards the solving first degree equations in one variable we need to understand about equations and their properties. 3.3.1 Equations and their Properties A statement of equality of two algebraic expressions which involves one or more variables and constants is called Equation. Following are the examples of equations. 1. 2. 3. 1. x = 5 2. x + 2y2 = 10 Standard form of linear equation in one variable is ax + b = 0 a 6= 0 3.3.2 Types of Equations Equations are basically of three types. 1. Identity equation 2. Conditional equation 3. False statement (contradiction) 1. Identity equation: In which we have the form 3(x + y) = 3x + 3y 2. Conditional equation: Which is true for special values of the variable for example 2x = 5 46

61. 3. False statement (contradiction):Not true for any value of variable x e.g. x = x + 1 1 = 1 + 1 1 6= 2 Example 1 Solve the given equation 2x − 3 = 7 Solution 2x − 3 = 7 Add 3 on both sides of the given equation 2x − 3 + 3 = 7 + 3 2x = 10 Now dividing by 2 the above equation we get x = 10 2 x = 5 Example 2 Solve the given equation x 2 − 1 = x 3 + 1 Solution Arrange the given equation and then solve x 2 − x 3 = 1 + 1 3x − 2x 6 = 2 x 6 = 2 x = 6 × 2 x = 12 Example 3 A builder makes concrete by mixing 1 part clay 3 part sand and 5 part crushed stone. If 765 cubic feet of concrete is needed, how much of each ingredient is needed? 47

62. Solution Assume amount of clay = x Amount of sand = 3x Amount of crushed stone = 5x According to the given condition of the question we have. x + 3x + 5x =765 9x =765 x = 765 9 = 85 Hence amount of clay = x = 85 Amount of sand = 3x = 3(85) = 255 Amount of crushed stone = 5x = 5(85) = 425 3.4 SOLVING SECOND-DEGREE EQUA- TIONS IN ONE VARIABLE An equation which contains exactly one variable that is squared. A quadratic equation is an equation of the second-degree in one variable. Standard form of quadratic equation is ax2 + bx + c = 0, a 6= 0 3.4.1 Methods to Solve Quadratic Equations Quadratic equation can have two real roots, one real root or no real roots such as Discriminant: b2 − 4ac 1. If b2 − 4ac 0 then two distinct and real roots 2. If b2 − 4ac = 0 then two equal and real roots 3. If b2 − 4ac 0 then no real roots 48

63. There are number of different procedures that can be used to determine the roots of quadratic equation. Now we will discuss methods to solve quadratic equation. Following are the methods that are used to solve quadratic equation. 1. Quadratic Formula 2. Factorization Method 3. Completing Square Example 4 Solve the given equation using quadratic formula x2 +2x−3 = 0 Solution Using the quadratic formula i.e. x = −b ± √ b2 − 4ac 2a Here a = 1, b = 2 and c = −3 substitute these values in quadratic formula we get. x = −2 ± p 22 − 4(1)(−3) 2(1) x = −2 ± √ 4 + 12 2 x = −2 ± √ 16 2 x = −2 ± 4 2 x = −2 + 4 2 or −2 − 4 2 x = 2 2 = 1 or −6 2 = −3 Solution set = {1, −3} Example 5 Solve the given equation using factorization method x2 +2x− 3 = 0 Solution 49

64. Firstly, factorize the given equation for this we will observe the coefficient of 1st and 3rd term. 1 × (−3) = −3 3x − x = 2x ⇒ x2 + 3x − x − 3 = 0 x(x + 3) − 1(x + 3) = 0 (x + 3)(x − 1) = 0 (x + 3) = 0 or (x − 1) = 0 x = −3 or x = 1 Solution set = {1, −3} Example 6 Solve the given equation using completing square method x2 + 2x − 3 = 0 Solution Firstly, make the given equation in the form of completing square by adding subtracting 1 in the given equation i.e. x2 + 2(x)(1) + (1)2 − (1)2 − 3 = 0 (x + 1)2 − 1 − 3 = 0 (x + 1)2 − 4 = 0 (x + 1)2 = 4 Taking square root on both sides we get x + 1 = ±2 ⇒ x + 1 = 2 or x + 1 = −2 x = 2 − 1 or x = −2 − 1 x = 1 or x = −3 Solution set = {1, −3} Example 7 A railing is to enclose a rectangular area of 1800 square feet. The length of the plot is twice with width. How much railing must be 50

65. used? Solution Assume width of plot = y length of plot = 2y Area of rectangle = length × width= (2y)(y) According to the given condition of the question we have (2y)(y) =1800 2y2 =1800 y2 = 1800 2 = 900 y = √ 900 = 30 Hence the width of plot is 30 ft. and length of the plot is 60 ft. Example 8 The price of a bag is q dollars each. Assume that a manufacturer will supply 6q2 +5q units of bag to the market and consumers demand is 5q2 + 24 units. Find the value of q for which the supply will equal the demand. Solution According to the given condition of the question 6q2 + 5q =5q2 + 24 6q2 + 5q − 5q2 − 24 =0 q2 + 5q − 24 =0 q2 + 8q − 3q − 24 =0 q(q + 8) − 3(q + 8) =0 (q − 3)(q + 8) =0 ⇒ (q − 3) = 0 and (q + 8) = 0 Since q = −8 and negative numbers of bags is not possible. So for q = 3 supply will equal to the demand. 51

66. 3.5 INEQUALITIES AND THEIR SOLU- TIONS An inequality is a mathematical sentence that uses symbols such as , , ≤ or ≥ and which is used to compare two quantities when they are not equal. 3.5.1 Inequalities Inequalities are the conditions in which two quantities are not equal. They may be greater or less each other. There are special symbols which show that in what way quantities are not equal. Consider a and b as two quantities. • a b shows that a is less than b • a b shows that a is greater than b 3.5.2 Types of Inequalities There are three types of inequalities: 1. Absolute Inequality: Which is always true e.g. 2 6 2. Conditional Inequality: Which is true for certain values of x e.g. 2x 6 If we take x = 1 then this inequality is true but it is not true for x = 5. 3. Double Inequality: In which we bound x from both sides 2 x 6 52

67. 3.6 INTERVAL NOTATION Let x be a set of real numbers that lie between two numbers a and b is called an interval. This can be specified using the following notation. (a, b) = {x : a x b} There are four main ways to represent interval notation. • Open interval: Represent as (a, b) it has endpoints a and b but this interval does not include endpoints. e.g. (2, 5) which can be represented as {x : 2 x 5} • Closed interval: Represent as [a, b] it has endpoints a and b, closed interval which includes endpoints as well. e.g. [2, 5] which can be represented as {x : 2 ≤ x ≤ 5} • Half-open interval: It includes one endpoint while not the other. e.g. (2, 5] which can be represented as {x : 2 x ≤ 5} • Half-closed interval: It also includes one endpoint while not the other. e.g. [2, 5) which can be represented as {x : 2 ≤ x 5} 3.6.1 Solving Inequalities Example 9 Determine the values of x which satisfy the inequality 2x−5 ≥ 7 Solution Add 5 on both sides of the given inequality. 2x − 5 + 5 ≥ 7 + 5 2x ≥ 12 Now dividing by 2 on the both sides 2x 2 ≥ 12 2 ⇒ x ≥ 6 53

68. For all values of x greater than and equal to 6, x satisfies the given inequality. Example 10 A company manufacturers shoes that has a unit selling price of $30 and unit cost price of $25. If fixed costs are $700, 000 then determine the least number of units that must be sold for the company to have a profit. Solution Assume number of units sold = x Total revenue = 30x Total cost = 25x + 700, 000 According to the given condition of the question we have Total Revenue Total cost 30x − 25x − 700, 000 0 5x 700, 000 x 700, 000 5 x 140000 Hence at least more than 140000 units should be sold to make a profit. 3.6.2 Second-Degree Inequalities Example 11 Solve the quadratic inequality x2 + 4x − 12 ≤ 0 Solution x2 + 4x − 12 ≤ 0 To solve the quadratic inequality, we must break the middle term using factorization method i.e. x2 + 6x − 2x − 12 ≤ 0 x(x + 6) − 2(x + 6) ≤ 0 (x + 6)(x − 2) ≤ 0 54

69. This implies we have two cases x + 6 ≥ 0 and x − 2 ≤ 0 or x + 6 ≤ 0 and x − 2 ≥ 0 Now we will solve both the cases one by one. Case I x + 6 ≥ 0 and x − 2 ≤ 0 x ≥ −6 and x ≤ 2 On the number line, it can be represented as Figure 3.1: Representation on the Number Line Here we have some common area between -6 and 2 so Solution Set = {x : −6 ≤ x ≤ 2} = [−6, 2] Case II x + 6 ≤ 0 and x − 2 ≥ 0 x ≤ −6 and x ≥ 2 On the number line, it can be represented as Figure 3.2: Representation on the Number Line As x ≤ −6 and x ≥ 2 we dont have common area between these two inequalities hence there is no solution set in this case. So Solution Set {x : x ≤ −6 and x ≥ 2} = (−∞, −6] and [2, ∞) 55

70. Therefore by combining the first and second case we have the solution set [−6, 2] Solution can be represented on the number line as, 3.7 ABSOLUTE VALUE Absolute value describes the distance of a number on the number line. Absolute value of a number cannot be negative and it is always greater than or equal to zero. Absolute value of c is denoted by |c|. Let us consider a figure the absolute value of |3| = 3. Its like a person walking from 0 towards −3. He covered 3 units distance from 0 since the movement is in negative direction towards −3, this does not mean that distance covered by person is −3 because distance can never be negative. Figure 3.3: Absolute Value Representation 56

71. Definition For any real number a, we define its absolute value as, |a| = ( a if a ≥ 0 −a if a 0 3.7.1 Some Properties of Absolute Values Following are the properties of absolute values. 1. For any real number a, |a| ≥ 0 2. Absolute value of any negative number is positive, | − a| = |a| 3. For any two real numbers x and y, |x − y| = |y − x| 4. For any two real numbers x and y, |xy| = |x||y| 5. For any two real numbers x and y,

75. x y

79. = |x| |y| 3.7.2 Solving equations and inequalities involving absolute values Example 12 Solve the equation |x + 2| = 7 Solution Firstly, to clear the absolute value we split the given equation into its two parts: x + 2 = 7 and x + 2 = −7 ⇒ x = 7 − 2 and x = −7 − 2 Hence solution is x = 5 and x = −2 To check the solution, we substitute these two values into the given equation and get. |5 + 2| = 7 and | − 9 + 2| = 7 |7| = 7 and | − 7| = 7 7 = 7 and 7 = 7 57

80. Figure 3.4: Representation on the Number Line We can represent the solution on number line as, Example 13 Solve the inequality |x − 3| 5 Solution Firstly, we will write the given equation in two possible ways. |x − 3| 5 ⇒ −5 x − 3 5 ⇒ −5 + 3 x − 3 + 3 5 + 3 ⇒ −2 x 8 ⇒ −2 x and x 8 Solution Set {x : −2 x 8} = (−2, 8) Solution of the given inequality can also be represented on the number line as, Figure 3.5: Representation on the Number Line Example 14 Solve the inequality |x − 3| 5 Solution 58

81. Firstly, we will write the given equation in two possible ways. |x − 3| 5 ⇒ −5 x − 3 5 ⇒ −5 + 3 x − 3 + 3 5 + 3 ⇒ −2 x 8 ⇒ x −2 and x 8 Solution Set {x : x −2 and x 8} = (−∞, −2) and (8, ∞) Solution of the given inequality can also be represented on the number line as, Figure 3.6: Representation on the Number Line Example 15 Solve the inequality |x + 5| ≥ 7 Solution Firstly, to clear the absolute bars we must split the equation into its two parts: x + 5 ≤ −7 and x + 5 ≥ 7 x ≤ −7 − 5 and x ≥ 7 − 5 x ≤ −12 and x ≥ 2 Solution Set {x : x ≤ x − 12 and x ≥ 2} = (−∞, −12] and [2, ∞) Solution can be represented on the number line as, Figure 3.7: Representation on the Number Line 59

82. 3.8 RECTANGULAR COORDINATE SYS- TEM 3.8.1 Applications In Economics, we use math widely for analysis and managing. Its very important to learn about coordinate system because in this area the Lorenz Curve is an obvious representation of the cdf cumulative distribution function of our probability distribution of wealth or income, and it was first studied by Max O.Lorenz in 1905 for analyzing inequality of the wealth distribution and the calculations are done in rectangular coordinate system. Figure 3.8: Lorenz Curve for Cumulative Income and Household 3.8.2 Rectangular Coordinates It is a system which specifies each point uniquely in the plane by a pair of coordinates. Usually we take x − y plane that is the horizontal line is 60

83. called x−axis which is perpendicular to vertical line called y−axis. Now any point on the plane is represented by (x, y) where x measures the signed distance to the point along x−axis and similarly y measures the signed distance to the point along y−axis. Also called Cartesian Coordinate System. It has four quadrants. Figure 3.9: Cartesian Plane or Cartesian Coordinate System 3.8.3 The Cartesian Coordinates We use the cartesian coordinate system, to plot points and graph lines. The horizontal line is named as the horizontal axis and vertical line is named as the vertical axis. These two axes together are called coordinate axes. The plane containing the coordinate axes is named as the coordinate plane or the cartesian plane. Location of any point specified by ordered pair of values (x, y) here x is called Abscissa or the x−coordinate and y is called Ordinate or y−coordinate. In cartesian coordinates we have four quadrants. 61

84. Figure 3.10: Four Quadrants of the Cartesian Plane 3.8.4 The Midpoint Formula In cartesian coordinate system suppose we have two points (x1, y1)and (x2, y2) then we can find the midpoint of these points using the midpoint formula. x1 + x2 2 , y1 + y2 2 Example 16 Find the midpoint of the points A = (2, 6) and B = (4, 8). Solution Using midpoint formula x1 + x2 2 , y1 + y2 2 2 + 4 2 , 6 + 8 2 6 2 , 14 2 Midpoint is (3, 7) 3.8.5 The Distance Formula The distance between two points (x1, y1)and (x2, y2) can be calculated using the Pythagoras Theorem. d(A, B) = p (x2 − x1)2 + (y2 − y1)2 62

85. Figure 3.11: Distance Formula Using Pythagoras Theorem Example 17 Find the distance between the points A = (1, 4) and B = (4, 0). Solution Using distance formula d(A, B) = p (x2 − x1)2 + (y2 − y1)2 d(A, B) = p (4 − 1)2 + (0 − 4)2 d(A, B) = p (3)2 + (−4)2 d(A, B) = √ 9 + 16 d(A, B) = √ 25 = 5 Therefore the distance between the given points is 5 units. 63

86. 3.9 SELF ASSESMENT QUESTIONS 1. Solve the following first-degree equations. a 8x − 6 = 5x + 3 b −15 + 35x = 8x − 9 c (x + 9) − (−6 + 4x) + 4 = 0 d y 8 − 10 = y 4 − 9 2. Solve the following quadratic equations using the factorization method. a t2 + 4t = 21 b 4z2 + 18z − 10 = 0 3. Solve the following quadratic equations using the quadratic formula a 4x2 + 3x − 1 = 0 b 4t2 − 64 = 0 4. Solve the following quadratic equations using the completing square method. a 2y2 + 5y − 3 = 0 b s2 + s 4 − 3 4 = 0 5. Solve the following inequalities. a z + 6 ≥ 10 − z b 3y + 6 ≤ 3y − 5 6. Solve the following second degree inequalities a 6t2 + t − 12 0 b 2s2 − 3s − 2 0 64

87. 7. Solve the following absolute values a |2y + 5| = |y − 4| b |y| = |−y + 7| c |z2 − 8| ≤ 8 d |z2 − 2| ≥ 2 8. Find the midpoint of the line segment connecting the following points. a (3, 8) and (5, 5) b (−2, −4) and (2, 4) c (6, 6) and (−3, −3) d (0, 9) and (9, 0) 9. Find the distance between the following points. a (12, 0) and (0, −8) b (4, 4) and (−5, −8) c (−2, 4) and (1, 0) d (−7, −2) and (1, −4) 10. Find the length of the line segment connecting points C and D located at (2, 4) and (4, 8), respectively. 65

88. Unit-4 LINEAR EQUATIONS WRITTEN BY: DR NASIR REHMAN REVIEWED BY: DR BABAR AHMED 66

89. 4.1 Introduction In our business when we make some calculations, when we deliver different types of goods, when we pay our workers etc., we have been using equations. Equations are actually all the mathematical statements where we see the symbol =. Equations are of many types but in this chapter we will explore only linear and quadratic equations. These equations involve one or morevariables. And solution of an equation or equations means that we have to find the exact values of these variables. Equations play an important role to solve the different problems of Business, Economics, Banking, Stock Market and Agriculture etc. In this chapter students will be able to learn about 1. Characteristics of linear equations 2. Representation of linear equations and linear equations with n variables 3. Graphical representation of linear equations 4. The concept of intercepts, slope, two point formula and slope intercept form 5. Determining the equation of straight line 6. Linear equation more than two variables. 4.2 OBJECTIVES After reading this unit the students will be able to 1. understand the concept of algebraic and graphical characteristics of linear equations 67

90. 2. understand the notion of slope and different forms of equations for solving practical problems 3. illustrate the applications of linear equations 68

91. 4.3 CHARACTERISTICS OF LINEAR EQUA- TIONS Standard form of linear equation with two variables is, ax + by = c, where a, b and c are constants, x and y are variables also a and b cannot be equal to zero. Linear equation is an equation between two variables that gives a straight line when plotted on a graph. Linear equations are first degree equations i.e. power of the variables involved is exactly one. Examples Some examples of linear equations are, y = 2x + 1 5x + 3y = 6 y 2 + x = 3 4.4 REPRESENTATION USING LINEAR EQUATIONS Solution set for the linear equation ax + by = c is the set of all ordered pairs (x, y) which satisfy the equation. Set Notation S = {(x, y)|ax + by = c} Example 1 Solve the linear equation y = 2x + 1 Solution First, we take some values of x and fix the corresponding values of y with the help of given linear equation. We construct the following table: Now draw these points on the graph. 69

92. x y = 2x + 1 (x, y) -1 -1 (-1,-1) 0 1 (0,1) 1 3 (1,3) 2 5 (2,5) Table 4.1: Solution Set of Linear Equation y = 2x + 1 Figure 4.1: Graph of the Linear Equation y = 2x + 1 Example 2 A company has fixed costs of $7,000 for plant and equip- ment and variables costs $600 for each unit of output. What is the total cost at varying levels of output? Solution Let x = Units of output and C = Total cost then Total Cost = Fixed Cost + Variable Cost C = 7, 000 + 600x Cost at varying levels of output is, Output= x Total cost= C (x, C) 15 units 16,000 (15,16000) 30 units 25,000 (30,25000) Table 4.2: Total Cost at Varying Level of Output Graphically it can be represented as, 70

93. Figure 4.2: Graphical Representation of Linear Equation Example 3 Graph x + 2y = 7 Solution y intercept is found by letting x = 0 in the equation x + 2y = 7 0 + 2y = 7 y = 7 2 = 3.5 Similarly the x intercept is found by letting y = 0 in the equation x+2y = 7 x + 2(0) = 7 x = 7 The x−intercept and y−intercepts are 7 and 3.5 respectively. Thus, the graph goes through the points (7, 0) and (0, 3.5) Example 4 A company produces a product for which the fixed cost is $80,000 and variable cost per unit is $6 and. The selling price of each unit $10. Find the number of units that must be sold to make a profit of $60,000. Solution Let x represents the number of units that must be sold. Then the variable cost is 6x. 71

94. Figure 4.3: Graphical Representation of Linear Equation The total cost is the sum of fixed cost and variable cost. Total cost = variable cost + fixed cost = 6x + 80, 000 Profit = total revenue + total cost 60, 000 = 10x − (6x + 80, 000) 60, 000 = 4x − 80, 000 140, 000 = 4x x = 140, 000 4 = 35, 000 Thus, 35,000 units must be sold to produce a profit of $60,000 Example 5 Mary paid 8% sales tax and $68 for delivery when she bought a new TV for a total of $1886.5. Find the purchased price of a TV? Solution 72

95. Let x denote purchase price of the TV Sales tax = 0.08x Delivery charges = $68 Total cost = $1886.5 Total cost = Sales tax + Delivery charges + Purchase price 1886.5 = 0.08x + 68 + x 1886.5 − 68 = (0.08 + 1)x 1818.5 = 1.08x x = 1683.8 The purchase price of the TV is $1683.8 Example 6 An economist studied the supply and demand for aluminum siding and found that the price per unit p and demand q are related by the linear equation p = 60 − 3 4 q Find the demand at a price of 50 per unit? 73

96. Solution Let p = 50 50 = 60 − 3 4 q 50 − 60 = − 3 4 q −10 = − 3 4 q 40 3 = q 40 3 ≈ 13 units will be demanded at the price of $50 per unit. 4.5 LINEAR EQUATIONS WITH n VARI- ABLES A linear equation which has n number of variables in called linear equation with n variables. In linear equation with n variables each variable has power 1. General Form a1x1 + a2x2 + ... + anxn = b, where x1, x2, ..., xn are variables and a1, a2, ..., an are constants. Also a1, a2, ..., an cannot be equal to zero but b can be equal to zero. 4.6 GRAPHICAL CHARACTERISTICS 4.6.1 Graphing Two-Variable Equations Graph of linear equation involving two variables is a straight line. To sketch a graph of linear equations we have to consider following steps. 1. Identify and plot the coordinates of any two points which lie on the line 2. Connect the two points with a straight line 74

97. 3. Extend the straight line in both directions as far as necessary Example 7 Sketch the graph of the equation 2x − 3y = 12 Solution First, we take some values of x and find the corresponding values of y with the help of given linear equation. We construct the following table. x y = 2 3 x − 4 (x, y) 0 -4 (0,-4) 1 -3.33 (1,-3.33) 2 -2.66 (2,-2.66) 3 -2 (3,-2) 4 -1.33 (4,-1.33) 5 -0.66 (5,-0.66) 6 0 (6,0) Table 4.3: Solution Set of the Linear Equation 2x − 3y = 12 75

98. Figure 4.4: Graphical Representation of Linear Equation 4.7 INTERCEPTS Intercepts are the point where a graph crosses the x and y axes. Or we can say that intercepts are the point of intersection where a graph meets coordinate axes. 4.7.1 x−Intercept The x−coordinate of a given point where a graph intersects the x−axis is named as the x−intercept on the graphical representation and is obtained by setting y = 0 in the equation of the given graph. Take the general form of linear equation of two variables. ax + by = c If b = 0 then we have ax = c Dividing both sides by a the equation ax = c will become, x = c a 76

99. Since c and a are constants so we take c a equal to k which is also a constant i.e. x = k Graph of the equation x = k is vertical line crossing the x−axis at x = k. For these equations, there is x−intercept (k, 0) but no y−intercept unless k = 0 For example, x = 3. Figure 4.5: x− Intercept at x = 3 4.7.2 y−Intercept The y−coordinate of a given point where a graph intersects the y−axis is named as the y−intercept of the graphical representation and is obtained by setting x = 0 in the equation of the given graph. Take the general form of linear equation of two variables. ax + by = c If a = 0 then we have by = c 77

100. Dividing both sides by b the equation by = c will become, y = c b Since c and b are constants so we take c b equal to k which is also a constant i.e. y = k Graph of the equation y = k is horizontal line crossing the y−axis at y = k. Equation of these form have no x−intercept unless k = 0. For example, y = 3. Figure 4.6: y− Intercept at y = 3 4.8 SLOPE The inclination of a line whether it rises or goes up or falls down as we move from left to right along the axis and the rate at which the line rises or falls. The slope of a line may be positive, negative, zero, or undefined. 1. A line which has positive slope, rises from left to right 78

101. 2. A line which has negative slope, falls from left to right 3. Horizontal line has zero slope 4. Vertical slope is undefined Figure 4.7: Types of Slope 79

102. Generally, slope can be represented as Slope = 4y 4x , where 4y is change in y and 4x is change in x. Figure 4.8: Mathematical Representation of Slope 4.9 TWO POINT FORMULA We can determine slope by using two-point formula. The slope m of the straight line connecting two points (x1, y1) and (x1, y2) is, m = 4y 4x m = y2 − y1 x2 − x1 , where x1 6= x2 Example 8 Find the slope of the line segment connecting the points (1, 1) and (2, 4). Solution We will find slope using two-point formula i.e. m = y2 − y1 x2 − x1 , 80

103. we have x1 = 1, x2 = 2, y1 = 1 and y2 = 4 m = 4 − 1 2 − 1 m = 3 1 Slope of the line segment of (1, 1) and (2, 4) is 3. Graphical representation of the line segment passing through (1,1) and (2,4) is given below here we can see that slope is positive so the line segment rises from left to right. Figure 4.9: Representation of Slope 81

104. 4.10 SLOPE INTERCEPT FORM The equation of any straight line is called linear equation which can be written as, y = mx + c, where m is the slope of the line and b is the y−intercept of the line. For example y = 2x + 3, where y−intercept is 3 and slope of the given equation is m = 2 Example 9 A manufacturers total cost consist of a fixed overhead of $200 plus production costs of $50 per unit. (a) Express the total as a function of the number of units produced (b) Graph the equation (c) Identify the slope and C intercept Solution Let x represents the number of units produced then total cost is given by Total Cost = (Cost Per Unit) times(Number of Units) + Overheads Where Cost per unit = 50 Number of units produced = x Overheads = 200 therefore C(x) = 50x + 200, where C intercept is 200 and slope is 50. Graphical representation of the given problem is, 82

105. Figure 4.10: Graphical Representation of Linear Equation 4.11 DETERMINING THE EQUATION OF A STRAIGHT LINE 4.11.1 Slope and Intercept Form Slope intercept form of a linear equation is given by y = mx + k. If we know the slope m and y−intercept (0, k) of the line representing an equation then we simply substitute the value of m and k into the slope intercept form. Example 10 Find the equation of the straight line that has slope m = 4 and passes through the point (−1, −6). Solution Firstly, we will find y−intercept by putting the value of x and y into the slope−intercept form. y = mx + k −6 = 4(−1) + k k = −2. 83

106. Hence the y− intercept is = 2 now we will substitute the values of and into the slope−intercept form to get the equation of the straight line i.e. y = 4x − 2. 4.11.2 Point and Slope Form If we know the slope m and a point which lies on the line then we can find the y−intercept which is k by substituting m and the given point into the slope−intercept form i.e. y = mx + k Example 11 Write the equation of a line in slope−intercept form, with a slope −3 and goes through the point (3, −2). Solution Since m = −3 and the point (x, y) = (3, −2) we substitute these values into the slope−intercept form. y = mx + k −2 = −3(3) + k −2 = −9 + k k = 7. Hence the equation of a line in slope−intercept form is, y = −3x + 7 4.11.3 Point-Slope Formula Point−slope formula for a straight line is given for non-vertical straight line with slope m and containing the point (x1, y1).The slope of the line connecting (x1, y1) with any other point (x, y) on the line would be expressed as, m = y − y1 x − x1 , 84

107. by rearranging them we can get the point-slope formula. (y − y1) = m(x − x1) Example 12 Find the equation of the straight line that has slope m = 4 and passes through the point (1,-6). Solution Since m = 4, x1 = −1 and y1 = −6 substitute these values into point−slope formula. (y − y1) = m(x − x1) y − (−6) = 4(x − (−1)) y + 6 = 4x + 4 y = 4x − 2, hence the equation of the straight line using the point-slope formula is y = 4x − 2 Example 13 Find the equation of the line that has slope 2 3 and passes through (6, −2) Solution Using point slope form with m = 2 3 and (x1, y1) = (6, −2) (y − y1) = m(x − x1) y − (−2) = 2 3 (x − 6) y + 2 = 2 3 (x − 6) y = 2 3 x − 6 4.11.4 Two Points Form Assume that we are given the coordinates of two points which lie on a straight line. We can determine the slope of the line by using the two- point formula i.e. m = y2 − y1 x2 − x1 85

108. After getting the slope, we can determine the y−intercept using the point- slope formula i.e. (y − y1) = m(x − x1) Example 14 Write the equation of a line in slope-intercept form that goes through the two points (-1, 4) and (2,-2). Solution Firstly, we will find the slope of the line by using two-point formula. m = y2 − y1 x2 − x1 m = −2 − 4 2 − (−1) m = −2. Now we will find y−intercept by using point−slope formula. (y − y1) = m(x − x1), substituting m = −2 and the coordinate (−1, 4) into the point-slope formula. y − 4 = −2(x − (−1)) y − 4 = −2x − 2 y = −2x + 2, hence the equation of the line that goes through the points (-1, 4) and (2, -2) is y = −2x + 2, where slope m = −2 and y−intercept k = 2. Example 15 Find an equation of the line passing through (2, −3) and (4, 3)? Solution The line has slope m = y2 − y1 x2 − x1 m = 3 − (−3) 4 − (−2) = 6 2 m = 3 86

109. Using point slope form with m = 3 and (x1, y1) = (2, −3) y − (−3) = 3(x − 2) y + 3 = 3x − 6 y = 3x − 9 4.11.5 Parallel Lines and Perpendicular Lines • Two lines with the given slopes m1 and m2 are siad to be parallel to each other if and only if m1 = m2 • Two lines with the given slopes m1 and m2 are said to be perpendicular to each other if and only if m1 = − 1 m2 or m1m2 = 1 Example 16 Given that two lines are passing through the point (3, −2). If one line is perpendicular to y = 3x + 1 and other line is parallel to it. Find equations of these two lines. Solution The slope of y = 3x + 1 is 3. The slope of the line passing through (3, −2) and perpendicular to y = 3x + 1 must be − 1 3 . Thus by using point slope form, we get y − (−2) = − 1 3 (x − 3) y = − 1 3 x − 1 The line passing through (3, −2) and parallel to y = 3x + 1 has also slope 3. By using point slope form, we get y − (−2) = 3(x − 3) y = 3x − 11 87

110. 4.11.6 Supply and Demand Analysis The supply and demand for a product are usually related to its price. If the supply dominates the demand the price usually tends to go down. If the demand dominates the supply, the price usually tends to rise. Ulti- mately the price moves toward an equilibrium price at which the supply and demand become equal. Example 17 Suppose that at a price of $9 per box of apples, the supply and demand are 320, 000 and 200, 000 boxes respectively. At a price of $8.5 each box, the supply and demand is 170, 000 and 300, 000 respectively. a Establish a price−supply equation of the form p = mx + b when p is the given price in dollars and x is the given supply in thousands of boxes. b Establish a price−demand equation of the form p = mx + b when p is the given price in dollars and x is the given demand in thousands of boxes. c Graph the price−supply and price−demand equations in same coordinate system and find equilibrium point. Solution a For the price−supply equation, we need to find two points of the form (q, p) that are on supply line. The slope of the line passing through (320, 9) and (270, 8.5) are m = 8.5 − 9 270 − 320 = −0.5 −50 = 0.01 The equation of the line using point slope form is p − 9 = 0.01(x − 320) p = 0.01x − 3.2 + 9 p = 0.02x + 5.8 88

111. b For the price−demand equation, the slope of the line passing through (200, 9) and (300, 8.5) are m = 8.5 − 9 300 − 200 = − 0.5 100 = −0.005 The equation of the line using point slope form is p − 9 = −0.005(x − 200) p = −0.005x + 1 + 9 p = −0.005x + 10 c We equate price supply equation and price demand equation to find the point of intersection. 0.02x + 5.8 = −0.005x + 10 0.02x + 0.005x = 10 − 5.8 x = 280 Figure 4.11: Graphical Representation of Supply and Demand 89

112. 4.12 LINEAR EQUATIONS INVOLVING MORE THAN TWO VARIABLES When there are more than two variables in a linear equation for this the algebraic properties remain same but the only difference occur in graphical characteristics. With the increase in variable the graph axes are also increase. Like or two variables x and y axes of graph are x−axis and y−axis but if we have three variables x, y and z so the axis of graph will be x−axis, y−axis and z−axis. 4.12.1 Three-Dimensional Coordinate Systems Three-dimensional space can be described by using a three−dimensional coordinate system. In three dimensions, we use three coordinates which are x, y and z. These coordinates are perpendicular to each other and in- tersect each other at origin i.e. O(0, 0, 0) As the coordinate axes in two dimensions divides the plane into 4 quadrants. Similarly, the axes in three dimensions divide 3-space into 8 oc- tants. Figure 4.12: Three-Dimensional Coordinate System 90

113. 4.12.2 Equations Involving Three Variables The standard form of linear equation involving three variables is a1x1 + a2x2 + a3x3 = b, where a1, a2, a3 and b are constants and x1, x2 and x3 are variables. Number of variables in an equation determine the number of dimensions required to represent the equation graphically. Since the above equation have three variables so we need three-dimensional coordinate system to represent this equation graphically. Example 18 Given the equation 2x + 3y + 4z = 12, determine the coordinates of the x, y and z intercepts. Solution Firstly, we set any two variables equal to zero and then we will solve for the remaining variable. Let y = z = 0 then we get 2x + 0 + 0 = 12 ⇒ x = 6 So, the point on x−axis will be (6, 0, 0). Similarly let x = z = 0 then we get 0 + 3y + 0 = 12 ⇒ y = 4 So, the point on y−axis will be (0, 4, 0). And let x = y = 0 then we get 0 + 0 + 4z = 12 ⇒ z = 3 So, the point on z−axis will be (0, 0, 3). Graphical representation of the three points (6, 0, 0), (0, 4, 0) and (0, 0, 3) is given below. The colored shape shows us the graphical representation of the given equation in first octant. 4.12.3 Equations Involving more than three Variables When we have more than three variables in an equation then we need more than three dimensions to represent it graphically. We cannot visualize the graphical representation of such equations. The term hyperplane is used 91

114. Figure 4.13: Graphical Representation of Given Equation to describe the geometric representation of such equations. Mathematically this type of equation can be written as a1x1 + a2x2 + ... + anxn = b, where a1, a2, ..., an and b are constants and x1, x2, ..., x3 are variables. 92

115. 4.13 SELF ASSESMENT QUESTIONS 1. Graph each of the following linear equations: a 9y − 4x = 12 b y = −x c 2x + 7y = 14 d x = 2y + 3 e 5y − 3x = 13 f 2y = 3x 2. Suppose that the price and demand for a certain brand of soap are given by p = 15 − 7 6 q Where p is price measured in dollars and q is demand. a Determine the price for a demand of i) 2 units ii) 5 units iii) 10 units b Determine demand at a price of i) $3 ii) $10 iii) $20 c Graph p = 15 − 7 6 q d Suppose the price and supply of the soap are given by p = 2 3 q Where p and q represent the price and supply respectively. De- termine the supply when price is i) $0 ii) $5 iii) $30 e Graph p = 2 3 q on the same axis. f Determine the equilibrium supply and equilibrium supply. 93

116. 3. John paid 7.5% sales tax and $150 title and license fee when he bought a new car for a total of $25868.5. Find the purchased price of a car? 4. Find the slope of the straight line passing through the indicated points i) (4, 1), (7, 5) ii) (5, −3), (6, −4) iii) (1, 0), (0, 5) iv) (2, −4), (3, −4) 5. Find the equation of a line that has the indicated properties and sketch the line a passing through (−1, 3) and parallel to y = 4x − 5 b has slope 0 and y−intercept − 1 2 c passes through (3, −1) and (−2, −9) d passing through origin and has slope −5 e passing through (−5, 4) and perpendicular to the line 2y = x + 1 f perpendicular to y = 3x − 5 and passing through (3, 4) g passing through (2, −8) and parallel to x = −4 6. A small company produces chairs. The weekly fixed and variable costs per chair are $1100 and $40 respectively. Find the total weekly cost of manufacturing x chairs. How many chairs can be manufactured for a total weekly cost of $4, 500? 7. The sales of a company were $33, 000 in its third year of operation and $95, 000 in its sixth year. If y denote sales in year x. what were the sales in the fifth year? 94

117. Unit-5 MATRICES WRITTEN BY: MS MUBASHARA HAFEEZ REVIEWED BY: DR NASIR REHMAN 95

118. 5.1 INTRODUCTION The idea of Matrices was first presented in nineteenth century by Arthur Cayley, a famous Mathematician who first gave Theory of Matrices in 1857 and applied them in linear transformations. Matrices and Determi- nants are widely applied in the field of Mathematics, Physics, Statistics, Electronics and several other disciplines. In this unit students will be able to learn about 1. Introduction to matrices and purpose of studying matrix algebra 2. Various types of matrices 3. Apply basics operations of addition, subtraction and multiplication on matrices 4. Solve a system of linear equations and related real life problems. 5.2 OBJECTIVES After studying this unit students will be able to 1. understand the data representation using matrices 2. business applications of matrices problems 3. translate the practical problem linear equations into matrices 4. Solve a system of linear equations and the related real life problems. 96

119. 5.3 INTRODUCTION TO MATRICES A matrix is a collection of numbers that are arranged into rows and columns. In general, a matrix is called a rectangular array of elements. 5.3.1 Generalized Form of a Matrix Matrices are commonly written as A =    a11 ... a1n . . . ... . . . am1 ... amn    5.3.2 Purpose of Studying Matrix Algebra In most of the scientific fields we use applications of matrices. Matrices are suitable for the storage, display, and manipulation of data. Matrices are used in graphic software to process linear transformations to render images. 5.4 SPECIAL TYPES OF MATRICES Matrices are distinguished on the basis of their order, elements and other conditions. There are several types of matrices but the most commonly used are discussed below 5.4.1 Vector Any object that has magnitude and direction is called vector. A vector is a matrix having only one row or one column. 5.4.2 Row Vector A row vector is a matrix having only one row. A row vector R having n elements rij has dimension (1 × n) and the general form is R = r11 r12 ... r1n 97

120. 5.4.3 Column Vector A column vector is a matrix having only one column. A column vector C having m elements cij has dimension (m × 1) and the general form is. C =      c11 c21 . . . cm1      5.4.4 Square Matrices A matrix with the same number of rows and columns is known as square matrix. Following matrix is a square matrix of order (3 × 3).   a11 a12 a13 a21 a22 a23 a31 a32 a33   5.4.5 Identity Matrix Identity matrix is also known as unit matrix, is a square for which the elements in the diagonal are equal to 1 and all other elements are equal to 0 such as: I = 1 0 0 1 5.4.6 Transpose A matrix which is formed by converting all the rows of a matrix into columns or converting all the columns into rows. The transpose of the given matrix A is written as At . Example Find the transpose of matrix A = 1 2 3 4 5 6 To find the transpose of matrix A we must change rows into columns. At =   1 4 2 5 3 6   98

121. 5.5 MATRICES OPERATIONS In this section we will discuss about the some of operations of matrix algebra. 5.5.1 Matrix Addition and Subtraction Two matrices can be added or subtracted only if they have same dimensions. They must have same number of rows and columns. If A and B are matrices of (m × n) after addition or subtraction of these two matrices we have new matrix C of the same dimension as A and B i.e. Cij = Aij ± Bij Example 1 Given A = 4 8 3 7 and B = 1 0 5 2 lets find A + B Solution We can find the sum by adding the corresponding entries in matrices A and B. This is shown below. A + B = 4 8 3 7 + 1 0 5 2 A + B = 4 + 1 8 + 0 3 + 5 7 + 2 A + B = 5 8 8 9 Example 2 The quarterly sales of wheat, cotton and corn for the year 2010 and 2011 are represented below in the form of matrix A and B. A =   20 25 22 20 10 20 18 10 15 20 15 15   and B =   10 15 20 20 5 20 18 10 8 30 15 10   99

122. Find the total quarterly sales of wheat, cotton and corn for these two years. Solution A + B =   20 25 22 20 10 20 18 10 15 20 15 15   +   10 15 20 20 5 20 18 10 8 30 15 10   A + B =   30 40 42 40 15 40 36 20 23 50 30 25   5.5.2 Scalar Multiplication Multiplication of a matrix by a scalar number is called scalar multiplication. Where scalar is a real number. If k is a scalar and A be a matrix then scalar multiplication is given below. kA = k a11 a12 a21 a22 = ka11 ka12 ka21 ka22 Example 3 For a matrix A = 1 0 −2 0 3 −1 find 2A. Solution To find 2A we just must multiply by 2 every entry in the matrix. 2A = 2 1 0 −2 0 3 −1 2A = 2 × 1 2 × 0 2 × (−2) 2 × 0 2 × 3 2 × (−1) 2A = 2 2 0 −4 0 6 −2 5.5.3 Inner Product Inner product is the generalization of the dot product. Let A = a11 a12 ... a1n and B =      b11 b21 . . . bn1      then the inner product, written as A.B is A.B = a11b11 + a12d21 + ... + a1nbn1 100

123. From the definition of inner product, we must follow the following three points. • The inner product is defined only if the number of elements in row and column are same. • When a row vector is multiplied by a column vector the result of the inner product is a scalar quantity. • The inner product is computed by multiplying corresponding elements in the two vectors and algebraically summing. Example 4 Find the inner product of row and column vectors A = −5 3 0 2 and B =     3 −1 4 2     Solution The inner product of row and column vector is computed as A.B = −5 3 0 2 .     3 −1 4 2     A.B = ((−5)(3) + (3)(−1) + (0)(4) + (2)(2)) A.B = −14 5.5.4 Matrix Multiplication Matrix multiplication is multiplication of entire matrix by another entire matrix. • The product of a matrix A and B is defined if and only if the number of rows of matrix A is equal to the number of columns of matrix B. • For matrix multiplication, we must multiply the rows of A by the columns of B. 101

124. • Take the first row of A and first column of B multiply their entries. • The result of the matrix multiplication is in the form of matrix. Example 5 Given A = 2 −3 4 5 and B = 1 3 7 4 find the product AB Solution Here A is (2 × 2) and B is also (2) so the product is possible. Now AB = 2 −3 4 5 1 3 7 4 AB = (2)(1) + (−3)(7) (2)(3) + (−3)(4) (4)(1) + (5)(7) (4)(3) + (5)(4) AB = 2 − 21 6 − 12 4 + 35 12 + 20 AB = −19 −6 39 32 Example 6 Ali, Aftab and Danish bought cookies of different brands P, Q and R. Ali bought 10 packets of S, 7 packets of T and 3 packets of U. Aftab bought 4 packets of S, 8 packets of T and 10 packets of U. Danish bought 4 packets of S, 7 packets of T and 8 packets of U. If brand S costs Rs 4, T costs Rs 5 and U costs Rs 6 each, then using matrix operation, find the total amount of money spent by these three persons individually. Solution Let Q represents the matrix for the quantity of each brand of biscuit bought by S, T and U and let C is the matrix representing the cost of each brand of biscuit. Q =   10 7 3 4 8 10 4 7 8   C =   4 5 6   102

125. Since the number of the columns of matrix Q is equal to the number of rows of the matrix C so the matrix multiplication is possible. Q × C =   (10)(4) + (7)(5) + (3)(6) (4)(4) + (8)(5) + (10)(6) (4)(4) + (7)(5) + (8)(6)   Q × C =   40 + 35 + 18 16 + 40 + 60 16 + 35 + 48   Q × C =   93 116 99   Amount spent by Ali, Aftab and Danish is Rs 93, Rs 116 and Rs 99 respectively. 5.6 REPRESENTATION OF AN EQUATION A linear equation of the form a1x1 +a2x2 +....+anxn = b can be represented in matrix form as a1 a2 ... an      x1 x2 . . . xn      = b Example 7 An equation 5x1 + 8x2 − 3x3 can be represented by the inner product such as. 5 8 −3   x1 x2 x3   5.7 REPRESENTATION OF A SYSTEM OF EQUATIONS An (m × n) system of equations having the form a11x1 + a12x2 + ...a1nxn = b1 . . . am1x1 + am2x2 + ...amnxn = bm 103

126. can be represented by the matrix equation AX = B    a11 ... a1n . . . ... . . . am1 ... amn         x1 x2 . . . xn      =      b1 b2 . . . bn      Example 8 A company produces three products every day. Their total production on a certain day is 50 tons. Further It is observed that the production of the third product exceeds the production of the first product by 8 tons also it is observed that the total combined production of the first and third product is twice that of the second product. Write the given problem in equation and matrix form. Solution Let us denote the production level of the given three products be x, y and z respectively. Then, we can make the following equations. x + y + z = 50 z = x + 8 x + z = 2y By rearranging the equation, we have, x + y + z = 50 z − x = 8 x − 2y + z = 0 Now we can write the system of linear equations in matrix form   1 1 1 −1 0 1 1 −2 1     x y z   =   50 8 0   Example 9 Ms. Juliet and Mr. John work for a car agency that sells two models. The last month was August for the models of this year and the models of the next year were presented in September. The following matrices represents the gross dollar sales for each month. Find 104

127. a The joint dollar sales in August and September for each salesperson each model? b The increase in dollar sale from August to September? c On a gross dollar sales, if both salespersons take 6% commissions then find the commission received by each person in September sale for each model? Solution a A + B = $3600 $72, 000 $72, 000 0 + $144, 000 $288, 000 $180, 000 $216, 000 A + B = $147, 600 $360, 000 $252, 000 $216, 000 b B − A = $144, 000 $288, 000 $180, 000 $216, 000 − $3600 $72, 000 $72, 000 0 B − A = $140, 400 $216, 000 $108, 000 $216, 000 c 0.06 × B = (0.06)$144, 000 (0.06)$288, 000 (0.06)$180, 000 (0.06)$216, 000 0.06 × B = $8, 640 $17, 280 $10, 800 $12, 960 Example 10 The following amounts of pencils, erasers and sharpeners are sold in two days 105

128. Pencils Erasers Sharpeners Monday 40 8 6 Tuesady 50 11 5 The price of each pencil, eraser and sharpener is $0.3, $0.2 and $3 respectively. How much was made each day? Solution 40 8 6 50 11 5 ×   0.3 0.2 3.0   = (40)(0.3) + (8)(0.2) + (6)(3) (50)(0.3) + (11)(0.2) + (5)(3) = 12 + 1.6 + 18 15 + 2.2 + 1 = 31.6 32.2 Example 11 Suppose a vector A = 23.45 45.75 29.50 represents the prices (in dollars per unit) of three books. Let vector B =   330 450 275   represents the quantities of books ordered by a university bookstore. What is the total cost of the purchase? Solution A × B = 23.45 45.75 29.50 ×   330 450 275   A × B = [(23.45)(330) + (45.75)(450) + (29.50)(275)] A × B = [36438.5] 106

129. 5.8 SELF ASSESMENT QUESTIONS 1. Find the transpose of each of the following matrices: i) A =   −1 0 0 6 9 −2 −4 5 1   ii) B =   6 −6 4 9 −2 0   iii) C = 0 0 2 2. If A =   −8 0 7 5 2 −1 −4 0 3  , then verify that (At )t = A 3. If A =   1 2 3 −1 −2 3 3 4 5   and B =   4 −1 1 1 4 3 3 −3 −2   then find i) 4A − 3B ii) A + 2(B − A) iii) 2At + 3Bt iv) AB v) (A − 2Bt )t vi) B2 − 3B + 2I vii) A2 B3 also verify that: i) (A+B)t = At +Bt ii) A+(B+A) = 2A+B iii) A(B−C) = AB − AC iv) (AB)t = Bt At 4. Prove that if AB = BA, then (A + B)(A − B) = A2 − B2 5. A pet store has 6 kittens, 8 parrots and 12 puppies. If the price of each kitten is $28, each parrot is $30 and each puppy is $40. Use matrix multiplication to find the total value of the pet stores inventory. 6. Attendance for the first three football games of the season is described in the table. Adult tickets sold for $3.00, student tickets sold for $1.5. Use matrix multiplication to find the revenue for each. Adults Students Game 1 340 180 Game 2 250 195 Game 3 300 220 107

130. 7. Let A = a1 a2 a3 represents the prices of the products X, Y, Z. If 15% prices are suggested to be increased, by what scalar, A can be multiplied to obtain new prices? 8. Burger Barns three locations sell fries, burgers and soft drinks. Bran I sells 500 orders of fries, 150 burgers and 300 soft drinks on daily basis. Bran II and Bran III sell 900 and 700 orders of fries respectively, per day. Bran II sells 440 burgers and 830 soft drinks while Bran III sells 580 burgers and 1200 soft drinks each day. a Write a 3 × 3 matrix that represents daily sales for all locations. b If fries cost $0.92 per order, burger $1.54 each and soft drink $0.58 each. Write a matrix of order 1 × 3 that represents the prices. c What is the total daily income from all the locations? 108

131. Unit-6 DETERMINANTS AND INVERSES OF MATRICES WRITTEN BY: MS MUBASHARA HAFEEZ REVIEWED BY: DR NASIR REHMAN 109

132. 6.1 INTRODUCTION The idea of Matrices was first presented in nineteenth century by Arthur Cayley, a famous Mathematician who first gave Theory of Matrices in 1857 and applied them in linear transformations. Matrices and Determi- nants are widely applied in the field of Mathematics, Physics, Statistics, Electronics and several other disciplines. In this unit students will be able to learn about 1. Determinants and properties of determinants 2. Methods of cofactors 3. Cramer’s rule 4. Inverse of a matrix and system of equations 6.2 OBJECTIVES After learning this unit students will be able to learn about 1. the apply basic concept of cofactors 2. applications of determinants in business related problems 3. the algebra of equations and the corresponding systems for solutions 110

133. 6.3 THE DETERMINANT A number which can be calculated from a square matrix is known as determinant of a matrix. The determinant of a matrix A is denoted as |A|or det A. 6.3.1 Determinant of a (1 × 1) Matrix Determinant of a matrix (1 × 1) is the number itself in the matrix. For Example We have a matrix A = [9] the determinant of a matrix A will be, |A| = 9 i.e. number given in matrix itself. 6.3.2 Determinant of a (2 × 2) Matrix The determinant of matrix (2 × 2) is found by subtracting the products of its diagonals i.e. A = a11 a12 a21 a22 |A| = a11a22 − a12a21 Example 1 Find the determinant of a matrix A = 4 1 3 2 Solution The determinant of the given matrix is A = 4 1 3 2 |A| = (4)(2) − (1)(3) |A| = 8 − 3 |A| = 5 6.3.3 Determinant of (3 × 3) Matrix The determinant of a matrix A =   a11 a12 a13 a21 a22 a23 a31 a32 a33   can be found by the following procedure. 111

134. 1. Write the first two columns of the matrix to the right of the original matrix. 2. Multiply the down diagonals and add them together. 3. Multiply the up diagonals and add them together. 4. The determinant equals to the sum of the products of down diagonals minus the sum of the products of up diagonals. Algebraically the determinant is computed as. Example 2 Find the determinant of the matrix A =   5 −2 1 0 3 −1 2 0 7   Solution To find the determinant of the matrix A firstly we will write the first two columns of the matrix A at the right side of the matrix i.e. Now the determinant is computed as, |A| = [(5)(3)(7) + (−2)(−1)(2) + (1)(0)(0)] − [(1)(3)(2) + (5)(−1)(0) + (−2)(0)(7)] |A| = (105 + 4) − (6) |A| = 103 112

135. 6.4 METHOD OF COFACTORS The method of cofactors is the more generalized computational procedure which can be applied for all square matrices. Following are the steps to find the cofactors. 1. Firstly, we must cross off row i and column j in the original matrix. 2. Focus on the remaining terms of the matrix. The remaining terms of the matrix form a submatrix of the original matrix. 3. Find the determinant of the remaining matrix known as minor of the element aij. 4. The factor a 0 ij is found by multiplication the minor by either +1 or -1 depending on the position of the element aij. Formula for computing cofactors is a 0 ij = (−1)i+j the minor Example 3 Find all the cofactors if A =   2 −1 1 3 2 0 0 4 −5   Solution 113

136. A11 = (−1)1+1

140. 2 0 4 −5

144. = −10 A12 = (−1)1+2

148. 3 0 0 −5

152. = 15 A13 = (−1)1+3

156. 3 2 0 4

160. = 12 A21 = (−1)2+1

164. −1 1 4 −5

168. = −1 A22 = (−1)2+2

172. 2 1 0 −5

176. = −10 A23 = (−1)2+3

180. 2 −1 0 4

184. = −8 A31 = (−1)3+1

188. −1 1 2 0

192. = −2 A32 = (−1)3+2

196. 2 1 3 0

200. = 3 A33 = (−1)3+3

204. 2 −1 3 2

208. = 7 6.4.1 Method of Cofactors Expansion a) Select any row or column of the matrix. b) Multiply each element in the row (column) by its corresponding cofactor and sum these products to yield the determinants. The determinant of (m × m) matrix A can be calculated using the method of cofactor expansion i.e. |A| = ai1a 0 i1 + ai2a 0 i2 + ... + aima 0 im Example 4 Find the determinant of the matrix A =   6 1 1 4 −2 5 2 8 7   Solution The determinant of a given matrix can be found by the method of cofactor expansion i.e. |A| = ai1a 0 i1 + ai2a 0 i2 + ... + aima 0 im 114

209. By the formula the cofactor is computed as, 115

210. Now we will substitute these cofactors into the cofactors expansion i.e. |A| = a11a 0 11 + a12a 0 12 + a13a 0 13 |A| = 6(−54) − 1(18) + 1(36) |A| = −306 Example 5 If A =   1 −2 −3 2 0 1 −4 6 8  , find |A| Solution |A| =

216. 1 −2 −3 2 0 1 −4 6 8

222. We expand it along the first row i.e. |A| = 1.

226. 0 1 6 8

230. − (−2)

234. 2 1 −4 8

238. + (−3)

242. 2 0 −4 6

246. = 1(0 − 6) + 2(16 − (−4)) − 3(12 − 0) = −6 + 40 − 36 = 40 − 42 |A| = −2 If we expand it along the second column, then |A| =

252. 1 −2 −3 2 0 1 −4 6 8

258. 116

259. |A| = −2(−1)1+2

263. 2 1 −4 8

267. + 0(−)2+2

271. 1 −3 −4 8

275. + 6(−1)3+2

279. 1 −3 2 1

283. = (−2)(−1)(16 + 4) + 0 + (6)(−1)(1 + 6) = 2(20) − 6(7) = 40 − 42 |A| = −2 Thus, the determinant of a square matrix is unique and is independent of the choice of the row or column for its evaluation. 6.4.2 Adjoint of a Matrix of Order n ≥ 3 The adjoint of a square matrix A is obtained by taking the transpose of the cofactor matrix of a given original matrix. Let A =   a11 a12 a13 a21 a22 a23 a31 a32 a33   and it’s cofactor matrix is A =   A11 A12 A13 A21 A22 A23 A31 A32 A33   then AdjA =   A11 A12 A13 A21 A22 A23 A31 A32 A33   t =   A11 A21 A31 A12 A22 A32 A13 A23 A33   6.5 PROPERTIES OF DETERMINANTS i. If all elements of any row or column equal to zero then determinant of that matrix is equal to zero. ii. If any two rows or columns are interchanged, then the sign of the determinant changes. iii. If the matrix B is obtained from the matrix A by multiplying every element of matrix A by scalar k then |A| = k|B|. iv. If any multiple of one row/column is added to another row/column, the value of the determinant in unchanged. v. If two rows/columns are identical then the determinant is zero. 117

284. 6.6 CRAMER’s RULE Suppose we have a system of linear equations AX = B Where A represents (n × n) square matrix of coefficients. Cramer’s Rule gives us a method of solving the given system of equations by using different determinants. To solve for the value of the jth variable, from matrix by Ai replacing the ith column of A with the column vector of B. The value of the ith variable is determined as, xi = |Ai| |A| Example 6 Total 820 tickets were sold for a game for a total of $9,128. If adults tickets were sold for $12.00 and children tickets were sold for $8.00, how many of each kind of tickets were sold? Solution Let x be the number of adults tickets and y be the number of children tickets. Then we have the system of two linear equations in two unknowns. x + y = 820 12x + 8y = 9128 Now apply the Cramers rule to solve the system. AX = B 1 1 12 8 x y = 820 9128 |A| = (1)(8) − (1)(12) |A| = −4 118

285. Now interchange the first column of matrix A by the column of matrix B. A1 = 820 1 9128 8 |A1| = (820)(8) − (9128)(1) |A1| = −2568 x = |A1| |A| x = −2568 −4 x = 642 Now interchange the second column of matrix A by the column of matrix B. A2 = 1 820 12 9128 |A2| = (9128)(1) − (820)(12)− |A2| = −712 y = |A2| |A| y = −712 −4 y = 178 642 adult tickets and 178 children tickets were sold. Example 7 The equilibrium conditions for two related markets are given below where Pc is the price of chicken and Pb is the price of beef. 18Pb − Pc = 87 −2Pb + 36Pc = 98 Find the equilibrium price for each market (Pc, Pb). Solution The following system of linear equations in the matrix form. AX = B 18 −1 −2 36 Pb Pc = 87 98 119

286. Apply the Cramers rule to solve the system. A = 18 −1 −2 36 |A| = (18)(36) − (−1)(−2) |A| = 646 Now interchange the first column of matrix A by the column of matrix B. A1 = 87 −1 98 36 |A1| = (87)(36) − (−1)(98)− |A1| = 3230 Pb = |A1| |A| Pb = 3230 646 Pb = 5 Now interchange the second column of matrix A by the column of matrix B. A2 = 18 87 −2 98 |A2| = (18)(98) − (−2)(87)− |A2| = 1938 Pc = |A2| |A| Pc = 1938 646 Pc = 3 The equilibrium price for each market is Pc = 3 and Pb = 5. Example 8 The sum of three numbers is 26. The third number is twice the second and1 less than three times the first. Find these three numbers? Solution Let us represent the first number by x The second number by y 120

287. and the third number by z Then x + y + z = 26 z = 2y ⇒ −2y + z = 0 z = 3x − 1 ⇒ −3x + z = −1 Let A =   1 1 1 0 −2 1 −3 0 1   , X =   x y z   and B =   26 0 −1   We will solve it using Cramers Rule |A| =

293. 1 1 1 0 −2 1 −3 0 1

299. |A| = 1(−2) − 1(3) + 1(−6) = −2 − 3 − 6 = −11 x =

305. 26 1 1 0 −2 1 −1 0 1

311. 11 = − 1 11 (26(−2) − 1(1) + 1(−2)) = − 1 11 (−55) = 5 y =

317. 1 26 1 0 0 1 −3 −1 1

323. 11 = − 1 11 (1(1) − 26(3) + 1(0)) = − 1 11 (−77) = 7 z =

329. 1 1 26 0 −2 0 −3 0 −1

335. 11 = − 1 11 (1(2) − 1(0) + 26(−6)) = − 1 11 (−154) = 14 6.7 INVERSE OF A MATRIX The inverse of any matrix A is denoted as A−1 with the property that is, where I is the identity matrix. Note that here A−1 does not mean that 1 A . AA−1 = A−1 A = I 121

336. 6.7.1 Some Facts Regarding the Inverse i. For a matrix A to have an inverse it must be a square matrix. ii. Inverse of matrix A will also be square and of the same dimension as A. iii. Not every square matrix has an inverse. 6.7.2 Determining the Inverse There are several methods for determining the inverse of a matrix. One of them is Gaussian elimination procedure. 6.7.3 Gaussian Reduction Procedure To determine the inverse of an (m × m) matrix A. i. Augment the matrix A with an (m × m) identity matrix resulting in (A|I). ii. Perform row operations on the entire augmented matrix to transform A into an (m × m) identity matrix. Then the matrix will have the form (I|A−1 ). iii. Where A−1 can be read to the right of the vertical line. Example 9 Lets determine the inverse of the matrix A = 1 2 3 4 Solution Use the Gauss elimination method to transform (A|I) into (I|A−1 ).We will do some row operation to make the matrix A into identity matrix as a result the changes occurs in identity matrix will give us inverse of a matrix A. Example 10 Find the inverse of a given matrix using Gaussian Reduction Method. A =   1 0 −3 2 −4 −6 3 2 1   122

337. Solution Hence A−1 =       − 1 5 3 20 3 10 1 2 − 1 4 0 − 2 5 1 20 1 10       6.7.4 Finding the Inverse Using the Cofactors The cofactor method for finding the inverse of a square matrix A is as follows. i. Determine the matrix of cofactors Ac for the matrix A. 123

338. ii. Determine the adjoint matrix Aj which is the transpose of the matrix Ac. Aj = At c iii. The inverse of the matrix A can be found by dividing the determinant of matrix A with the adjoint matrix i.e. A−1 = 1 A Aj Example 11 Find the inverse of the given matrix A using cofactors. A =   2 1 0 −1 −5 2 0 1 3   Solution We first find the cofactors of the elements of A. A11 = (−1)1+1

342. −5 2 1 3

346. = −17 A12 = (−1)1+2

350. −1 2 0 3

354. = 3 A13 = (−1)1+3

358. −1 −5 0 1

362. = −1 A21 = (−1)2+1

366. 1 0 1 3

370. = −3 A22 = (−1)2+2

374. 2 0 0 3

378. = 6 A23 = (−1)2+3

382. 2 1 0 1

386. = −2 A31 = (−1)3+1

390. 1 0 −5 2

394. = 2 A32 = (−1)3+2

398. 2 0 −1 2

402. = −4 A33 = (−1)3+3

406. 2 1 −1 −5

410. = −9 124

411. The cofactor matrix of A =   A11 A12 A13 A21 A22 A23 A31 A32 A33   =   −17 3 −1 −3 6 −2 2 −4 −9     −17 3 −1 −3 6 −2 2 −4 −9   t =   −17 −3 2 3 6 −4 −1 −2 −9   |A| = a11A11 + a12A12 + a13A13 |A| = 2(−17) + 1(3) + 0(−1) |A| = −34 + 3 + 0 = −31 6= 0 A−1 = AdjA |A| A−1 = − 1 31   −17 −3 2 3 6 −4 −1 −2 −9   6.7.5 The Inverse and System of Equations The matrix inverse can be used to determine the solution set for a system of equations. Given a system of equations of the form AX = B where A is a square matrix if we multiply both sides with inverse of matrix A then we have X = A−1 B that is the solution vector for the system of equations. Example 12 A company utilizes three kinds of steels s1, s2 and s3 (in tons) for manufacturing three kinds of cars c1, c2, c3. The requirement of steel for each kind of a car is given by Figure 6.1: Cars Manufacturing Industry (Courtesy Budnick) 125

412. Find the quantity of cars of all kind that can be manufactured by using 30, 20, 10 tons of steel of three kinds respectively. 126

413. Solution Let x, y, z represent the number of cars that can be produced, then 4x + 5y + 6z = 30 2x + 2y + 4z = 20 5x + 4y + 2 = 10 This system can be written in matrix form as   4 5 6 2 2 4 5 4 2     x y z   =   30 20 10   AX = B |A| =

419. 4 5 6 2 2 4 5 4 2

425. |A| = 4(4 − 16) − 5(4 − 20) + 6(8 − 10) |A| = 4(−12) − 5(−16) + 6(−2) |A| = −48 + 80 − 12 = 20 6= 0 We can write AX = B as X = A−1 B. To find inverse we use Gaussian Reduction method. 127

426. From X = A−1 B we have Thus, x = 0, y = 0, z = 5 Example 13 The input-output matrix for a three-industry economy is If nonindustrial demands are respectively $100,000,000, $60,000,000, and 150,000,000. Then determine the equilibrium levels of output for three industries. Demands =   100, 000, 000 60, 000, 000 150, 000, 000   = D Solution 128

427. The equilibrium levels of output can be determined by using the formula. (I − A)−1 D = X (I − A) =   1 0 0 0 1 0 0 0 1   −   0.25 0.30 0.20 0.20 0.30 0.20 0.40 0.10 0.25   (I − A) =   0.75 −0.30 −0.20 −0.20 0.7 −0.20 −0.40 −0.10 0.75   (I − A) =   0.75 −0.30 −0.20 1 0 0 −0.20 0.7 −0.20 0 1 0 −0.40 −0.10 0.75 0 0 1   129

428. Hence the Industry should produce $198800000, $321860000, and $446050000 130

429. 6.8 SELF ASSESMENT QUESTIONS 1. Evaluate the following determinants. Use properties of determinants if possible. (a)

435. 2 3 −1 1 1 0 2 −3 5

441. (b)

447. 2a a a b 2b b c c 2c

453. (c)

459. 1 2 3x 2 3 6x 3 5 9x

465. (d)

471. a − b b − c c − a b − c c − a a − b c − a a − b b − c

477. (e)

485. 2 3 −1 1 3 0 2 5 0 8 6 3 7 −4 5 0

493. (f)

501. 1 0 −2 3 0 0 0 −5 4 2 1 4 1 5 −4 3

509. 2. Let A =   1 3 −1 5 2 3 −2 1 0   and B =   −1 −2 0 3 5 1 −2 −3 4   (a) Find A−1 , B−1 , A−1 B−1 , B−1 A−1 and (AB)−1 (b) Is it true that (AB)−1 = A−1 B−1 ? (c) Is it true that (AB)−1 = B−1 A−1 ? (d) Does the matrix A + B have an inverse? If so, what is it? (e) Does the matrix A − B have an inverse? If so, what is it? 3. Find the inverse, if exist, of the given matrices by using both Cofactor approach and Gaussian reduction method? (a)   6 0 −1 1 3 5 −3 1 0   (b)   1 2 3 1 3 5 1 5 12   (c)   −1 2 −3 0 1 5 2 4 7   (d)   8 0 −2 1 −2 1 −3 1 1   4. Solve the following system of linear equations using Matrices as well as Cramers rule. 131

510. (a) 2x − y + z = 5 4x + 2y + 3z = 8 3x − 4y − z = 3 (b) x + y = 2 2x − z = 1 2y − 3z = −1 (c) 2x + y + 3z = 3 x + y − 2z = 0 −3x − y + 2z = −4 5. An electronic company manufactures resistors, transistors and computer chips. Each resistor requires 3 units of copper, 2 units of zinc and 1 unit of glass. Each transistor requires 3, 2, 1 units of each material respectively and each computer chip requires 2, 1 and 2 units of these materials respectively. Using the following amounts of materials, how many of each item can be prepared? (a) 800 units of copper, 400 units of zinc and 500 units of glass 6. In a pottery factory, a machine takes 3 minutes to make a glass and 2 minutes to make a plate. The material for a glass costs $0.25 and the material for a plate costs $0.20. If the machine runs for 10 hours and exactly $50 is used up for material. How many glass and plates can be produced? 132

511. Unit-7 DIFFERENTIATION WRITTEN BY: MS FOUZIA REHMAN REVIEWED BY: DR BABAR AHMED 133

512. 7.1 INTRODUCTION The concept of limit is related to finding the areas of plane region and finding the tangent lines to the curve. The problem in which we deal with the tangent problem is called differentiation. In limit and differentiation we deal with physical phenomena which involve changing quantities the speed of a rocket, the number of bacteria in a culture, the inflation of a currency, the shock of an earthquake, the voltage of an electric signal and a relation exist between tangent and rates of change. In this unit students will be able to learn about 1. Limits and test for existence of a limit 2. Properties of limit and continuity 3. Horizontal and vertical asymptote 4. Average rate of change 5. Derivative and rules of differentiation 6. Higher order of derivatives 7.2 OBECTIVES After going through this unit students will be able to learn about 1. applied concepts of limit and continuity 2. rate of change of certain business related terminologies 3. real life problems related to derivatives in applied business 4. Average rate of change of certain quantities 5. Application of derivatives 134

513. 7.3 LIMITS In differentiation limit is the most important concept. Limits are used to define continuity, derivatives and integrals. In this section we will discuss about limits of function, test for existence of limits and it’s properties. 7.3.1 Limits of Function Limit is the value that a function or sequence approaches as the input approaches to some value. This limiting value when it exists is called limit. It is denoted as lim x→a f(x) = L 7.3.2 Test for Existence of a Limit If the value of the function approaches the same number L as x approaches a from positive direction right side or negative direction left side then the limit equals L i.e. If limx→a− f(x) = L and limx→a+ f(x) = L then, lim x→a f(x) = L Example 1 Determine the limx→4 f(x) where f(x) = −x2 + 2x + 2 Solution Let’s construct a table of assumed values of x and corresponding values for f(x). limx→4− 3.9 3.99 3.999 f(x) -5.41000 -5.94010 -5.99400 limx→4+ 4.1 4.01 4.001 f(x) -6.61000 -6.06010 -6.00600 Note that the value x = 4 has been approaches from both the left and the right side and approaching the same value of f(x) i.e. -6 hence lim x→4+ f(x) = −6 and lim x→4− f(x) = −6 ⇒ lim x→4 f(x) = −6 135

514. Example 2 Find the indicated limit (if exist) by constructing a table lim x→0 1 x2 Solution Let f(x) = 1 x2 Figure 7.1: Graphical Representation of Limit It can be seen from the table and graph that as x gets closer and closer to 0 from both left and right, f(x) increases without bound. Thus lim x→0 1 x2 does not exist Example 3 Find limx→1 x3 − 1 x − 1 Solution As x approaches to 1 both numerator and denominator approaches to zero gives 0 0 form which has no meaning. The limit can be determined by factoring the quotient i.e., x3 − 1 x − 1 = (x − 1)(x2 + x + 1) x − 1 = (x2 + x + 1) 136

515. Thus lim x→1 x3 − 1 x − 1 = lim x→1 (x2 + x + 1) = 3 Figure 7.2: Graphical Representation of Limit Example 4 Let f(x) =      x − 2; x 0 x2 0 ≤ x ≤ 2 2x x 2 Find (a)limx→0 f(x) (b) limx→1 f(x) (c) limx→2 f(x) Solution (a) lim x→0− f(x) = lim x→0− (x − 2) = −2 lim x→0+ f(x) = lim x→0+ (x2 ) = 0 lim x→0− f(x) 6= lim x→0+ f(x) 137

516. Thus limx→0 f(x) does not exist. (b) lim x→1 f(x) = lim x→1 (x2 ) = 1 (c) lim x→2− f(x) = lim x→2− (x2 ) = 4 lim x→2+ f(x) = lim x→2+ (2x) = 4 lim x→2− f(x) = lim x→2+ f(x) = 4 Thus limx→2 f(x) exist. Example 5 Suppose that the production costs per week for producing x widgets in a factory is given by C(x) = 500 + 350x − 0.09x2 Calculate the cost to produce the 301st widget at x = 300? Solution We must compute the cost to produce 301st for this firstly we will compute C(301) as that is the cost of producing 301 widgets and then we will minus it with C(300) which is the cost of producing 300 widgets. C(301) = 500 + 350(301) − 0.09(301)2 C(301) = 97695.91 C(300) = 500 + 350(300) − 0.09(300)2 C(301) = 97400 C(301) − C(300) = 97695.91 − 97400 = 295.91 Hence the cost of producing the 301st widget is 295.91 Example 6 After s days on the job training of a company training program a new employ on average can do p(s) pieces of work per day. Where P(s) = 60s s + 4 Find (a) P(1) (b) P(11) (c) limx→11 P(s) 138

517. Solution (a) P(1) = 60(1) 1 + 4 P(1) = 60 5 = 12 (b) P(11) = 60(11) 11 + 4 P(11) = 660 15 = 44 (c) lim x→11 P(s) = lim x→11 60s s + 4 lim x→11 P(s) = 60(11) 11 + 4 lim x→11 P(s) = 660 15 = 44 7.4 PROPERTIES OF LIMIT Properties of limits and continuity are defined below. 7.4.1 Some Properties of Limit i. If f(x) = c where c is a real number then, lim x→a (c) = c ii. If f(x) = xn where n is a positive integer then, lim x→a (xn ) = an 139

518. iii. If f(x) has a limit as x → a and c is a real number then, lim x→a (c)f(x) = c lim x→a f(x) iv. If limx→a f(x) and limx→a g(x) exist then, lim x→a [f(x) ± g(x)] = lim x→a f(x) ± lim x→a g(x) v. If limx→a f(x) and limx→a g(x) exist then, lim x→a [f(x).g(x)] = lim x→a f(x). lim x→a g(x) vi. If limx→a f(x) and limx→a g(x) exist then, lim x→a f(x) g(x) = limx→a f(x) limx→a g(x) Provided lim x→a g(x) 6= 0 7.4.2 Limit at Infinity In limits at infinity we are looking in the behavior of function when is becomes very large in either the positive or negative sense. For example lim x→∞ f(x) = lim x→∞ f(x) 2x4 − x2 + 8x −5x4 + 7 Multiply and divide by x4 lim x→∞ f(x) = lim x→∞ f(x) 2 − 1 x2 + 8 x3 −5 + 7 x4 lim x→∞ f(x) = 2 + 0 + 0 −5 + 0 lim x→∞ f(x) = − 2 5 As x approaches to infinity the function f(x) approaches to −2 5 7.4.3 Continuity Over an Interval A function f(x) is said to be continuous at x = a if i. limx→a f(x) exists 140

519. ii. f(a) is defined iii. limx→a f(x) = f(a) Example 7 Given the graph of function f(x) determine if f(x) is continuous at x = −2, x = 0 and x = 3 Figure 7.3: Graphical Representation of f(x) Solution At x = −2 lim x→−2+ f(x) = −1 and lim x→−2− f(x) = 2 Therefore limx→−2 f(x) does not exist. At x = 0 f(0) = 0 lim x→0 f(x) = 1 The function is continuous at this point since the function and limit have the same value. At x = 3 f(3) = −1 lim x→3 f(x) = 0 The function is not continuous at x = 3. Although the limit exists but not equal to the value of the function at x = 3. 141

520. 7.5 HORIZONTAL AND VERTICAL ASYM- POTOTES Asymptotes are the curves or lines that approaches a given curve arbi- trarily closely. Or we can say that an asymptote is a line that a curve is about to approaches infinity. Now further we will discuss about horizontal asymptote and vertical asymptote. 7.5.1 Horizontal Asymptote The line y = a is a horizontal asymptote of the graph of f if and only if lim x→∞ f(x) = a or lim x→−∞ f(x) = a 7.5.2 Vertical Asymptote The line x = a is a vertical asymptote of the graph of f if and only if lim x→a− f(x) = ∞ or (−∞) or lim x→a f(x) = ∞ or (−∞) Figure 7.4: Horizontal and Vertical Asymptotes 142

521. 7.6 AVERAGE RATE OF CHANGE The average rate of change is the change of of function over an interval. For a function we can find the average rate of change if we divide the change in one quantity by the change in other quantity. 7.6.1 Average Rate of Change and Slope We can calculate the slope of a straight line by applying the two-point formula i.e. m = ∆y ∆x m = y2 − y1 x2 − x1 Example 8 An object is dropped from a bridge which is 576 feet high. The height of the object can be determined as a function of time according to the function. h(t) = 576 − 16t2 Where h(t) is height measured in feet and t is measured in seconds. Deter- mine the average rate of change in height between t = 0 and t = 1, between t = 0 and t = 2, and between t = 0 and t = 4. Solution The average rate of change means average velocity which is computed as Distance travelled Time travelled h(t) = 576 − 16t2 Average rate of change between t = 0 and t = 1 will be ∆h ∆t = h(1) − h(0) 1 − 0 ∆h ∆t = (576 − 16) − (576) 1 ∆h ∆t = 560 − 576 = −16 143

522. Average rate of change between t = 0 and t = 2 will be ∆h ∆t = h(2) − h(0) 2 − 0 ∆h ∆t = (576 − 16(2)2 ) − (576) 2 ∆h ∆t = 512 − 576 2 = −32 Average rate of change between t = 0 and t = 4 will be ∆h ∆t = h(4) − h(0) 4 − 0 ∆h ∆t = (576 − 16(4)2 ) − (576) 2 ∆h ∆t = 320 − 576 2 = −128 7.7 DERIVATIVES In this definition we will discuss about the concept of derivatives. We learn that how the derivative of a function of a single variable at a chosen input value is related to the slope of the tangent line when it exists, in the graph of the function. 7.7.1 Instantaneous Rate of Change At a moment, the rate of change is known as instantaneous rate of change. The slope of tangent line is same as the instantaneous rate of change a point. The Derivatives The derivative of the function y = f(x) is dy dx = lim δx→0 f(x + δx) − f(x) δx provided this limit exists. 7.7.2 Limit Approach for the Derivatives To find the derivative by limit approach we will do the following two steps. 144

523. i. Determine the difference quotient for f using equation ∂y ∂x = f(x + δx) − f(x) δx ii. Find the limit of the difference quotient as δx → 0 using dy dx = lim δx→0 f(x + δx) − f(x) δx Example 9 Find the derivative of f(x) = ax2 Solution dy dx = lim δx→0 f(x + δx) − f(x) δx dy dx = lim δx→0 a(x + δx)2 − ax2 δx dy dx = lim δx→0 a(x2 + δx2 + 2xδx) − ax2 δx dy dx = lim δx→0 ax2 + aδx2 + 2axδx − ax2 δx dy dx = lim δx→0 aδx2 + 2axδx δx dy dx = lim δx→0 δx(aδx + 2ax) δx dy dx = lim δx→0 (aδx + 2ax) dy dx = lim δx→0 aδx + 2ax = 2ax Example 10 The production costs per day for some widget is given by, C(x) = 2500 − 10x − 0.01x2 + 0.0002x3 What is the marginal cost when x = 200, x = 300 and x = 400 Solution To find the marginal cost we need to compute the derivative of the given 145

524. function i.e. C 0 (x) = −10 − 0.02x + 0.0006x2 C 0 (200) = −10 − 0.02(200) + 0.0006(200)2 C 0 (200) = 10 C 0 (300) = −10 − 0.02(300) + 0.0006(300)2 C 0 (300) = 38 C 0 (400) = −10 − 0.02(400) + 0.0006(400)2 C 0 (400) = 78 The marginal cost at x = 200, x = 300 and x = 400 is $10, $38 and $78 respectively. Example 11 The position of a particle moving along a straight line at time t is given by s = f(t) = 2t2 +7 where t and s are measured in seconds and meters respectively. Find (a) Find average velocity over the interval [10, 10.5] (b) Find the velocity at t = 12 seconds Solution 146

525. (a) t = 10 t + ∆t = 10.5 ∆t = 10.5 − 10 = 0.5 vave = ∆s ∆t = f(t + ∆t) − f(t) ∆t vave = f(10.5 + 0.5) − 10 0.5 vave = (2(10.5)2 + 7) − (2(10)2 + 7) 0.5 vave = 2(110.25) + 7 − 2(100) − 7 0.5 vave = 220.5 + 7 − 200 − 7 0.5 vave = 20.5 0.5 = 41m/s (b) at t = 12 vinst = ds dt = lim t→0 ∆s ∆t vinst = lim t→0 f(t + ∆t) − f(t) ∆t vinst = lim t→0 f(12 + ∆t) − f(12) ∆t vinst = lim t→0 2(12 + ∆t)2 + 7 − 2(12)2 − 7 ∆t vinst = lim t→0 2(144 + (∆t)2 + 24∆t) − 2(144) ∆t vinst = lim t→0 288 + 2(∆t)2 + 48∆t − 288 ∆t vinst = lim t→0 ∆t(2∆t + 48) ∆t = 48m/s 7.8 RULES OF DIFFERENTIATION Differentiation is a process of finding derivatives. There exists many functions for which the derivatives does not exist but in this section our concern is to study those functions which are differentiable. There are rules of differentiation exists for finding the derivatives of several common functions. 147

526. Rule 1: Constant Function If f(x) = c, where c is any constant then, f 0 (x) = 0 Where f 0 (x) (read as f prime x) represents the derivative of the function f at x. Rule 2: Power Rule If f(x) = xn where n is real number then. f 0 (x) = nx(n−1) Rule 3: Constant times a Function If f(x) = cg(x), where c is some scalar number or constant and the function g is differentiable. the we can write it as f 0 (x) = c.g 0 (x) Rule 4: Sum or Difference of Function If f(x) = u(x) ± v(x), where u and v are differentiable. f 0 (x) = u 0 (x) ± v 0 (x) Rule 5: Product Rule If f(x) = u(x).v(x), where it is given that both u and v are differentiable, then f 0 (x) = u 0 (x).v(x) + u(x).v 0 (x) Rule 6: Quotient Rule 148

527. If f(x) = u(x) v(x) , where it is given that both u and v are differentiable and also it is supposed that v(x) 6= 0, then f 0 (x) = u 0 (x)v(x) − u(x)v 0 (x) [v(x)]2 Example 12 Differentiate the following function f(x) = 15x100 − 3x12 + 5x − 46 Solution To differentiate the given function, we need to remember the first and second rule of differentiation i.e. constant function and power rule. f 0 (x) = 15 × 100x(100−1) − 3 × 12x(12−1 ) + 5 × 1x(1−1) − 0 f 0 (x) = 1500x99 − 36x11 + 5 Example 13 The position of an object at any instant t (in seconds) is given by s(t) = 2t3 − 21t2 + 60t − 10 Determine the time when this object is moving to the right and also when this object is moving to the left. Solution To determine the direction of the object we need to determine the velocity of the object. If the velocity is positive this implies the direction of the object is moving off to the right and if the velocity is negative this implies the direction of the object is moving to the left. To determine the velocity, we derivate the given function. s 0 (t) = 2 × 3t(3−1) − 21 × 2t(2−1) + 60 × 1t(1−1) s 0 (t) = 6t2 − 42t + 60 s 0 (t) = 6(t − 2)(t − 5) The intervals at which velocity is positive or negative is given by: 1st Positive at ∞ t 2 and 5 t ∞ i.e.(−∞, 2) ∪ (5, ∞) 149

528. 2nd Negative at 2 t 5 i.e. (2, 5) Hence the object is moving to the right over the interval (−∞, 2) ∪ (5, ∞) and the object is moving to the left over the interval (2, 5). Example 14 Differentiate the given function f(x) = (6x3 − x)(10 − 20x) Solution To differentiate the given function, we apply the rule 5 that is product rule. f 0 (x) = u 0 (x).v(x) + u(x).v 0 (x) f 0 (x) = (6 × 3x(3−1) − 1 × 1x(1−1) )(10 − 20x) + (6x3 − x)(−20 × 1x(1−1) ) f 0 (x) = (18x2 − 1)(10 − 20x) + (6x3 − x)(−20) f 0 (x) = −480x3 + 180x2 + 40x − 10 Example 15 Differentiate the following function f(x) = 3x + 9 2 − x Solution To differentiate the given function, we apply the rule 6 that is quotient rule. f 0 (x) = u 0 (x).v(x) − u(x).v 0 (x) [v(x)]2 f 0 (x) = (3 × 1x(1−1) )(2 − x) − (3x + 9)(−1 × 1x(1−1) ) [(2 − x)2 f 0 (x) = 3(2 − x) − (3x + 9)(−1) [(2 − x)2 f 0 (x) = 15 [(2 − x)2 7.8.1 Additional Rules of Differentiation In this section we will discuss about some more rules of differentiation. Rule 7: Power of a Function If f(x) = [u(x)]n where u is a differentiable function and n is a real number 150

529. then f 0 (x) = n.[u(x)](n−1) .u 0 (x) Rule 8: Base e Exponential Function If f(x) = eu(x) , where u is the differentiable then f 0 (x) = u 0 (x)eu(x) Rule 9: Natural Logarithm Function If f(x) = ln u(x), where u is differentiable then f 0 (x) = u 0 u(x) Rule 10: Chain Rule If the function y = f(u) is differentiable function and u = g(x) is an other differentiable function then dy dx = dy du . du dx Example 16 Use the chain rule to differentiate the given function y(x) = √ 5x − 8 Solution Using the chain rule, we need to identify the two functions that we needed for composition i.e. f(x) = √ x g(x) = 5x − 8 f 0 (x) = 1 2 √ x g 0 (x) = 5 Now using the chain rule, we get. y 0 (x) = f 0 (g(x))g 0 (x) y 0 (x) = f 0 (5x − 8)g 0 (x) y 0 (x) = 1 2 (5x − 8)−1 2 (5) y 0 (x) = 5 2 √ 5x − 8 151

530. Example 17 The demand equation of a certain product is given by p = 6 − 1 2 x dollars. Find the level of production of this product which results in the maximum revenue. The Revenue function R(x) is R(x) = xp = x(6 − 1 2 x) R(x) = 6x − 1 2 x2 Solution The marginal revenue is given by R 0 (x) = 6 × 1x1−1 − 1 2 × 2x2−1 R 0 (x) = 6 − x The graph of Revenue function R(x) is a parabola that opens downward. a hor So we have a horizontal tangent at x for which R 0 (x) = 0 that is for those points x for which the marginal revenue is zero is at x = 6. Hence the corresponding value of Revenue is given by R(6) = (6)(6) − 1 2 (6)2 R(6) = 18$ Hence, the rate of production which results in maximum revenue is given by x = 6, which results in total revenue of 18 dollars. 7.8.2 Average Cost If y = C(x) represents the total cost of producing x unit of a product. The rate of change of y with respect to x is called marginal cost function i.e. C 0 (x). The average cost per item is obtained by dividing the total cost by number of items produced and is defined as Average cost function = C(x) = C(x) x We call the derivative of average cost function as Marginal Average Cost i.e., C 0 (x) 152

531. Figure 7.5: Graphical Representation of Maximum Revenue Example 18 The total cost of manufacturing x items is given by C(x) = −x3 + 3x2 + 7x + 1000 Find (a) The average cost (b) The marginal average cost. Solution (a) The average cost is obtained by dividing the total cost by the number of items i.e., C = C(x) x = −x3 + 3x2 + 7x + 1000 x (b) The rate of change of average cost is the marginal average cost i.e., d dx C(x) x = d dx −x3 + 3x2 + 7x + 1000 x 153

532. Using Quotient Rule we will find the above derivative C 0 = x d dx (−x3 + 3x2 + 7x + 1000) − (−x3 + 3x2 + 7x + 1000) d dx (x) x2 ! = x(−3x2 + 6x + 7)(−x3 + 3x2 + 7x + 1000)(1) x2 = −3x3 + 6x2 + 7x + x3 − 3x2 − 7x − 1000 x2 = −2x3 + 3x2 − 1000 x2 7.8.3 Consumption Function In economic analysis consumption function plays a vital role. It represents the relationship between the total national income I and total national consumption C i.e., C = f(I). The rate at which consumption changes with respect to income is called marginal propensity to consume. Marginal Propensity to consume = dC dI If saving S is considered to be the difference between I and C, then S = I − C The derivative of S with respect to I is called marginal propensity to save which represents how fast saving changes w.r.t income. Marginal Propensity to save = dS dI = dI dI − dC dI = 1 − dC dI = 1 − Marginal Propensity to consume Example 19 Suppose the consumption function is given as C = 3(4 √ I3 + 5) 2I + 3 Where I is measured in billion dollars. Find the marginal propensity to consume and the marginal propensity to save when income I = 200. Solution Marginal propensity to consume = dC dI = 3 d dI 4 √ I3 + 5 2I + 3 ! 154

533. Apply quotient rule to find dC dI dC dI = 3 (2I + 3) d dI (4 √ I3 + 5) − (4 √ I3 + 5) d dI (2I + 3) (2I + 3)2 = 3 h (2I + 3)6I 1 2 − (4 √ I3 + 5)2 i (2I + 3)2 = 12I 3 2 + 18I 1 2 − 8I 3 2 − 10 (2I + 3)2 = 6 h 2I 3 2 + 9I 1 2 − 5 i (2I + 3)2 = 6 h 2(200) 3 2 + 9(200) 1 2 − 5 i (2(200) + 3)2 = 6(18377.86) (404)2 ≈ 0.679 Marginal Propensity to save = 1 − dC dI = 1 − 0.679 = 0.321 This means that if current income of $200 billion increases by $1 billion, then the nation consumes and saves approximately 67.9% and 32.1% respectively of that income. 7.8.4 Marginal Revenue Product Marginal revenue product approximates the change in revenue that results when a manufacturer hires an additional employee. Consider R = px, where R is the total revenue that manufacturer obtain for selling x unit of a product per day and p is the price per unit. Suppose a manufacturer hires n number of employees then x can be assumed to be the function of n. Also the demand function p = f(x). Thus R can be considered as a function of number of employees n. To develop the formula for marginal revenue product dR dn , apply product rule for derivative on R = px dR dn = p. dx dn + x. dp dn (7.8.1) 155

534. By using chain rule dp dn = dp dx . dx dn Equation (7.6.1) will become dR dn = p. dx dn + x. dp dx . dx dn = dx dn p + x. dp dx Thus, Marginal revenue product = dR dn = dx dn p + x. dp dx Example 20 Suppose the demand function for a product p = 500 √ x . Find the marginal revenue product if the number of employees n = 10 and x = 2n2 + 3 √ n2 + 5 Solution dR dn = dx dn p + x. dp dx p = 500 √ x dp dx = 500 1 2 x−3 2 = 250x−3 2 Also from x = 2n2 + 3 √ n2 + 5 and n = 10 we have x = 2(10)2 + 3 p (10)2 + 5 x = 203 10.25 = 19.8 p = 500 √ 10 = 158.11 dp dx = 250(19.8)−3 2 = 2.84 dx dn = √ n2 + 5 4n − (2n2 + 3) 2n 2 √ n2+5 n2 + 5 = 4n √ n2 + 5 − n(2n2+3) √ n2+5 n2 + 5 156

535. = 4n(n2 + 5) − 2n3 − 3n (n2 + 5) √ n2 + 5 = 2n3 − 17n (n2 + 5) √ n2 + 5 When n = 10 dx dn = 2(10)3 − 17(10) ((10)2 + 5) p (10)2 + 5 = 1830 1075.93 = 1.7 So equation (7.6.1) implies dR dn = 1.7(158.11 + (19.8)(2.84)) = 368.38 This means that if 11th employee is hired, revenue will increase by approximately $368.38 per day. Example 21 Suppose the saving S and income I of a country is related by the equation S = ln 4 3 + 2e−2I Find the marginal propensity to consume? Solution Marginal Propensity to save = 1 − Marginal Propensity to consume Marginal Propensity to consume = dC dI = 1 − Marginal Propensity to save dC dI = 1 − d dI ln 4 3 + 2e−2I = 1 − 1 4 3+e−2I d dI 4 3 + e−2I = 1 − 3 + e−2I 4 −4 (3 + e−2I)2 d dI (3 + e−2I ) = 1 − 3 + e−2I 4 −4 (3 + e−2I)2 (e−2I ) = 1 − e−2I 3 + e−2I Example 22 Find the value of positive constant α if d dx (αx − xα ) = 0 157

536. when x = 1 Solution d dx (αx − xα ) = 0 d dx (αx ) − d dx (xα ) = 0 αx ln α − αxα−1 = 0 when x = 1 α1 ln α − α1α−1 = 0 α ln α − α = 0 α ln α = α ln α = 1 α = e 7.9 INSTANTANEOUS RATE OF CHANGE AND THE INTERPRETATION The derivatives can be interpreted as Instantaneous Rate of Change Inter- pretation. Now the average rate of change for the function y = f(x) from x = c to x = d is given by. Average Rate of Change = f(d) − f(c) d − c Example 23 Suppose a companys total cost in dollars to produce the x units of a certain product is given by the equation C(x) = 0.01x2 + 25x + 1500 Calculate the average rate of change of total cost for a) The first 100 units i.e. which are produced from x = 0 to x = 100 b) The second 100 units which are produced from x = 101 to x = 200 158

537. Solution a) We know that the average rate of change of total cost from x = 0 to x = 100 units will be C(100) − C(0) 100 − 0 = 0.01(100)2 + 25(100) + 1500 − (1500) 100 C(100) − C(0) 100 = 4100 − 1500 100 C(100) − C(0) 100 = 2600 100 = 26 therefore the average rate of the change of total cost from x = 0 to x = 100 units is 26 dollars per unit. b) Similarly the average rate of change of total cost from x = 101 to x = 200 units is given by C(200) − C(100) 100 − 0 = 0.01(200)2 + 25(200) + 1500 − (4100) 100 C(200) − C(100) 100 = 2800 100 C(200) − C(100) 100 = 28 Therefore the average rate of change of total cost from x = 100 to x = 200 units is 28 dollars per unit. 7.10 HIGHER ORDER DERIVATIVES In this section we will discuss about higher order derivatives and their interpretation. For a given function f there are other derivatives which can be defined. 7.10.1 The Second Derivative The derivative f 0 of the function is referred as first order derivative of the function. The derivative of the function f 0 (denoted by f 00 ) is referred as second derivative. 159

538. 7.10.2 Third and Higher Order Derivatives The derivative of the function of second order derivative is referred as third derivative. The nth −order derivative of f denoted by fn is found by differentiating the derivatives of order n − 1 i.e. at x, fn (x) = d dx (f(n−1) (x) Example 24 Find three higher order derivatives of the given function f(x) = 16x3 − 4x2 Solution The derivatives of f are f 0 (x) = 48x2 − 8x f 00 (x) = 96x − 8 f 000 (x) = 96 7.11 DIFFERENTIATION OF EXPLICIT AND IMPLICIT FUNCTIONS The explicit function is defined as a function in which the dependent variable is given explicitly in terms of the independent variable. The Implicit function is defined as a function in which the dependent variable is not given explicitly in terms of the independent variable. 7.11.1 Explicit Function Function which is given in terms of independent variable is known as explicit function. For example, y = x2 + 3x Here y is dependent variable and is given in terms of the independent variable x. 160

539. 7.11.2 Implicit Function A function which is given in terms of both dependent and independent variables is known as implicit function. For example, y + x2 + 3x + 2 = 0 Sometimes it is not convenient to express a function explicitly such as x2 + y2 = 4 since it could be written as y = √ 4 − x2 or y = − √ 4 − x2 Example 25 Find dy dx if y2 = x2 + sin xy Solution y2 = x2 + sin xy, differentiating both sides with respect to x. d dx (y2 ) = d dx (x2 ) + d dx (sin xy), treating y as a function of x and use the chain rule 2y dy dx = 2x + (cos xy) d dx (xy), treat xy as a product and apply product rule. 2y dy dx = 2x + (cos xy) y + x dy dx , collect the terms of dy dx 2y dy dx − (cos xy) x dy dx = 2x + y cos xy, now factor out dy dx (2y − x cos xy) dy dx = 2x + y cos xy, dy dx will be dy dx = 2x + y cos xy 2y − x cos xy 161

540. Example 26 Find dy dx if y = 5x + cos x Solution y = 5x + cos x, differentiate both sides with respect to x dy dx = d dx (5x) + d dx (cos x) dy dx = 5 d dx (x) − sin x, hence dy dx will be dy dx = 5 − sin x Example 27 Suppose that P, the proportion of people affected by a certain disease is given by ln P 1 − P = 0.5t Where t represents time in months. Determine the rate of change of P with respect to time. Solution d dt ln P 1 − P = 0.5 d dt t 1 P 1−P d dt ln P 1 − P = 0.5 1 − P P (1 − P) d dt P − P d dt (1 − P) (1 − P)2 = 0.5 (1 − P)dP dt + P dP dt P(1 − P) = 0.5 dP dt (P + 1 − P) = 0.5(1 − P) dP dt = 0.5P(1 − P) 162

541. 7.12 MAXIMA AND MINIMA WITH AP- PLICATIONS In this section we will discuss about maxima and minima. The maxima and minima are the plurals of maximum and minimum collectively known as extrema. Maxima is the highest point and minima is the lowest point. 7.12.1 Absolute Maximum For a function f with domain D, we say that this function f has an absolute maximum value on D at a point c if f(x) ≤ f(c) for all x in D 7.12.2 Absolute Minimum Let f be a function with domain D. Then f has an absolute minimum value on D at a point c if f(x) ≥ f(c) for all x in D Example 28 Let a ball is thrown in the air then its height at any instant t is represented by the equation h = 3 + 14t − 5t2 What is the maximum height? Solution Using the derivative, we can find the slope of the given function. h = 3 + 14t − 5t2 , now derivate with respect to t d dt h = 0 + 14 − 10t, 163

542. the slope of the function at any time t is d dt h = 14 − 10t, now find when the slope is zero i.e. 14 − 10t = 0 10t = 14 t = 14 10 = 1.4 The slope will be zero at t = 1.4 and the corresponding height at that instant will be h = 3 + 14 × 1.4 − 5(1.4)2 h = 3 + 19.6 − 9.8 h = 12.8 The maximum height is 12.8m at t = 1.4s Example 29 Find the absolute extrema (maximum and minimum values) of f(x) = x2 on [−2, 1]. Solution The given function is differentiable over the entire domain, so only the critical point where f 0 (x) = 2x will be zero is at x = 0 Consider the given domain of the function i.e. [−2, 1] at x = 0 f(0) = 0 at x = −2 f(−2) = 4 at x = 1 f(1) = 1 Therefore the function will have an absolute maximum value at x = −2 and also an absolute minimum value at x = 0 164

543. 7.13 APPLICATIONS OF DERIVATIVES There are many applications of derivatives we use derivatives to determine the maximum and minimum values of a particular function. Deriva- tives helps us to solve several types of real world problem or example cost, amount of material used in a building strength, profit and loss etc.Derivatives are used in many engineering and science problem particularly when studying the behavior of moving objects. Example 30 Suppose we make a cubical box which is open from top side from a tin sheet having dimensions 12-inch by 12-inch by cutting small congruent squares from the corners of that sheet of tin and bending up the sides. What size of the squares should be cut from the corners be to make the box in such a way that it hold as much as possible? Solution Let the x be the side of the box. The volume of the box is a function of variable x i.e. V (x) = x(12 − 2x)2 V (x) = 144x − 48x2 + 4x3 (7.13.1) It is given that the sides of the sheet of tin are 12 in. long, x ≤ 6 and the domain of V is the interval 0 ≤ x ≤ 6. The function V (x) will be zero at x = 0 and x = 6 and a maximum near x = 2. Now we will derivate the equation (7.11.1) i.e. dV dx = 144 − 96x + 12x2 dV dx = 12(12 − 8x + x2 ) dV dx = 12(2 − x)(6 − x) At x = 2 and x = 6 the dV dx = 0 Only x = 2 lies in the interior of the function’s domain and makes the 165

544. critical point. The values of V at the x = 2 and at the domain 0 ≤ x ≤ 6 are at x = 2 V (2) = 128 at x = 0 V (0) = 0 at x = 6 V (6) = 0 The maximum volume is 128in3 . The cutout squares should be 2in on a side. 166

545. 7.14 SELF ASSESMENT QUESTIONS 1. Estimate the following limits by constructing a table (a) limx→2 2 x − 3 (b) limx→1 ln x x − 1 (c) limx→−2 x2 − 4 x + 2 2. Find the indicated limits: (a) limx→3 x4 − 81 x2 − x − 6 (b) limx→∞ 4 − 3x3 x3 − 1 (c) limx→−7− x2 + 1 √ x2 − 49 (d) limx→∞ 2 x − x2 x2 − 1 (e) Find the value of the constant such that limx→2 f(x) exist if f(x) = (√ 2 − x; x 2 x3 + k(x + 1); x ≥ 2 (f) Graph of f(x) use the graph to find each of the following limits if exists. (i) limx→−1 f(x) (ii) limx→0 f(x) (iii) limx→1 f(x) where f(x) =      x2 + 1; x 1 3x + 2; −1 ≤ x ≤ 1 2 − x; x ≥ 1 3. The population of a certain village t years from now on is expected to be N = 15000 + 5000 (t + 3)2 Determine limt→∞ N i.e., population in the long run. 4. The cost for manufacturing a certain item is C(x) = 10000 + 6x where x represents the number of items produced. The average cost per item is represented by C(x) and is obtained by dividing C(x) by x. Determine the following. (a) C(1500) (b) C(10, 000) (c) limx→5000 C(x) 167

546. 5. The cost of manufacturing x units of a certain item is C(x) = x2 − 5x + 80 (a) Find the average rate of change with respect to the number of items produced between 10 and 15. (b) Find the instantaneous rate of change when 12 items are produced. 6. Find the derivatives of the following functions: (a) y = 9 − 15x + 43 − 7x5 x4 (b) f 0 (8) if f(t) = −4 5 √ t3 − 1 3 4 √ t2 + 1 (c) f(x) = x3 + 2x2 + 3 (5x − 2)(3x2 + 2) (d) y = (3 √ x − 5)(2 √ x + x3 + 7) (e) y = 5 + 2x − 7x2 (6x3 − x2 + 3x + 5)2 (f) y = 5 r 8x2 − 5 x2 − 3 (g) y = ln r x4 + 2x2 + 10 x4 + 4 ! (h) y = e(3x3+9x2+2x+1) (i) y = ln([(x2 + 4)(x3 + 8x − 3)2 ] (j) (x) = 5(x2 ln x) find f 0 (1), f 0 (5) 7. Find dy dx , d2 y dx2 and d3 y dx3 (a) ln(xy) + x = 4x2 (b) y2 = ln(x + y) (c) (1 + e3x )2 = 3 + ln(x + y) (d) If x p y + x2 = x p x + y2 find dy dx at (3,3) 8. Consider the consumption function of a country is estimated by C = 0.8 √ I − 0.3I + 10 √ I3 2 + 3 √ I 168

547. Where C and I are measured in billions of dollars. Find Marginal propensity to consume and marginal propensity to save when I = 25 billion dollars. 9. The cost function is given as y = C(x). If C 0 (x) = 0 then show that the marginal cost function and average cost function are equal. 10. If the average cost function is given as C0.03x + 5.8 − 3 x3 + 0.2x2 Find (a) The marginal cost (b) The average marginal cost. 11. Find the marginal revenue product if there are 20 employees who produce x unit of a product per day, where x = 2n(3n + 1)2 And the demand function for a product is given by p = 535 (x + 5)2 12. The saving of a country and the national income are related by the equation S3 + 1 5 I2 = s2 I + 2I2 Where S and I both are measured in billions of dollars. Find marginal propensity to consume when I = 16 and S = 12 and interpret the result. 169

548. Unit-8 PARTIAL DERIVATIVES WRITTEN BY: MS FOUZIA REHMAN REVIEWED BY: DR NASIR REHMAN 170

549. 8.1 INTRODUCTION In this chapter the concept of derivatives extend from function of oner variable totwo or more variables commonly called functions of several variables. We will discuss about limits and continuity for function of two or more variables. Partial derivatives are type of derivatives for multi-variable functions. Partial derivatives are denote by symbol ∂. Marquis de Con- dorcet is the first person who use this symbol in mathematics. Partial derivatives appear in thermodynamic equations and quantum mechanics. It plays important role in business and economics, most of the functions de- scribing economic behavior postulate that behavior depends on more than one variable. There are many cases which describes the amount spent on consumer goods is depending on both income and wealth. In this unit students will be able to learn about 1. Functions or more than one variable and their partial derivatives 2. Geometric interpretation of partial derivatives 3. Maxima and minima for multi variable functions 4. Critical and saddle point 8.2 OBJECTIVES After studying this unit students will be able to learn about 1. Geometric interpretation of partial derivatives 2. minimization and maximization of real life problems 3. Application of partial derivatives 171

550. 8.3 GEOMETRIC INTERPRETATION OF PARTIAL DERIVATIVES In previous unit, we have seen that when y = f(x), then f 0 x0 can be interpreted as either slope of the tangent line to the curve y = f(x) at x = x0 or rate of change of f(x) with respect to x at x = x0. Partial derivatives have similar interpretations. Consider the surface z = f(x, y) and suppose that C1 and C2 are the intersections of the surface z = f(x) with the plane y = y0 and x = x0 respectively. Thus fx(x0, y0) is the rate of change of f with respect to x along the curve C1 when y is held fixed at the value y0 and fy(x0, y0) is the rate of change of z with respect to y along the curve C2 when x is held fixed at the value x0. Geometrically, fx(x0, y0) and fy(x0, y0) can be interpreted as the slope of a tangent line to the curves C1 and C2 respectively. Figure 8.1: Geometric Interpretation of Partial Derivatives 8.4 PARTIAL DERIVATIVES WITH RE- SPECT TO x At the point (x0, y0) the partial derivative of f(x, y) with respect to x is ∂f ∂x = lim h→0 f(x0 + h, y0) − f(x0, y0) h 172

551. Provided that the limit exists. 8.5 PARTIAL DERIVATIVES WITH RE- SPECT TO y At the point (x0, y0) the partial derivative of f(x, y) with respect to y is ∂f ∂y = lim h→0 f(x0, y0 + h) − f(x0, y0) h Provided that the limit exists. Example 1 Find the values of ∂f ∂x and ∂f ∂y at the point (4, −5) if f(x, y) = x2 + 3xy + y − 1 To find the ∂f ∂x we treat y as a constant and differentiate with respect to x ∂f ∂x = ∂ ∂x x2 + 3xy + y − 1 = 2x + 3y The value of ∂f ∂x at (4,5) is ∂f ∂x = 2(4) + 3(−5) = −7 To find the ∂f ∂y we treat x as a constant and differentiate with respect to y ∂f ∂y = ∂ ∂y x2 + 3xy + y − 1 = 3x + 1 The value of ∂f ∂y at (4,5) is ∂f ∂y = 3(4) + 1 = 13 173

552. 8.6 MAXIMA AND MINIMA OF FUNC- TIONS OF MULTI-VARIABLES Let f(x, y) be a multivariable function defined on a region R and contains the point (a, b). Then i. f(a, b) will be a local maximum value of f if we have the inequality f(a, b) ≥ f(x, y) for all domain points (x, y) in an open disk centered at (a, b). ii. f(a, b) will be a local minimum value of f if we have the inequality f(a, b) ≤ f(x, y) for all domain points (x, y) in an open disk centered at (a, b). 8.6.1 The Second Partial Derivative Test Suppose that f be a function of two variables which has continuous second order partial derivatives in some open disk centered at a critical point (x0, y0) and also let D = fxx(x0, y0)fyy(x0, y0) − f2 xy(x0, y0) a) If we have D 0 and fxx(x0, y0) 0 , then the function f has a relative minimum at the point (x0, y0). b) If we have D 0 and fxx(x0, y0) 0 , then the function f has a relative maximum at the point (x0, y0). c) If D 0 , then f has a saddle point at (x0, y0). d) If we have D = 0 , then this test fails and no conclusion is to be taken. 174

553. 8.7 CRITICAL AND SADDLE POINTS 8.7.1 Critical Point We call x = c a critical of the function f(x) if f(c) exists and if either of the following are true. f 0 (c) = 0 or f 0 (c) does not exist 8.7.2 Saddle Point A point of a function which is stationary point having no extremum is known as saddle point. Example 2 If f(x, y) = 3x2 − 2xy + y2 − 8y find the extrema of f. Solution Firstly, we calculate the fx(x, y) and fy(x, y) i.e. fx(x, y) = 6x − 2y fy(x, y) = 2y − 2x − 8 The critical points of f satisfy the equations 6x − 2y = 0 2y − 2x − 8 = 0 Solving the above equations for x and y yields x = 2, y = 6 Hence (2, 6) is the only critical point. Now we will take the second order partial derivative. fxx(x, y) = 6 fyy(x, y) = 2 fxy(x, y) = −2 And the relative minimum value is given by D = fxx(x0, y0)fyy(x0, y0) − f2 xy(x0, y0) D = fxx(2, 6)fyy(2, 6) − f2 xy(2, 6) ⇒ fxx(2, 6) = 6 0 175

554. And the relative minimum value is given by f(2, 6) = 3(2)2 − 2(2)(6) + 62 − 8(6) f(2, 6) = −24 Hence f has relative minimum at (2, 6). Example 3 A rectangular box with no top and having a volume of 12ft3 is to be constructed. The cost per square foot of the material to be used is given by at the rate of $4 for the bottom, at the rate of $3 for two of the opposite sides, and at the rate of $2 for the remaining pair of opposite sides. Find the three dimensions of this rectangular box that will extremize i.e. minimize the cost. Solution Let x and y are the dimensions of the base and z be the altitude. Since there are two sides of area xz and two sides of area yz. The cost of the material is given by C = 4xy + 3(2xz) + 2(2yz) (8.7.1) Where x 6= 0, y 6= 0 and z 6= 0 Since volume = xyz = 12 ⇒ z = 12 xy Substitute the value of z in equation (8.4.1) we get C = 4xy + 72 y + 48 x Now we will calculate the extrema for this we need to calculate Cx and Cy first. Cx = 4y − 48 x2 = 0 Cy = 4x − 72 y2 = 0 The above equations imply that x2 y = 12 and xy2 = 18 x2 y = 12 ⇒ y = 12 x2 176

555. Now substitute the value of y in xy2 = 18 we get 18x3 = 144 x = 2 Substitute x = 2 in y = 12 x2 we get y = 3 Using the value of x and y we can obtain the value of z i.e. z = 12 xy z = 2 Hence, the minimum cost will occur if the dimensions to be considered are following For the base 3ft by 2ft and the altitude is 2ft. The longer sides should be made from the $2 material and the shorter sides out of the $3 material. Example 4 Let f(x, y) = xe(x2+y2) + 4y. Find (a) Slope of the surface z = f(x, y) in the x−direction at point (2, 1) (b) Slope of the surface z = f(x, y) in the y−direction at point (2, 1) Solution (a) Taking derivative of f with respect to x by treating y as a constant i.e., ∂f ∂x = ∂ ∂x (xe(x2+y2) + 4y) = ∂ ∂x (xex2+y2 ) + ∂ ∂x 4y = x ∂ ∂x (ex2+y2 ) + (ex2+y2 ) ∂ ∂x (x) + 0 = xe(x2+y2) (2x) + e(x2+y2) = e(x2+y2) (2x2 + 1) at (2,1) ∂f ∂x will be ∂f ∂x |(2,1) = e5 (2(4) + 1) = 9e5 177

556. (b) Taking derivative of f with respect to y by treating x as a constant i.e., ∂f ∂y = ∂ ∂y (xe(x2+y2) + 4y) = ∂ ∂y (xex2+y2 ) + ∂ ∂y 4y = xe(x2+y2) (2y) + 4 = 2xye(x2+y2) + 4 at (2,1) ∂f ∂y will be ∂f ∂y |(2,1) = 2(2)(1)e(22+1) + 4 = 4e5 + 4 = 4(e5 + 1) Example 5 A company manufactures two types of a certain product. The joint cost function for producing x units of a product A and y unit of a product B is c = f(x, y) = 80xy − 2xy2 + 30x2 + 2y3 + 1500 Find the marginal costs ∂c ∂x and ∂c ∂y when x = 70 and y = 50 and interpret the result. Solution The marginal costs are ∂c ∂x = ∂ ∂x (80xy − 2xy2 + 30x2 + 2y3 + 1500) = 80y − 2y2 + 60x at (70,50) ∂c ∂x will be 80(50) − 2(50)2 + 60(70) = 4000 − 5000 + 4200 = 3200 Now we will find ∂c ∂y ∂c ∂y = ∂ ∂y (80xy − 2xy2 + 30x2 + 2y3 + 1500) = 80x − 4xy + 6y2 178

557. at (70,50) ∂c ∂y will be 80(70) − 4(70)(50) + 6(50)2 = 5600 − 14000 + 15000 = 6600 This means that increasing the output of product A from 70 to 71 while maintaining the production of the product B at 50, increases cost by $3200. Similarly, by increasing the output of product B from 50 to 51 while holding the production of product A at 70, increases cost by $6600. 8.8 APPLICATIONS OF PARTIAL DERIVA- TIVES There is a very common use of partial derivative in our daily life. Par- tial derivatives are used in basic laws of physics for example Maxwell’s equations of electromagnetism, Einstein’s equation in general relativity and Newton’s law of linear motion. In economics partial derivatives are used to determine the behavior of other variables while taking one variable as constant. Example 6 Given the profit function P(x, y) = 140x + 200y − 4x2 + 2xy − 12y2 − 700 Verify that maximum profit occurs at the point (20, 10). Solution Calculate the first derivative of the given function i.e. Px = 140 − 8x + 2y Py = 200 + 2x − 24y At point (20,10) Px = 140 − 8(20) + 2(10) = 0 Py = 200 + 2(20) − 24(10) = 0 Since both first partial derivatives are equal to zero at given point. It is possible that maximum profit does occur at the point (20, 10) to verify 179

558. that we need to calculate the second derivative and evaluate them at point (20,10). Pxx = −8 Pyy = −24 Now we will find Pxy i.e. Pxy = 2 Since D = fxx(x0, y0)fyy(x0, y0) − f2 xy(x0, y0) D = (−8)(−24) − (2)2 = 220 We know that if D 0 and f(x0, y0) 0 , then f has a relative maximum at (x0, y0). Here D = 220 0 and fxx(x0, y0) = Pxx(20, 10) = −8 0 therefore the maximum profit occurs at the point (20,10). The maximum profit is given by P(20, 10) = 140(20) + 200(10) − 4(20)2 + 2(20)(10) − 12(10)2 − 700 P(20, 10) = 1700 Example 7 The cost in dollars to produce a batch of soft drink is approx- imated by C(x, y) = 2000 + 30x3 − 60xy + 10y2 Where x and y represents the number of kilograms of sugar and the number of grams of flavoring per batch respectively. (a) Determine the quantity of sugar and flavoring that result in lowest cost for a batch. (b) What is the minimum cost? Solution C(x, y) = 2000 + 30x3 − 60xy + 10y2 Cx = 90x2 − 60y and Cy = −60x + 20y 180

559. Set Cx and Cy equal to zero and solve for y. 90x2 − 60y = 0 y = 90 60 x2 y = 3 2 x2 (8.8.1) −60x + 20y = 0 y = 60 20 x y = 3x (8.8.2) From equation (8.6.1) and (8.6.2) we have 3 2 x2 = 3x x2 − 2x = 0 x(x − 2) = 0 x = 0 or x = 2 Substitute x = 0 in equation (8.6.2) y = 0 Substitute x = 2 in equation (8.6.2) y = 6 So the critical points are: (0,0)and (2,6) Cxx(x, y) = 180x, Cyy(x, y) = 20, Cxy(x, y) = −60 Critical Point(x0, y0) Cxx(x0, y0) Cxy(x0, y0) Cxy(x0, y0) D = CxxCyy − C2 xy (0,0) 0 20 -60 0-3600=-3600 (2,6) 360 20 -60 360(20) − 1(−60)2 = 3600 181

560. Since D 0 and Cxx(2, 6) 0, the cost is minimum at (2,6). (x, y) = 2000 + 30x3 − 60xy + 10y2 C(2, 6) = 2000 + 30(2)3 − 60(2)(6) + 10(6)2 = $80 The minimum cost for a batch is $80 182

561. 8.9 SELF ASSESMENT QUESTIONS 1. Find fx, fy, fx(3, 4) and fy(1, 2) (a)f(x, y) = x + 3xy2 x3 + y2 (b) f(x, y) = ye(5x+2y2) (c) f(x, y) = (x+1)2 +(y −3)3 +5xy3 −2 (d) f(x, y) = √ x2 + 4 x2y + y2x (e) f(x, y) = x4 ln(1 + xy 3 2 ) (f) f(x, y) = − 2 e(4x−xy2+y) 2. If z = x2 − y2 + 2xy then show that ∂2 z ∂x2 + ∂2 z ∂y2 = 0 3. Let f(x, y, z) = x3 y5 z7 + xy2 + y3 z − x2 z2 . Find (a) fxz (b) fyz (c) fxy (d) fxyz (e) fyzx (f) fxzy (g) fxxy (h) fxxz (i) fyyx (j) fyzy 4. Suppose z = f(x, y) represents the cost to construct a certain struc- ture, where x and y express the labor and material cost respectively. Explain what fx and fy represent. 5. The joint cost function for producing x units of product A and y units of product B is given by c = x2 (y3 + x) 1 2 17 + xy 1 2 + 600 Find the marginal costs ∂c ∂x and ∂c ∂y when x = 17 and y = 8 and interpret the result. 6. Find all points where functions have any relative extrema. (a) f(x, y) = (y2 −4)(ex −1) (b) f(x, y) = x2 +y2 +xy−9x+1 (c) f(x, y) = 4xy − x4 − y4 (d) f(x, y) = x2 + y2 + 32 xy (e) f(x, y) = xe(y+1) 183

562. 7. Suppose a production function P is given by P = f(l, k) = 0.8l3 − 0.02l2 + 2.5k2 − 0.05k3 Find the values of l and k so as to maximize output P. 8. Suppose that the profit of a certain company is given by P(x, y) = 1200 + 80x − 2x2 + 100y − y2 Where x and y represent the cost of unit of labor and unit of goods respectively. Find x and y that maximize the profit. Find the maximum profit. 9. The revenue in thousands of dollars obtained from selling the x spas and y solar heaters is given by R(x, y) = 15 + 80x + 81y − 2x2 − 3y2 − 4xy Find the quantity of each that must be sold to generate maximum revenue. Find the maximum revenue. 184

563. Unit-9 OPTIMIZATION WRITTEN BY: DR NASIR REHMAN REVIEWED BY: DR BABAR AHMED 185

564. 9.1 INTRODUCTION Optimization is the process of achieving the best possible result under given circumstances. In design, construction and maintenance. It has many applications in several fields. In mathematics optimization is useful to finding the minimum of a function of several variables under a prescribed set of constraints. In statistics optimization are used in the analysis of experi- mental data and in the construction of empirical models. In business and economics there are many applied problems that require optimization. In this chapter students will be able to learn about 1. Increasing and decreasing function 2. Concavity, first and second derivative test 3. Curve sketching 4. Application of revenue, cost and profit 9.2 OBJECTIVES After learning this unit students will be able to learn about 1. use of different functions 2. use of derivative tests 3. Applications in business 186

565. 9.3 INCREASING AND DECREASING FUNCTIONS Informally, a function is increasing on an interval if the graph of a function rises from left to right and a function is decreasing on an interval if the graph of a function falling from left to right over the interval. Figure 9.1: Graphical Representation of Increasing and Decreasing Func- tions Theorem Consider a function f has derivative at each point in an open interval then a) f is increasing on an interval if f 0 (x) 0 for every x in that interval b) f is decreasing on an interval if f 0 (x) 0 for every x in that interval c) f is constant on an interval if f 0 (x) = 0 for every x in that interval Example 1 Find the interval on which f is increasing and decreasing if f(x) = x3 − 12x − 5 Solution f 0 (x) = 3x2 − 12 = 3(x + 2)(x − 2) 187

566. To find critical point we set the derivative equal to 0 and solve the equation 3(x + 2)(x − 2) = 0 x = −2 or x = 2 x = −2 and x = 2 are critical points. Thus we have intervals (−∞, −2), (−2, 2) and (2, ∞) on which f either increasing or decreasing. We choose −3 as a test point in the interval (−∞, −2) i.e., f 0 (−3) = 3(−3 + 2)(−3 − 2) = 3(−1)(−5) = 15 0 So, f is increasing on this interval We choose 0 as a test point in the interval (−2, 2) i.e., f 0 (0) = 3(0 + 2)(0 − 2) = −12 0 So, f is decreasing on this interval. We choose 3 as a test point in the interval (2, ∞) i.e., f 0 (3) = 3(3 + 2)(3 − 2) = 3(5)(1) = 15 So, f is increasing on this interval We summarize the result in the following table Intervals Evaluated f 0 Sign of f 0 Behavior of f −∞ x −2 f 0 (−3) = 15 + Increasing −2 x −2 f 0 (0) = 12 − decreasing 2 x ∞ f 0 (3) = 15 + Increasing Table 9.1: Behaviour of the Function f(x) on Different Intervals 9.4 CONCAVITY The sign of derivative of f tells whether the function is increasing or decreasing but it does not tell the direction of curvature. If the graph 188

567. Figure 9.2: Graphical Representation of the Given Function of a function lies above the tangent line at every point on an interval or we can say f 0 is increasing on I then f is said to be concave up on that interval and f is said to be concave down if its graph lies below the tangent lines or we can say f 0 is decreasing on I. The point where function changes its concavity is called point of inflection. Figure 9.3: Concavity and Point of Inflection 189

568. Theorem If f is twice differentiable on an open interval I, then a) The function f is concave up on I if f 00 (x) 0 for each x in I b) The function f is concave down on I if f 00 (x) 0 for each x in I Example 2 Find all the intervals where f is increasing, decreasing, concave up and concave down. Find all the inflection points and verify that your result is consistent with the graph. If f(x) = x4 − 4x3 + 10 Solution f 0 (x) = 4x3 − 12x2 To find critical point we set the derivative equal to 0 and solve the equation 4x2 (x − 3) = 0 x = 0 or x = 3 x = 0 and x = 3 are critical points. Thus we have intervals (−∞, 0), (0, 3) and (3, ∞) on which f either increasing or decreasing. Intervals Evaluated f 0 Sign of f 0 Behavior of f −∞ x 0 f 0 (−1) = −16 − decreasing 0 x 3 f 0 (1) = −8 − decreasing 3 x ∞ f 0 (4) = 64 + Increasing Table 9.2: Behaviour of the Function f(x) on Different Intervals So, f is Decreasing on (−∞, 0), (0, 3) and increasing on (3, ∞) To find whether the function is concave up or concave down, take double derivative of f f 00 (x) = 12x2 − 24x = 12x(x − 2) 190

569. Set f 00 (x) = 0 12x(x − 2) = 0 x = 0 or x = 2 So we have intervals (−∞, 0), (0, 2) and (2, ∞) on which either concave up or concave down. Intervals Evaluated f 00 Sign of f 00 Behavior of f −∞ x 0 f 00 (−1) = 36 − concave up 0 x 2 f 00 (1) = −12 − concave down 2 x ∞ f 00 (4) = 96 + concave up Table 9.3: Behaviour of the function f(x) on Different Intervals So, f is concave up on (−∞, 0) and (2, ∞) and concave down on (0, 2) The inflection points occur at x = 0 and x = 2 because f changes concavity at these points.Thus the inflection points are (0, f(0)) = (0, 10) and (2, f(2)) = (2, −6). Figure 9.4: Graphical Representation of the Increasing, Decreasing, Con- cave up, Concave down and Point of Inflection 191

570. 9.5 FIRST DERIVATIVE TEST • First step is to find all critical values x∗ . • Calculate the value of f 0 (x) to the left (xl) and right (xr) of x∗ . a) If f 0 (xl) 0 and f 0 (xr) 0, there is a relative maximum for f at [x∗ , andisgivenbyf(x∗ )]. b) If f 0 (xl) 0 and f 0 (xr) 0, there is a relative minimum for f at [x∗ , andisgivenbyf(x∗ )]. c) If f 0 (x) has the same sign at both xl and xr , then we have an inflection point at [x∗ , f(x∗ )]. 9.6 SECOND DERIVATIVE TEST • Calculate all critical values x∗ such that f 0 (x∗ ) = 0. • Determine the value of f 00 (x∗ ) for any critical value x∗ . a) If f 00 (x∗ ) 0 then there is a relative minimum for f at [x∗ , f(x∗ )]. b) If f 00 (x∗ ) 0 then there is a relative maximum for f at [x∗ , f(x∗ )]. c) If f 00 (x∗ ) = 0 then no conclusion can be drawn at x∗ . 9.7 CURVE SKETCHING Curve sketching includes techniques that can be used to produce a idea of overall shape of a plane curve given its equation. One can get feeling for the general shape of the graph of a function without determining and plotting a large number of ordered pairs. In this section we will discuss about the some of the key determinants of 192

571. shape of the graph of a function and demonstrate curve sketching procedure. 9.7.1 Key Data Points To determine the general shape of the graph of a function the following attributes are the most significant. • Relative maxima and minima • Inflection points • x and y intercepts • Ultimate direction Example 3 Sketch the graph of the function f(x) = x3 3 − 4x2 + 12x + 5 Solution I. Relative Maxima and Minima To find the relative extrema on f firstly we calculate the first derivative. f 0 (x) = x2 − 8x + 12 Setting f 0 equal to 0 yields x2 − 8x + 12 = 0 (x − 2)(x − 6) = 0 Hence the critical values are x = 2 and x = 6 at x = 2 f(x) = 23 3 − 4(2)2 + 12(2) + 5 f(x) = 8 3 − 16 + 24 + 5 = 15.66 193

572. at x = 6 f(x) = 63 3 − 4(6)2 + 12(6) + 5 f(x) = 72 − 144 + 72 + 5 = 5 Hence the critical points exist at (2, 15.66) and (6, 5) graphically at these points zero slope conditions exist. The second derivative of f is f 00 (x) = 2x − 8 Now we test the nature of critical point (2, 15.66) by substituting x = 2 in second derivative of f. f 00 (2) = 2(2) − 8 = −4 0 Therefore, a relative minimum occurs at (6, 5). II. Inflection Points To find inflection point we set f 00 equal to 0 i.e. 2x − 8 = 0 x = 4 By substituting x = 4 into f f(4) = 43 3 − 4(4)2 + 12(4) + 5 f(4) = 10.33 Hence the inflection point occurs at (4,10.33). III. Intercepts The y intercept usually an easy point to locate in this case f(0) = 5 The y intercept occurs at (0,5). 194

573. IV. Ultimate Direction In the given function f the highest power term is x3 3 . To determine the behavior of x more and more positive we need to observe the behavior of x3 3 as x becomes more and more positive. As x → +∞ x3 3 → +∞, therefore as x → +∞ x3 3 − 4x2 + 12x + 5 → +∞, similarly, as x → −∞ x3 3 → −∞, and x → −∞ x3 3 − 4x2 + 12x + 5 → −∞, The final sketch of the given function is. Figure 9.5: Graphical Representation of Maxima, Minima and Point of Inflection 195

574. Example 4 Find all relative extrema for f(x) = 3x5 − 5x3 Solution f 0 (x) = 15x4 − 15x2 = 15x2 (x2 − 1) = 15x2 (x − 1)(x + 1) f 00 (x) = 60x3 − 30x = 30x(2x2 − 1) To find critical point, set f 0 (x) equals to zero 2x3 (3x2 − 4) = 0 x = 0, x = −1, x = 1 We use second derivative test to find the relative extrema. Stationary Point Evaluated f 00 Sign of f 00 2nd Derivative Test −1 f 00 (−1) = −30 − Relative Maxima 0 f 00 (0) = 0 0 Inconclusive 1 f 00 (1) = 30 + Relative Minima Table 9.4: Second Derivative Test of f Since test fails at x = 0, so we apply first derivative test at that point. Interval Evaluated f 0 Sign of f 0 −1 x 0 f 0 (−0.5) = −2.8 − 0 x 1 f 0 (0.5) = −2.8 − Table 9.5: First Derivative Test of f Since there is no sign change in f 0 (x) at x = 0. Thus f has neither relative maxima nor relative minima at this point. 9.8 REVENUE, COST AND PROFIT APPLICATIONS IN BUSINESS In this section we will discuss about revenue, cost and profit and it’s application in business. 196

575. Figure 9.6: Graphical Representation of Relative Extrema 9.8.1 Revenue Revenue is the total payment received from selling a good or per- forming a service. The revenue function, R(x), reflects the revenue from selling x amount of output items at a price of p per item. Total Revenue = (Price per unit)(Quantity sold) Example 5 A firm has determined that total revenue is a function of the price charged for its product. Specifically, the total revenue function is R = f(p) = −10p2 + 1750p, where p equals the price in dollars. a) Determine the price p which results in maximum total revenue b) What is the maximum value for total revenue? Solution a) Firstly, we will calculate the first derivative of the revenue function f 0 (p) = −20p + 1750, 197

576. now set f 0 (p) = 0 −20p + 1750 = 0 p = 87.5 Hence relative maximum occurs at p = 87.5. We can verify this by using the second derivative test i.e. f 00 (p) = −20 f 00 (87.5) = −20 0, therefore, a relative maximum off occurs at p = 87.5 b) The maximum value of R is found by substituting p = 87.5 into f f(87.5) = −10(87.5)2 + 1750(87.5) f(87.5) = 76562.5 Thus, total revenue is expected to be maximized at $76562.5 when the firm charges $87.5 per unit. The graphical representation of the given function is Figure 9.7: Graphical Representation of Quadratic Revenue Function 9.8.2 Cost Cost is the total cost of producing output. The cost function consists of two different types of cost. 198

577. i. Variable cost ii. Fixed cost 9.8.3 Variable Cost Variable cost varies/changes with output (i.e. the number of units produced). The total variable cost can be expressed by taking the product of variable cost per unit and number of units produced. If more items are produced cost is more. 9.8.4 Fixed Cost Fixed Cost normally do not vary with output. In general, it is supposed that these costs must be incurred whether the items are produced or not. C(x) = F + V (x), where C is total cost, F is fixed cost, V , is variable cost per unit and x is number of units produced and sold. Example 6 The total cost of producing q units of a certain product is described by the function C = 1000, 000 + 1, 500q + 0.2q2 , where C is the total cost stated in dollars. Determine the number of units of q that should be produced to minimize the average cost per unit. Solution Average cost per unit C is calculated by dividing the total cost by the number of units produced i.e. C = f(q) = C q = 1000, 000 q + 1 , 500 + 0.2q 199

578. The first derivative of the average cost function is. f 0 (q) = −1000, 000q(−2) + 0.2, set f 0 (q) = 0 1000, 000q(−2) + 0.2 = 0 q2 = 1000, 000 0.2 q2 = 500, 000 Since q are number of units so round it i.e. q = ±707(units) Since the production must be positive so the value q = −707 is mean- ingless. Now we test the nature of the critical point by taking second derivative. f 00 (q) = 2000, 000 q3 At q = 707 f 00 (707) = 2000, 000 (707)3 = 0.00056 0 The second derivative is positive which means that the minimum value of f occurs when q = 707. The minimum average cost per unit is. f(707) = 1000, 000 707 + 1500 + 0.2(707) f(707) = 141.42 + 1500 + 141.4 f(707) = $1782.82 Sketch of the average cost function is. 9.8.5 Profit Function The profit function P(x) is the difference between the revenue function R(x) and the total cost function C(x). Profit = Revenue − Cost P(x) = R(x) − C(x) 200

579. Figure 9.8: Graphical Representation of Quadratic Revenue Function 9.8.6 Marginal Approach to Profit Maximization Given a level of output q∗ where R 0 (q) = C 0 (q) producing q∗ will result in profit maximization if R 00 (q) C 00 (q) Example 7 The total cost and total revenue functions for a product are C(q) = 500 + 100q + 0.5q2 R(q) = 500q a) Using the marginal approach determine the profit maximizing level of output. b) What is the maximum profit? Solution The Profit function using the equation is given by P(q) = R(q) − C(q), differentiate the above equation P 0 (q) = R 0 (q) − C 0 (q), 201

580. set P 0 (q) = 0 and solve for q i.e. R 0 (q) = C 0 (q), let q∗ be the value where R 0 (q) = C 0 (q). The second derivative of P is P 00 (q) = R 00 (q) − C 00 (q), by the second derivative test profit will be maximized at q = q∗ provided P 00 (q∗ ) 0 R 00 (q∗ ) − C 00 (q∗ ) 0 R 00 (q∗ ) C 00 (q∗ ) a) Firstly, we derivate the given equations C(q) and R(q). C 0 (q) = 100 + q R 0 (q) = 500, therefore C 0 (q) = R 0 (q) 100 + q = 500 q∗ = 400 Now again derivate C 0 (q) and R 0 (q) C 00 (q) = 1 R 00 (q) = 0, since R 00 (q∗ ) = 0 and C 00 (q∗ ) = 1, hence R 00 (q∗ ) C 00 (q∗ ) 0 1, there is a relative maximum on the profit function when q = 400 202

581. b) The maximum profit is given by P = f(q) = R(q) − C(q) P = 500q − 500 − 100q − 0.5q2 P = 400q − 0.5q2 − 500, at q = 400 P = 400(400) − 0.5(400)2 − 500 P = 160000 − 80000 − 500 P = 79500 Hence the maximum profit is 79500. Example 8 The total cost c of producing x units of a product is given by c = x2 4 + 3x + 400 At what level of output will average cost per item be minimum? What is this minimum? Solution The quantity to be minimized is average cost c. The average cost per item is defined as c = x2 4 + 3x + 400 x = x 4 + 3 + 400 x (9.8.1) d dx c = 1 4 − 400 x2 = x2 − 1600 4x2 To find critical point, we set dc dx = 0 and solve the equation. x2 − 1600 4x2 = 0 (x − 40)(x + 40) 4x2 = 0 x = 0, x = 40, x = −40 are critical points. To apply 2nd derivative test we use only those critical points that are stationary points. Here 203

582. stationary points are x = 40, x = −40 Since x 0 so the only critical point we use is x = 40 d2 c dx2 = 800 x3 Thus, x = 40 is only relative extremum. Stationary Point Evaluated d2 c dx2 Sign of d2 c dx2 2nd Derivative Test 40 c 00 (40) = 0.0125 + Relative Minima Table 9.6: Second Derivative Test of c Put x = 40 in equation (9.8.1) which gives minimum average cost c = 23 204

583. 9.9 SELF ASSESMENT QUESTIONS (a) Find the interval on which f is increasing, decreasing, concave up, concave down and find the location of the point of inflection. (a) f(x) = x2 − 3x + 8 (b) f(x) = x 4 3 − x 1 3 (c) f(x) = x x2 + 2 (d) f(x) = x4 − 5x3 + 9x2 (b) Use both first and second derivative test to find the relative extrema of the function f (a) f(x) = x4 − 2x3 + 4x2 (b) f(x) = x2 (x − 2)2 (c) f(x) = 3x + 2x 1 3 (d) f(x) = x2 x4 + x2 − 1 (e) f(x) = x + 4 x − 3 (c) Sketch the graph of the functions (a) f(x) = 2x3 − 3x2 − 12x + 1 (b) f(x) = 3x2 x2 + 5 (c) f(x) = x − 1 x (d) f(x) = x − 1 x + 1 (e) f(x) = x3 (d) A manufacturer sells an product with the following cost and revenue functions, where x represents the number of items sold. C(x) = 2.2x − 0.0002x2 0 ≤ x ≤ 1200 R(x) = 4.5x − 0.004x2 0 ≤ x ≤ 1200 Find the interval on which profit function is increasing. (e) The demand function is given by p(x) = 300 − 2x Determine when is marginal revenue increasing? (f) The total cost c of producing x unit of a product is given by c = 0.03x2 + 8x + 300 At what level of output will average cost per unit be minimum 205

584. (g) The demand p and cost function c for a product is given by p = 52 − 0.02x c(x) = 400 + 40x At what level of output will profit be maximized? At what price does this profit arise and what is the profit? (h) A TV cable company has 500 subscribers who are paying $20 per month. There will be 200 more subscribers for each $0.6 decrease in monthly fee. At what rate will maximum revenue be obtained and what will this revenue be? 206

585. BIBLIOGRAPHY (a) Applied Mathematics for Business, Economics and the Social Sciences, Franck S. Budnick, Forth Edition University of Rhodes Island. (b) Probability with Applications, Muhammad Kazim Khan, Sec- ond Edition Kent State University Ohio, United States of Amer- ica USA. (c) Introduction to Mathematical Statistics, HOGG, Mckean and Craig, Sixth Edition, University of Iowa, United States of Amer- ica USA. (d) Business Mathematics, Nadeem Akhtar Siddiqui and Tariq Ali Khan, Level One Revised Azeem Academy, 22-Urdu Bazar La- hore. (e) Introductory Mathematical Analysis for Business, Economics and the Life and Social Sciences, Nineth Edition, Ernest F. Haeusster and Richard S. Paul. (f) Calculus, Early Transcedentals Anton, Bivens and Davis, Tenth Edition. (g) Mathematics with Applications in the Management, Natural and Social Sciences, Lial, Hungerford and Miller, Sixth Edition. 207

586. (h) Calculus and Analytic Geometry, George B. Thomas, Jr. and Ross L. Finney, Eleventh Edition. Also the following websites have been consulted for the contents development guidance and inclusion (i) www.lib.vcomsats.edu.pk (j) www.coursehero.com (k) www.scribd.com (l) www.es.scribd.com (m) www.fr.scribd.com (n) www.docstoc.com (o) www.documents.com (p) www.algebra.com 208

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