Linear Programming Mathematics in the Modern World

MODULE 8
MODULE 8
 Graphing Systems of Linear
Graphing Systems of Linear
Inequalities in Two Variables
Inequalities in Two Variables
 Linear Programming Problems
Linear Programming Problems
 Graphical Solutions of Linear
Graphical Solutions of Linear
Programming Problems
Programming Problems
Linear Programming: A Geometric Approach
Linear Programming: A Geometric Approach

8.1
8.1
Graphing Systems of Linear Inequalities
in Two Variables
in Two Variables
x
x
y
y
4
4x
x + 3
+ 3y
y =
= 12
12
12 12
7 7
( , )
P 12 12
7 7
( , )
P
x
x –
– y
y =
= 0
0
4 3 12
0
x y
x y
 
 
4 3 12
0
x y
x y
 
 
4
4
3
3
2
2
1
1
–
–1
1 1
1 2
2 3
3

Graphing Linear Inequalities
Graphing Linear Inequalities
 Note that a linear
Note that a linear equation
equation in two variables
in two variables x
x and
and y
y
has a solution set that may be exhibited graphically as points
has a solution set that may be exhibited graphically as points
on a straight line in the
on a straight line in the xy
xy-plane.
-plane.
 There is also a simple graphical representation for linear
There is also a simple graphical representation for linear
inequalities
inequalities of two variables:
of two variables:
0
ax by c
  
0
ax by c
  
0
ax by c
  
0
ax by c
  
0
ax by c
  

Procedure for Graphing Linear Inequalities
Procedure for Graphing Linear Inequalities

Example 1
Example 1
 Determine the solution set for the inequality
Determine the solution set for the inequality 2
2x
x + 3
+ 3y
y 
 6
6.
.
Solution
Solution
 Replacing the inequality
Replacing the inequality 
 with an equality
with an equality =
=, we obtain
, we obtain
the equation
the equation 2
2x
x + 3
+ 3y
y = 6
= 6, whose graph is:
, whose graph is:
x
x
y
y
7
7
5
5
3
3
1
1
–
–1
1
–
–5
5 –
–3
3 –
–1
1 1
1 3
3 5
5
2
2x
x + 3
+ 3y
y = 6
= 6

Example 1
Example 1
 Determine the solution set for the inequality
Determine the solution set for the inequality 2
2x
x + 3
+ 3y
y 
 6
6.
.
Solution
Solution
 Picking the origin as a test point, we find
Picking the origin as a test point, we find 2(0) + 3(0)
2(0) + 3(0) 
 6
6,
,
or
or 0
0 
 6
6, which is false.
, which is false.
 Thus, the solution set is:
Thus, the solution set is:
x
x
y
y
7
7
5
5
3
3
1
1
–
–1
1
–
–5
5 –
–3
3 –
–1
1 1
1 3
3 5
5
2
2x
x + 3
+ 3y
y = 6
= 6
2
2x
x + 3
+ 3y
y 
 6
6
(0, 0)
(0, 0)

–
–5
5 –
–3
3 1
1 3
3 5
5
Example 2
Example 2
 Graph
Graph x
x – 3
– 3y
y > 0
> 0.
.
Solution
Solution
 Replacing the inequality
Replacing the inequality >
> with an equality
with an equality =
=, we obtain
, we obtain
the equation
the equation x
x – 3
– 3y
y = 0
= 0, whose graph is:
, whose graph is:
x
x
y
y
3
3
1
1
–
–1
1
–
–3
3
x
x – 3
– 3y
y = 0
= 0

Example 2
Example 2
 Graph
Graph x
x – 3
– 3y
y > 0
> 0.
.
Solution
Solution
 We use a dashed line to indicate the line itself will
We use a dashed line to indicate the line itself will not
not be
be
part of the solution, since we are dealing with a strict
part of the solution, since we are dealing with a strict
inequality
inequality >
>.
.
x
x
y
y
x
x – 3
– 3y
y = 0
= 0
–
–5
5 –
–3
3 1
1 3
3 5
5
3
3
1
1
–
–1
1
–
–3
3

–
–5
5 –
–3
3 1
1 3
3 5
5
3
3
1
1
–
–1
1
–
–3
3
Example 2
Example 2
 Graph
Graph x
x – 3
– 3y
y > 0
> 0.
.
Solution
Solution
 Since the origin lies on the line, we cannot use the origin
Since the origin lies on the line, we cannot use the origin
as a testing point:
as a testing point:
x
x
y
y
x
x – 3
– 3y
y = 0
= 0
(0, 0)
(0, 0)

Example 2
Example 2
 Graph
Graph x
x – 3
– 3y
y > 0
> 0.
.
Solution
Solution
 Picking instead
Picking instead (3, 0)
(3, 0) as a test point, we find
as a test point, we find (3) – 2(0) > 0
(3) – 2(0) > 0,
,
or
or 3 > 0
3 > 0, which is true.
, which is true.
 Thus, the solution set is:
Thus, the solution set is:
y
y
x
x – 3
– 3y
y = 0
= 0
x
x – 3
– 3y
y > 0
> 0
–
–5
5 –
–3
3 1
1 3
3 5
5
3
3
1
1
–
–1
1
–
–3
3
x
x
(3, 0)
(3, 0)

 The solution set of a system of linear inequalities in two
The solution set of a system of linear inequalities in two
variables
variables x
x and
and y
y is the set of all points
is the set of all points (
(x
x,
, y
y)
) that satisfy
that satisfy
each inequality of the system.
each inequality of the system.
 The graphical solution of such a system may be obtained
The graphical solution of such a system may be obtained
by graphing the solution set for each inequality
by graphing the solution set for each inequality
independently and then determining the region in
independently and then determining the region in
common with each solution set.
common with each solution set.

Example 3
Example 3
 Determine the solution set for the system
Determine the solution set for the system
Solution
Solution
 The intersection of the solution regions of the two
The intersection of the solution regions of the two
inequalities represents the solution to the system:
4 3 12
0
x y
x y
 
 
x
x
y
y
4
4
3
3
2
2
1
1
4
4x
x + 3
+ 3y
y 
 12
12
4
4x
x + 3
+ 3y
y = 12
= 12
–
–1
1 1
1 2
2 3
3

Example 3
Example 3
Solution
Solution
4 3 12
0
x y
x y
 
 
x
x
y
y
x
x –
– y
y 
 0
0 x
x –
– y
y = 0
= 0
4
4
3
3
2
2
1
1
–
–1
1 1
1 2
2 3
3

Example 3
Example 3
Solution
Solution
4 3 12
0
x y
x y
 
 
x
x
y
y
4
4x
x + 3
+ 3y
y = 12
= 12
x
x –
– y
y = 0
= 0
4 3 12
0
x y
x y
 
 
4
4
3
3
2
2
1
1
–
–1
1 1
1 2
2 3
3
12 12
7 7
( , )
P

Bounded and Unbounded Sets
Bounded and Unbounded Sets
 The solution set of a system of linear inequalities
The solution set of a system of linear inequalities
is bounded if it can be enclosed by a circle.
is bounded if it can be enclosed by a circle.
 Otherwise, it is unbounded.
Otherwise, it is unbounded.

Example 3
Example 3
 The solution to the problem we just discussed is
The solution to the problem we just discussed is
unbounded, since the solution set cannot be
unbounded, since the solution set cannot be
enclosed in a circle:
enclosed in a circle:
x
x
y
y
4
4x
x + 3
+ 3y
y = 12
= 12
12 12
7 7
( , )
P
x
x –
– y
y = 0
= 0
4 3 12
0
x y
x y
 
 
4
4
3
3
2
2
1
1
–
–1
1 1
1 2
2 3
3

7
7
5
5
3
3
1
1
–
–1
1 1
1 3
3 5
5 9
9
Example 4
Example 4
Solution
Solution
 The intersection of the solution regions of the four
The intersection of the solution regions of the four
6 0 2 8 0 0 0
x y x y x y
       
x
x
y
y
2 8 0
x y
  
6 0
x y
  
(2,4)
P

Example 4
Example 4
Solution
Solution
 Note that the solution to this problem is bounded, since it
Note that the solution to this problem is bounded, since it
can be enclosed by a circle:
can be enclosed by a circle:
6 0 2 8 0 0 0
x y x y x y
       
–
–1
1 1
1 3
3 5
5 9
9
x
x
y
y
7
7
5
5
3
3
1
1
6 0
x y
  
(2,4)
P
2 8 0
x y
  
Hence, the solution set is
the region bounded by
polygon with vertices (0, 0),
(0, 6), (2, 4), and (4, 0).

EXERCISE 1: Graph the given inequalities.
EXERCISE 1: Graph the given inequalities.
EXERCISE 2: Graph the solution to the given system of
EXERCISE 2: Graph the solution to the given system of
inequalities and find all the vertices.
inequalities and find all the vertices.

8.2
8.2
0
0
x
y


1.2
P x y
 
3 300
x y
 
2 180
x y
 
Maximize
Subject to

Linear Programming Problem
Linear Programming Problem
 A linear programming problem consists of a
A linear programming problem consists of a
linear objective function to be maximized or
linear objective function to be maximized or
minimized subject to certain constraints in the
minimized subject to certain constraints in the
form of linear equations or inequalities.
form of linear equations or inequalities.

Applied Example 1:
Applied Example 1: A Production Problem
A Production Problem
 Ace Novelty wishes to produce two types of souvenirs:
Ace Novelty wishes to produce two types of souvenirs:
type-A will result in a profit of
type-A will result in a profit of $1.00
$1.00, and type-B in a
, and type-B in a
profit of
profit of $1.20
$1.20.
.
 To manufacture a type-A souvenir requires
To manufacture a type-A souvenir requires 2
2 minutes on
minutes on
machine I and
machine I and 1
1 minute on machine II.
minute on machine II.
 A type-B souvenir requires
A type-B souvenir requires 1
1 minute on machine I and
minute on machine I and 3
3
minutes on machine II.
minutes on machine II.
 There are
There are 3
3 hours available on machine I and
hours available on machine I and 5
5 hours
hours
available on machine II.
available on machine II.
 How many souvenirs of each type should Ace make in
How many souvenirs of each type should Ace make in
order to maximize its profit?
order to maximize its profit?

Applied Example 1:
Solution
Solution
 Let’s first tabulate the given information:
Let’s first tabulate the given information:
 Let
Let x
x be the number of type-A souvenirs and
be the number of type-A souvenirs and y
y the number
the number
of type-B souvenirs to be made.
of type-B souvenirs to be made.
Type-A
Type-A Type-B
Type-B Time Available
Time Available
Profit/Unit
Profit/Unit $1.00
$1.00 $1.20
$1.20
Machine I
Machine I 2 min
2 min 1 min
1 min 180 min
180 min
Machine II
Machine II 1 min
1 min 3 min
3 min 300 min
300 min

Applied Example 1:
Solution
Solution
 Then, the total profit (in dollars) is given by
Then, the total profit (in dollars) is given by
which is the objective function to be maximized.
which is the objective function to be maximized.
1.2
P x y
 
Type-A
Type-A Type-B
Time Available
Profit/Unit
Profit/Unit $1.00
$1.00 $1.20
$1.20
Machine I
Machine I 2 min
2 min 1 min
1 min 180 min
180 min
Machine II
Machine II 1 min
1 min 3 min
3 min 300 min
300 min

Applied Example 1:
Solution
Solution
 The total amount of time that machine I is used is
The total amount of time that machine I is used is
and must not exceed
and must not exceed 180
180 minutes.
minutes.
 Thus, we have the inequality
Thus, we have the inequality
2x y

2 180
x y
 
Type-A
Type-A Type-B
Time Available
Profit/Unit
Profit/Unit $1.00
$1.00 $1.20
$1.20
Machine I
Machine I 2 min
2 min 1 min
1 min 180 min
180 min
Machine II
Machine II 1 min
1 min 3 min
3 min 300 min
300 min

Applied Example 1:
Solution
Solution
 The total amount of time that machine II is used is
The total amount of time that machine II is used is
and must not exceed
and must not exceed 300
300 minutes.
minutes.
 Thus, we have the inequality
Thus, we have the inequality
3
x y

3 300
x y
 
Type-A
Type-A Type-B
Time Available
Profit/Unit
Profit/Unit $1.00
$1.00 $1.20
$1.20
Machine I
Machine I 2 min
2 min 1 min
1 min 180 min
180 min
Machine II
Machine II 1 min
1 min 3 min
3 min 300 min
300 min

Applied Example 1:
Solution
Solution
 Finally, neither
Finally, neither x
x nor
nor y
y can be negative, so
can be negative, so
0
0
x
y


Type-A
Type-A Type-B
Time Available
Profit/Unit
Profit/Unit $1.00
$1.00 $1.20
$1.20
Machine I
Machine I 2 min
2 min 1 min
1 min 180 min
180 min
Machine II
Machine II 1 min
1 min 3 min
3 min 300 min
300 min

Applied Example 1:
Solution
Solution
 In short, we want to maximize the objective function
In short, we want to maximize the objective function
subject to the system of inequalities
 We will discuss the solution to this problem in the
We will discuss the solution to this problem in the
proceeding section.
proceeding section.
0
0
x
y


1.2
P x y
 
3 300
x y
 
2 180
x y
 

Applied Example 2:
Applied Example 2: A Nutrition Problem
A Nutrition Problem
 A nutritionist advises an individual who is suffering from
A nutritionist advises an individual who is suffering from
iron and vitamin B deficiency to take at least
iron and vitamin B deficiency to take at least 2400
2400
milligrams (mg) of iron,
milligrams (mg) of iron, 2100
2100 mg of vitamin B
mg of vitamin B1
1, and
, and 1500
1500
mg of vitamin B
mg of vitamin B2
2 over a period of time.
over a period of time.
 Two vitamin pills are suitable, brand-A and brand-B.
Two vitamin pills are suitable, brand-A and brand-B.
 Each brand-A pill costs
Each brand-A pill costs 6
6 cents and contains
cents and contains 40
40 mg of iron,
mg of iron,
10
10 mg of vitamin B
mg of vitamin B1
1, and
, and 5
5 mg of vitamin B
mg of vitamin B2
2.
.
 Each brand-B pill costs
Each brand-B pill costs 8
8 cents and contains
cents and contains 10
10 mg of iron
mg of iron
and
and 15
15 mg each of vitamins B
mg each of vitamins B1
1 and B
and B2
2.
.
 What combination of pills should the individual purchase
What combination of pills should the individual purchase
in order to meet the minimum iron and vitamin
requirements at the lowest cost?
requirements at the lowest cost?

Applied Example 2:
A Nutrition Problem
Solution
Solution
 Let
Let x
x be the number of brand-A pills and
be the number of brand-A pills and y
y the number of
the number of
brand-B pills to be purchased.
brand-B pills to be purchased.
Brand-A
Brand-A Brand-B
Brand-B Minimum Requirement
Minimum Requirement
Cost/Pill
Cost/Pill 6
6¢
¢ 8
8¢
¢
Iron
Iron 40 mg
40 mg 10 mg
10 mg 2400 mg
2400 mg
Vitamin B
Vitamin B1
1 10 mg
10 mg 15 mg
15 mg 2100 mg
2100 mg
Vitamin B
Vitamin B2
2 5mg
5mg 15 mg
15 mg 1500 mg
1500 mg

Applied Example 2:
A Nutrition Problem
Solution
Solution
 The cost
The cost C
C (in cents) is given by
(in cents) is given by
and is the objective function to be minimized.
and is the objective function to be minimized.
Brand-A
Brand-A Brand-B
Minimum Requirement
Cost/Pill
Cost/Pill 6
6¢
¢ 8
8¢
¢
Iron
Iron 40 mg
40 mg 10 mg
10 mg 2400 mg
2400 mg
Vitamin B
Vitamin B1
1 10 mg
10 mg 15 mg
15 mg 2100 mg
2100 mg
Vitamin B
Vitamin B2
2 5mg
5mg 15 mg
15 mg 1500 mg
1500 mg
6 8
C x y
 

Applied Example 2:
A Nutrition Problem
Solution
Solution
 The amount of iron contained in
The amount of iron contained in x
x brand-A pills and
brand-A pills and y
y
brand-B pills is given by
brand-B pills is given by 40
40x
x + 10
+ 10y
y mg, and this must be
mg, and this must be
greater than or equal to
greater than or equal to 2400
2400 mg.
mg.
 This translates into the inequality
This translates into the inequality
Brand-A
Brand-A Brand-B
Minimum Requirement
Cost/Pill
Cost/Pill 6
6¢
¢ 8
8¢
¢
Iron
Iron 40 mg
40 mg 10 mg
10 mg 2400 mg
2400 mg
Vitamin B
Vitamin B1
1 10 mg
10 mg 15 mg
15 mg 2100 mg
2100 mg
Vitamin B
Vitamin B2
2 5mg
5mg 15 mg
15 mg 1500 mg
1500 mg
40 10 2400
x y
 

Applied Example 2:
A Nutrition Problem
Solution
Solution
 The amount of vitamin B
The amount of vitamin B1
1 contained in
contained in x
x brand-A pills and
brand-A pills and
y
y brand-B pills is given by
10x
x + 15
+ 15y
greater or equal to
greater or equal to 2100
2100 mg.
mg.
Brand-A
Brand-A Brand-B
Minimum Requirement
Cost/Pill
Cost/Pill 6
6¢
¢ 8
8¢
¢
Iron
Iron 40 mg
40 mg 10 mg
10 mg 2400 mg
2400 mg
Vitamin B
Vitamin B1
1 10 mg
10 mg 15 mg
15 mg 2100 mg
2100 mg
Vitamin B
Vitamin B2
2 5mg
5mg 15 mg
15 mg 1500 mg
1500 mg
10 15 2100
x y
 

Applied Example 2:
A Nutrition Problem
Solution
Solution
 The amount of vitamin B
The amount of vitamin B2
2 contained in
contained in x
x brand-A pills and
brand-A pills and
y
y brand-B pills is given by
5x
x + 15
+ 15y
greater or equal to
greater or equal to 1500
1500 mg.
mg.
Brand-A
Brand-A Brand-B
Minimum Requirement
Cost/Pill
Cost/Pill 6
6¢
¢ 8
8¢
¢
Iron
Iron 40 mg
40 mg 10 mg
10 mg 2400 mg
2400 mg
Vitamin B
Vitamin B1
1 10 mg
10 mg 15 mg
15 mg 2100 mg
2100 mg
Vitamin B
Vitamin B2
2 5mg
5mg 15 mg
15 mg 1500 mg
1500 mg
5 15 1500
x y
 

Applied Example 2:
A Nutrition Problem
Solution
Solution
 In short, we want to minimize the objective function
In short, we want to minimize the objective function
 We will discuss the solution to this problem in the
We will discuss the solution to this problem in the
proceeding section.
proceeding section.
5 15 1500
x y
 
6 8
C x y
 
40 10 2400
x y
 
10 15 2100
x y
 
0
0
x
y



8.3
8.3
Graphical Solutions
Graphical Solutions
of Linear Programming Problems
of Linear Programming Problems
200
200
100
100
100
100 200
200 300
300
x
x
y
y
S
S
10 15 2100
x y
 
10 15 2100
x y
 
40 10 2400
x y
 
40 10 2400
x y
 
5 15 1500
x y
 
5 15 1500
x y
 
C
C(120, 60)
(120, 60)
D
D(300, 0)
(300, 0)
A
A(0, 240)
(0, 240)
B
B(30, 120)
(30, 120)

Feasible Solution Set and Optimal Solution
Feasible Solution Set and Optimal Solution
 The constraints in a linear programming problem form a
The constraints in a linear programming problem form a
system of linear inequalities, which have a solution set
system of linear inequalities, which have a solution set S
S.
.
 Each point in
Each point in S
S is a candidate for the solution of the linear
is a candidate for the solution of the linear
programming problem and is referred to as a feasible
programming problem and is referred to as a feasible
solution.
solution.
 The set
The set S
S itself is referred to as a feasible set.
itself is referred to as a feasible set.
 Among all the points in the set
Among all the points in the set S
S, the point(s) that
, the point(s) that
optimizes the objective function of the linear programming
optimizes the objective function of the linear programming
problem is called an optimal solution.
problem is called an optimal solution.

Theorem 1
Theorem 1
Linear Programming
Linear Programming
 If a linear programming problem has a solution,
If a linear programming problem has a solution,
then it must occur at a vertex, or corner point,
then it must occur at a vertex, or corner point,
of the feasible set
of the feasible set S
S associated with the problem.
associated with the problem.
 If the objective function
If the objective function P
P is optimized at
is optimized at two
two
adjacent vertices of
adjacent vertices of S
S, then it is optimized at
, then it is optimized at every
every
point
point on the line segment joining these vertices, in
on the line segment joining these vertices, in
which case there are infinitely many solutions to
which case there are infinitely many solutions to
the problem.
the problem.

Theorem 2
Theorem 2
Existence of a Solution
Existence of a Solution
 Suppose we are given a linear programming
Suppose we are given a linear programming
problem with a feasible set
problem with a feasible set S
S and an objective
and an objective
function
function P
P =
= ax
ax +
+ by
by.
.
a.
a. If
If S
S is bounded, then
is bounded, then P
P has both a maximum and
has both a maximum and
a minimum value on
a minimum value on S
S.
.
b.
b. If
If S
S is unbounded and both
is unbounded and both a
a and
and b
b are
are
nonnegative, then
nonnegative, then P
P has a minimum value on
has a minimum value on S
S
provided that the constraints defining
provided that the constraints defining S
S include
include
the inequalities
the inequalities x
x 
 0
0 and
and y
y 
 0
0.
.
c.
c. If
If S
S is the empty set, then the linear
is the empty set, then the linear
programming problem has no solution: that is,
programming problem has no solution: that is, P
P
has neither a maximum nor a minimum value.
has neither a maximum nor a minimum value.

The Method of Corners
The Method of Corners
1.
1. Graph the feasible set.
Graph the feasible set.
2.
2. Find the coordinates of all corner points
Find the coordinates of all corner points
(vertices) of the feasible set.
(vertices) of the feasible set.
3.
3. Evaluate the objective function at each corner
Evaluate the objective function at each corner
point.
point.
4.
4. Find the vertex that renders the objective
Find the vertex that renders the objective
function a maximum or a minimum.
function a maximum or a minimum.
✦ If there is only one such vertex, it constitutes a
If there is only one such vertex, it constitutes a
unique solution to the problem.
unique solution to the problem.
✦ If there are two such adjacent vertices, there
If there are two such adjacent vertices, there
are infinitely many optimal solutions given by
are infinitely many optimal solutions given by
the points on the line segment determined by
the points on the line segment determined by
these vertices.
these vertices.

Applied Example 1:
 Recall the previous
Recall the previous Applied Example 1
Applied Example 1 which required us
which required us
to find the optimal quantities to produce of type-A and
to find the optimal quantities to produce of type-A and
type-B souvenirs in order to maximize profits.
type-B souvenirs in order to maximize profits.
 We restated the problem as a linear programming problem
We restated the problem as a linear programming problem
in which we wanted to maximize the objective function
in which we wanted to maximize the objective function
 We can now solve the problem graphically.
We can now solve the problem graphically.
0
0
x
y


1.2
P x y
 
3 300
x y
 
2 180
x y
 

200
200
100
100
100
100 200
200 300
300
Applied Example 1:
 We first graph the feasible set
We first graph the feasible set S
S for the problem.
for the problem.
✦ Graph the solution for the inequality
Graph the solution for the inequality
considering only positive values for
considering only positive values for x
x and
and y
y:
:
2 180
x y
 
x
x
y
y
2 180
x y
 
(90, 0)
(90, 0)
(0, 180)
(0, 180)
2 180
x y
 

200
200
100
100
Applied Example 1:
S for the problem.
for the problem.
x and
and y
y:
:
3 300
x y
 
100
100 200
200 300
300
x
x
y
y
3 300
x y
 
(0, 100)
(0, 100)
(300, 0)
(300, 0)
3 300
x y
 

200
200
100
100
Applied Example 1:
S for the problem.
for the problem.
✦ Graph the intersection of the solutions to the inequalities,
Graph the intersection of the solutions to the inequalities,
yielding the feasible set
yielding the feasible set S
S.
.
(Note that the feasible set
(Note that the feasible set S
S is bounded)
is bounded)
100
100 200
200 300
300
x
x
y
y
S
S 3 300
x y
 
2 180
x y
 

200
200
100
100
Applied Example 1:
 Next, find the vertices of the feasible set
Next, find the vertices of the feasible set S
S.
.
✦ The vertices are
The vertices are A
A(0, 0)
(0, 0),
, B
B(90, 0)
(90, 0),
, C
C(48, 84)
(48, 84), and
, and D
D(0, 100)
(0, 100).
.
100
100 200
200 300
300
x
x
y
y
S
S
C
C(48, 84)
(48, 84)
3 300
x y
 
2 180
x y
 
D
D(0, 100)
(0, 100)
B
B(90, 0)
(90, 0)
A
A(0, 0)
(0, 0)

200
200
100
100
Applied Example 1:
 Now, find the values of
Now, find the values of P
P at the vertices and tabulate them:
at the vertices and tabulate them:
100
100 200
200 300
300
x
x
y
y
S
S
C
C(48, 84)
(48, 84)
3 300
x y
 
2 180
x y
 
D
D(0, 100)
(0, 100)
B
B(90, 0)
(90, 0)
A
A(0, 0)
(0, 0)
Vertex
Vertex P
P =
= x
x + 1.2
+ 1.2 y
y
A
A(0, 0)
(0, 0) 0
0
B
B(90, 0)
(90, 0) 90
90
C
C(48, 84)
(48, 84) 148.8
148.8
D
D(0, 100)
(0, 100) 120
120

200
200
100
100
Applied Example 1:
 Finally, identify the vertex with the highest value for
Finally, identify the vertex with the highest value for P
P:
:
✦ We can see that
We can see that P
P is maximized at the vertex
is maximized at the vertex C
C(48, 84)
(48, 84)
and has a value of
and has a value of 148.8
148.8.
.
100
100 200
200 300
300
x
x
y
y
S
S 3 300
x y
 
2 180
x y
 
D
D(0, 100)
(0, 100)
B
B(90, 0)
(90, 0)
A
A(0, 0)
(0, 0)
Vertex
Vertex P
P =
= x
x + 1.2
+ 1.2 y
y
A
A(0, 0)
(0, 0) 0
0
B
B(90, 0)
(90, 0) 90
90
C
C(48, 84)
(48, 84) 148.8
148.8
D
D(0, 100)
(0, 100) 120
120
C
C(48, 84)
(48, 84)

Applied Example 1:
 Finally, identify the vertex with the highest value
Finally, identify the vertex with the highest value
for
for P
P:
:
✦ We can see that
We can see that P
P is maximized at the vertex
is maximized at the vertex
C
C(48, 84)
(48, 84) and has a value of
and has a value of 148.8
148.8.
.
✦ Recalling what the symbols
Recalling what the symbols x
x,
, y
y, and
, and P
P
represent, we conclude that ACE Novelty
represent, we conclude that ACE Novelty
would maximize its profit at
would maximize its profit at $148.80
$148.80 by
by
producing
producing 48
48 type-A souvenirs and
type-A souvenirs and 84
84 type-B
type-B
souvenirs.
souvenirs.

Applied Example 2:
A Nutrition Problem
 Recall the previous
Recall the previous Applied Example 2
Applied Example 2, which asked us to
, which asked us to
determine the optimal combination of pills to be purchased
determine the optimal combination of pills to be purchased
requirements at the lowest cost.
requirements at the lowest cost.
 We restated the problem as a linear programming problem
We restated the problem as a linear programming problem
in which we wanted to minimize the objective function
in which we wanted to minimize the objective function
 We can now solve the problem graphically.
We can now solve the problem graphically.
5 15 1500
x y
 
6 8
C x y
 
40 10 2400
x y
 
10 15 2100
x y
 
, 0
x y 

200
200
100
100
Applied Example 2:
A Nutrition Problem
S for the problem.
for the problem.
x and
and y
y:
:
100
100 200
200 300
300
x
x
y
y
40 10 2400
x y
 
40 10 2400
x y
 
(60, 0)
(60, 0)
(0, 240)
(0, 240)

200
200
100
100
Applied Example 2:
A Nutrition Problem
S for the problem.
for the problem.
x and
and y
y:
:
100
100 200
200 300
300
x
x
y
y
10 15 2100
x y
 
10 15 2100
x y
 
(210, 0)
(210, 0)
(0, 140)
(0, 140)

200
200
100
100
Applied Example 2:
A Nutrition Problem
S for the problem.
for the problem.
x and
and y
y:
:
100
100 200
200 300
300
x
x
y
y
5 15 1500
x y
 
5 15 1500
x y
 
(300, 0)
(300, 0)
(0, 100)
(0, 100)

200
200
100
100
Applied Example 2:
A Nutrition Problem
S for the problem.
for the problem.
✦ Graph the intersection of the solutions to the inequalities,
Graph the intersection of the solutions to the inequalities,
yielding the feasible set
yielding the feasible set S
S.
.
(Note that the feasible set
(Note that the feasible set S
S is unbounded)
is unbounded)
100
100 200
200 300
300
x
x
y
y
S
S
10 15 2100
x y
 
40 10 2400
x y
 
5 15 1500
x y
 

200
200
100
100
Applied Example 2:
A Nutrition Problem
 Next, find the vertices of the feasible set
Next, find the vertices of the feasible set S
S.
.
✦ The vertices are
The vertices are A
A(0, 240)
(0, 240),
, B
B(30, 120)
(30, 120),
, C
C(120, 60)
(120, 60), and
, and
D
D(300, 0)
(300, 0).
.
100
100 200
200 300
300
x
x
y
y
S
S
10 15 2100
x y
 
40 10 2400
x y
 
5 15 1500
x y
 
C
C(120, 60)
(120, 60)
D
D(300, 0)
(300, 0)
A
A(0, 240)
(0, 240)
B
B(30, 120)
(30, 120)

Applied Example 2:
A Nutrition Problem
 Now, find the values of
Now, find the values of C
C at the vertices and tabulate them:
at the vertices and tabulate them:
200
200
100
100
100
100 200
200 300
300
x
x
y
y
S
S
10 15 2100
x y
 
40 10 2400
x y
 
5 15 1500
x y
 
C
C(120, 60)
(120, 60)
D
D(300, 0)
(300, 0)
A
A(0, 240)
(0, 240)
B
B(30, 120)
(30, 120)
Vertex
Vertex C
C = 6
= 6x
x + 8
+ 8y
y
A
A(0, 240)
(0, 240) 1920
1920
B
B(30, 120)
(30, 120) 1140
1140
C
C(120, 60)
(120, 60) 1200
1200
D
D(300, 0)
(300, 0) 1800
1800

Applied Example 2:
A Nutrition Problem
 Finally, identify the vertex with the lowest value for
Finally, identify the vertex with the lowest value for C
C:
:
✦ We can see that
We can see that C
C is minimized at the vertex
is minimized at the vertex B
B(30, 120)
(30, 120)
and has a value of
and has a value of 1140
1140.
.
200
200
100
100
100
100 200
200 300
300
x
x
y
y
S
S
10 15 2100
x y
 
40 10 2400
x y
 
5 15 1500
x y
 
C
C(120, 60)
(120, 60)
D
D(300, 0)
(300, 0)
A
A(0, 240)
(0, 240)
Vertex
Vertex C
C = 6
= 6x
x + 8
+ 8y
y
A
A(0, 240)
(0, 240) 1920
1920
B
B(30, 120)
(30, 120) 1140
1140
C
C(120, 60)
(120, 60) 1200
1200
D
D(300, 0)
(300, 0) 1800
1800
B
B(30, 120)
(30, 120)

Applied Example 2:
A Nutrition Problem
 Finally, identify the vertex with the lowest value
Finally, identify the vertex with the lowest value
for
for C
C:
:
✦ We can see that
We can see that C
C is minimized at the vertex
is minimized at the vertex
B
B(30, 120)
(30, 120) and has a value of
and has a value of 1140
1140.
.
✦ Recalling what the symbols
Recalling what the symbols x
x,
, y
y, and
, and C
C
represent, we conclude that the individual
represent, we conclude that the individual
should purchase
should purchase 30
30 brand-A pills and
brand-A pills and 120
120
brand-B pills at a minimum cost of
brand-B pills at a minimum cost of $11.40
$11.40.
.

EXERCISE 2
EXERCISE 2
 A housing contractor designed two styles of
A housing contractor designed two styles of
houses – bungalows and 2-story houses for 100
houses – bungalows and 2-story houses for 100
available lots. The bungalows style requires
available lots. The bungalows style requires
P1.5M capital and produces a profit of P200,000
P1.5M capital and produces a profit of P200,000
when sold. A 2-story house requires P2M capital
when sold. A 2-story house requires P2M capital
and will earn a profit of P400,000. If the available
and will earn a profit of P400,000. If the available
capital on hand is P180M, how many houses of
capital on hand is P180M, how many houses of
each type should be built to gain a maximum
each type should be built to gain a maximum
profit?
profit?

Linear Programming Mathematics in the Modern World

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