A Optimization Techniques

1
W E B C H A P T E R
W E B C H A P T E R P R E V I E W Normative
economic decision analysis involves determining
the action that best achieves a desired goal or objec-
tive. This means finding the action that optimizes
(that is, maximizes or minimizes) the value of an
objective function. For example, in a price-output
decision-making problem, we may be interested in
determining the output level that maximizes profits.
In a production problem, the goal may be to find the
combination of inputs (resources) that minimizes
the cost of producing a desired level of output. In a
capital budgeting problem, the objective may be to
select those projects that maximize the net present
value of the investments chosen. There are many
techniques for solving optimization problems such
as these. This chapter (and appendix) focuses on the
use of differential calculus to solve certain types of
optimization problems. In Web Chapter B, linear-
programming techniques, used in solving con-
strained optimization problems, are examined.
Optimization techniques are a powerful set of tools
that are important in efficiently managing an enter-
prise’s resources and thereby maximizing share-
holder wealth.
A
Optimization Techniques
PHOTO
ON
PAGE
1
AND
2:
©
DANIEL
MACKIE/STONE.

2 WEB CHAPTER A Optimization Techniques
TYPES OF OPTIMIZATION TECHNIQUES
In Chapter 1 we defined the general form of a problem that managerial economics at-
tempts to analyze. The basic form of the problem is to identify the alternative means of
achieving a given objective and then to select the alternative that accomplishes the ob-
jective in the most efficient manner, subject to constraints on the means. In program-
ming terminology, the problem is optimizing the value of some objective function, sub-
ject to any resource and/or other constraints such as legal, input, environmental, and
behavioral restrictions.
In 1990 the U.S. Air Force publicly unveiled its
newest long-range strategic bomber—the B-2 or
“Stealth” bomber. This plane is characterized by a
unique flying wing design engineered to evade
detection by enemy radar. The
plane has been controversial
because of its high cost. However,
a lesser-known controversy
relates to its fundamental design.
The plane’s flying wing
design originated from a secret
study of promising military tech-
nologies that was undertaken at
the end of World War II. The
group of prominent scientists
who undertook the study con-
cluded that a plane can achieve
maximum range if it has a design
in which virtually all the volume
of the plane is contained in the
wing. A complex mathematical
appendix was attached to the
study that purported to show
that range could be maximized
with the flying wing design.
However, a closer examination of the technical
appendix by Joseph Foa, now an emeritus professor
of engineering at George Washington University,
discovered that a fundamental error had been made
in the initial report. It turned out that the original
researchers had taken the first derivative of a com-
plex equation for the range of a plane and found
that it had two solutions. The original researchers
mistakenly concluded that the
all-wing design was the one that
maximized range, when, in fact,
it minimized range.
In this chapter we introduce
some of the same optimization
techniques applied to an analy-
sis of the Stealth bomber project.
We develop tools designed to
maximize profits or minimize
costs. Fortunately, the mathe-
matical functions we deal with
in this chapter and throughout
the book are much simpler than
those that confronted the origi-
nal “flying wing” engineers. We
introduce techniques that can be
used to check whether a func-
tion, such as profits or costs, is
being minimized or maximized
at a particular level of output.
1
This Managerial Challenge is based primarily on W. Biddle,
“Skeleton Alleged in the Stealth Bomber’s Closet,” Science, 12
May 1989, pp. 650–651.
M A N A G E R I A L C H A L L E N G E
A Skeleton in the Stealth Bomber’s Closet1

WEB CHAPTER A Optimization Techniques 3
Mathematically, we can represent the problem as
Optimize y ⫽ f(x1, x2, . . . , xn) [A.1]
subject to gj(x1, x2, . . . , xn) bj j ⫽ 1, 2, . . . , m [A.2]
where Equation A.1 is the objective function and Equation A.2 constitutes the set of con-
straints imposed on the solution. The xi variables, x1, x2, . . ., xn, represent the set of de-
cision variables, and y ⫽ f(x1, x2, . . ., xn) is the objective function expressed in terms of
these decision variables. Depending on the nature of the problem, the term optimize
means either maximize or minimize the value of the objective function. As indicated in
Equation A.2, each constraint can take the form of an equality (⫽) or an inequality (ⱕ
or ⱖ) relationship.
Complicating Factors in Optimization
Several factors can make optimization problems fairly complex and difficult to solve.
One such complicating factor is the existence of multiple decision variables in a problem.
Relatively simple procedures exist for determining the profit-maximizing output level
for the single-product firm. However, the typical medium- or large-size firm often pro-
duces a large number of different products, and as a result, the profit-maximization
problem for such a firm requires a series of output decisions—one for each product. An-
other factor that may add to the difficulty of solving a problem is the complex nature of
the relationships between the decision variables and the associated outcome. For example,
in public policy decisions on government spending for such items as education, it is ex-
tremely difficult to determine the relationship between a given expenditure and the ben-
efits of increased income, employment, and productivity it provides. No simple rela-
tionship exists among the variables. Many of the optimization techniques discussed here
are only applicable to situations in which a relatively simple function or relationship can
be postulated between the decision variables and the outcome variable. A third compli-
cating factor is the possible existence of one or more complex constraints on the decision
variables. For example, virtually every organization has constraints imposed on its deci-
sion variables by the limited resources—such as capital, personnel, and facilities—over
which it has control. These constraints must be incorporated into the decision problem.
Otherwise, the optimization techniques that are applied to the problem may yield a so-
lution that is unacceptable from a practical standpoint. Another complicating factor is
the presence of uncertainty or risk. In this chapter, we limit the analysis to decision mak-
ing under certainty, that is, problems in which each action is known to lead to a specific
outcome. Chapter 2 examines methods for analyzing decisions involving risk and un-
certainty. These factors illustrate the difficulties that may be encountered and may ren-
der a problem unsolvable by formal optimization procedures.
Constrained versus Unconstrained Optimization
The mathematical techniques used to solve an optimization problem represented by
Equations A.1 and A.2 depend on the form of the criterion and constraint functions. The
simplest situation to be considered is the unconstrained optimization problem. In such a
problem no constraints are imposed on the decision variables, and differential calculus can
be used to analyze them. Another relatively simple form of the general optimization prob-
lem is the case in which all the constraints of the problem can be expressed as equality
再ⱕ
⫽
ⱖ
冎

(⫽) relationships. The technique of Lagrangian multipliers can be used to find the opti-
mal solution to many of these problems.
Often, however, the constraints in an economic decision-making problem take the
form of inequality relationships (ⱕ or ⱖ) rather than equalities. For example, limitations
on the resources—such as personnel and capital—of an organization place an upper
bound or budget ceiling on the quantity of these resources that can be employed in max-
imizing (or minimizing) the objective function. With this type of constraint, all of a
given resource need not be used in an optimal solution to the problem. An example of a
lower bound would be a loan agreement that requires a firm to maintain a current ratio
(that is, ratio of current assets to current liabilities) of at least 2.00. Any combination of
current assets and current liabilities having a ratio greater than or equal to 2.00 would
meet the provisions of the loan agreement. Such optimization procedures as the La-
grangian multiplier method are not suited to solving problems of this type efficiently;
however, modern mathematical programming techniques have been developed that can
efficiently solve several classes of problems with these inequality restrictions.
Linear-programming problems constitute the most important class for which efficient
solution techniques have been developed. In a linear-programming problem, both the ob-
jective and the constraint relationships are expressed as linear functions of decision vari-
ables.2
Other classes of problems include integer-programming problems, in which some
(or all) of the decision variables are required to take on integer values, and quadratic-
programming problems, in which the objective relationship is a quadratic function of the
decision variables.3
Generalized computing algorithms exist for solving optimization
problems that meet these requirements.
The remainder of this chapter deals with the classical optimization procedures of dif-
ferential calculus. Lagrangian multiplier techniques are covered in the Appendix. Linear
programming is encountered in Web Chapter B.
CONSTRAINED OPTIMIZATION: OPTIMIZING FLIGHT CREW
SCHEDULES AT AMERICAN AIRLINES
4
One problem that faces major airlines, such as American Airlines, is the development of
a schedule for airline crews (pilots and flight attendants) that results in a high level of
crew utilization. This scheduling problem is rather complex because of Federal Aviation
Administration (FAA) rules designed to ensure that a crew can perform its duties with-
out risk from fatigue. Union agreements require that flight crews receive pay for a con-
tractually set number of hours of each day or trip. The goal for airline planners is to con-
struct a crew schedule that meets or exceeds crew pay guarantees and does not violate
FAA rules. With the salary of airline captains at $140,000 per year or more, it is impor-
tant that an airline make the maximum possible usage of its crew personnel. Crew costs
are the second largest direct operating cost of an airline.
The nature of the constrained optimization problem facing an airline planner is to
minimize the cost of flying the published schedule, subject to the following constraints:
1. Each flight is assigned only one crew.
2
A linear relationship of the variables x1, x2,. . ., xn is a function of the form:
a1x1 ⫹ a2 x2 ⫹ . . . ⫹ an xn
where all the x variables have exponents of 1.
3
A quadratic function contains either squared terms (xi
2
) or cross-product terms (xixj).
4
Ira Gershkoff, “Optimizing Flight Crew Schedules,” Interfaces, July/August 1989, pp. 29–43.
Example
http://
Decision science modeling
at American Airlines is
described at
http://www.amrcorp.com/
corpcomm.htm

2. Each pairing of a crew and a flight must begin and end at a home “crew base,”
such as Chicago or Dallas for American.
3. Each pairing must be consistent with union work rules and FAA rules.
4. The number of jobs at each crew base must be within targeted minimum and
maximum limits as specified in American’s personnel plan.
American was able to develop a sophisticated constrained optimization model that saved
$18 million per year compared with previous crew allocation models used.
DIFFERENTIAL CALCULUS
In Chapter 2, marginal analysis was introduced as one of the fundamental concepts of
economic decision making. In the marginal analysis framework, resource-allocation de-
cisions are made by comparing the marginal benefits of a change in the level of an ac-
tivity with the marginal costs of the change. A change should be made as long as the mar-
ginal benefits exceed the marginal costs. By following this basic rule, resources can be
allocated efficiently and profits or shareholder wealth can be maximized.
In the profit-maximization example developed in Chapter 2, the application of the
marginal analysis principles required that the relationship between the objective (profit)
and the decision variable (output level) be expressed in either tabular or graphic form.
This framework, however, can become cumbersome when dealing with several decision
variables or with complex relationships between the decision variables and the objec-
tive. When the relationship between the decision variables and criterion can be ex-
pressed in algebraic form, the more powerful concepts of differential calculus can be
used to find optimal solutions to these problems.
Relationship between Marginal Analysis and Differential Calculus
Initially, let us assume that the objective we are seeking to optimize, Y, can be expressed
algebraically as a function of one decision variable, X,
Y ⫽ f(X) [A.3]
Recall that marginal profit is defined as the change in profit resulting from a one-unit
change in output. In general, the marginal value of any variable Y, which is a function of
another variable X, is defined as the change in the value of Y resulting from a one-unit
change in X. The marginal value of Y, My, can be calculated from the change in Y, ⌬Y,
that occurs as the result of a given change in X, ⌬X:
[A.4]
When calculated with this expression, different estimates for the marginal value of Y
may be obtained, depending on the size of the change in X that we use in the computa-
tion. The true marginal value of a function (e.g., an economic relationship) is obtained
from Equation A.4 when ⌬X is made as small as possible. If ⌬X can be thought of as a con-
tinuous (rather than a discrete) variable that can take on fractional values,5
then in calcu-
lating My by Equation A.4, we can let ⌬X approach zero. In concept, this is the approach
My ⫽
⌬Y
⌬X
5
For example, if X is a continuous variable measured in feet, pounds, and so on, then ⌬X can in theory take
on fractional values such as 0.5, 0.10, 0.05, 0.001, 0.0001 feet or pounds. When X is a continuous variable,
⌬X can be made as small as desired.

taken in differential calculus. The derivative, or more precisely, first derivative,6
dY/dX, of
a function is defined as the limit of the ratio ⌬Y/⌬X as ⌬X approaches zero; that is,
[A.5]
⌬X→0
Graphically, the first derivative of a function represents the slope of the curve at a
given point on the curve. The definition of a derivative as the limit of the change in
Y (that is, ⌬Y) as ⌬X approaches zero is illustrated in Figure A.1(a). Suppose we are
interested in the derivative of the Y ⫽ f(X) function at the point X0. The derivative
dY/dX measures the slope of the tangent line ECD. An estimate of this slope, albeit a
dY
dX
⫽ limit
⌬Y
⌬X
6
It is also possible to compute second, third, fourth, and so on, derivatives. Second derivatives are discussed
later in this chapter.
Y0
Y1
Y2
Y
X0 X1 X2 X
C
E
D
B
A
Y = f(X)
Y0
Y
X0 X
C
E
D
Y = f(X)
∆X
∆Y
(a) Marginal change in Y = f(X) as ∆X approaches 0
(b) Measurement of the slope Y = f(X) at point C
F I G U R E A.1
First Derivative of a
Function
Derivative
Measures the marginal effect
of a change in one variable
on the value of a function.
Graphically, it represents the
slope of the function at a
given point.

poor estimate, can be obtained by calculating the marginal value of Y over the inter-
val X0 to X2. Using Equation A.4, a value of
is obtained for the slope of the CA line. Now let us calculate the marginal value of Y using
a smaller interval, for example, X0 to X1. The slope of the line C to B, which is equal to
gives a much better estimate of the true marginal value as represented by the slope of the
ECD tangent line. Thus we see that the smaller the ⌬X value, the better the estimate of the
slope of the curve. Letting ⌬X approach zero allows us to find the slope of the Y ⫽ f(X)
curve at point C. As shown in Figure A.1(b), the slope of the ECD tangent line (and the
Y ⫽ f(X) function at point C) is measured by the change in Y, or rise, ⌬Y, divided by the
change in X, or run, ⌬X.
Process of Differentiation
The process of differentiation—that is, finding the derivative of a function—involves de-
termining the limiting value of the ratio ⌬Y/⌬X as ⌬X approaches zero. Before offering
some general rules for finding the derivative of a function, we illustrate with an exam-
ple the algebraic process used to obtain the derivative without the aid of these general
rules. The specific rules that simplify this process are presented in the following section.
PROCESS OF DIFFERENTIATION: PROFIT MAXIMIZATION
AT ILLINOIS POWER
Suppose the profit, ␲, of Illinois Power can be represented as a function of the output
level Q using the expression
␲ ⫽ ⫺40 ⫹ 140Q ⫺ 10Q2
[A.6]
We wish to determine d␲/dQ by first finding the marginal-profit expression ⌬␲/⌬Q
and then taking the limit of this expression as ⌬Q approaches zero. Let us begin by ex-
pressing the new level of profit (␲ ⫹ ⌬␲) that will result from an increase in output to
(Q ⫹ ⌬Q). From Equation A.6, we know that
(␲ ⫹ ⌬␲) ⫽ ⫺40 ⫹ 140(Q ⫹ ⌬Q) ⫺ 10(Q ⫹ ⌬Q)2
[A.7]
Expanding this expression and then doing some algebraic simplifying, we obtain
(␲ ⫹ ⌬␲) ⫽ ⫺40 ⫹ 140Q ⫹ 140⌬Q ⫺ 10[Q2
⫹ 2Q⌬Q ⫹ (⌬Q)2
]
⫽ ⫺40 ⫹ 140Q ⫺ 10Q2
⫹ 140⌬Q ⫺ 20Q⌬Q ⫺ 10(⌬Q)2
[A.8]
Subtracting Equation A.6 from Equation A.8 yields
⌬␲ ⫽ 140⌬Q ⫺ 20Q⌬Q ⫺ 10(⌬Q)2
[A.9]
Forming the marginal-profit ratio ⌬␲/⌬Q, and doing some canceling, we get
⫽140 ⫺ 20Q ⫺ 10⌬Q
[A.10]
⌬␲
⌬Q
⫽
140⌬Q ⫺ 20Q⌬Q ⫺ 10冠⌬Q冡2
⌬Q
M⬙
y ⫽
⌬Y
⌬X
⫽
Y1 ⫺ Y0
X1 ⫺ X0
M⬘
y ⫽
⌬Y
⌬X
⫽
Y2 ⫺ Y0
X2 ⫺ X0
Example

Taking the limit of Equation A.10 as ⌬Q approaches zero yields the expression for the
derivative of Illinois Power’s profit function (Equation A.6)
⌬Q→0
⫽ 140 ⫺ 20Q
[A.11]
If we are interested in the derivative of the profit function at a particular value of Q,
Equation A.11 can be evaluated for this value. For example, suppose we want to know
the marginal profit, or slope of the profit function, at Q ⫽ 3 units. Substituting Q ⫽ 3
in Equation A.11 yields
Rules of Differentiation
Fortunately, we do not need to go through this lengthy process every time we want the
derivative of a function. A series of general rules, derived in a manner similar to the
process just described, exists for differentiating various types of functions.7
Constant Functions A constant function can be expressed as
Y ⫽ a [A.12]
where a is a constant (that is, Y is independent of X). The derivative of a constant func-
tion is equal to zero:
[A.13]
For example, consider the constant function
Y ⫽ 4
which is graphed in Figure A.2(a). Recall that the first derivative of a function (dY/dX)
measures the slope of the function. Because this constant function is a horizontal
straight line with zero slope, its derivative (dY/dX) is therefore equal to zero.
Power Functions A power function takes the form of
Y ⫽ aXb
[A.14]
where a and b are constants. The derivative of a power function is equal to b times a,
times X raised to the (b ⫺ 1) power:
[A.15]
A couple of examples are used to illustrate the application of this rule. First, consider
the function
Y ⫽ 2X
dY
dX
⫽ b · a · Xb⫺1
dY
dX
⫽ 0
Marginal
profit
⫽
d␲
dQ
⫽ 140 ⫺ 20冠3冡 ⫽ $80 per unit
d␲
dQ
⫽ limit 关140 ⫺ 20Q ⫺ 10⌬Q兴
http://
You can connect to Illinois
Power Company, a subsidiary
of Illinova, at
http://www.illinova.com/
Clicking on the “inside
Illinova” button will give
you access to financial infor-
mation.
7
A more expanded treatment of these rules can be found in any introductory calculus book such as André L.
Yandl, Applied Calculus (Belmont, Calif.: Wadsworth, 1991).

which is graphed in Figure A.2(b). Note that the slope of this function is equal to 2 and
is constant over the entire range of X values. Applying the power function rule to this
example, where a ⫽ 2 and b ⫽ 1, yields
⫽ 2
Note that any variable to the zero power, e.g., X0
, is equal to 1.
Next, consider the function
Y ⫽ X2
which is graphed in Figure A.2(c). Note that the slope of this function varies depending on
the value of X. Application of the power function rule to this example yields (a ⫽ 1, b ⫽ 2):
⫽ 2X
As we can see, this derivative (or slope) function is negative when X ⬍ 0, zero when
X ⫽ 0, and positive when X ⬎ 0.
Sums of Functions Suppose a function Y ⫽ f(X) represents the sum of two (or more)
separate functions, f1(X), f2(X), that is,
Y ⫽ f1(X) ⫹ f2(X) [A.16]
The derivative of Y with respect to X is found by differentiating each of the separate
functions and then adding the results:
[A.17]
This result can be extended to finding the derivative of the sum of any number of
functions.
dY
dX
⫽
df1 冠X冡
dX
⫹
df2 冠X冡
dX
dY
dX
⫽ 2 · 1 · X2⫺1
dY
dX
⫽ 1 · 2 · X1⫺1
⫽ 2X0
2
Y
X
(a) Constant function
0 2
Y = 4
4
6
8
10
4 6 8
2
Y
X
(b) Linear function
0 2
Y = 2X
4
6
8
10
4 6 8
∆Y = 2
∆X = 1
2
Y
X
(c) Quadratic function
–2
Y = X2
4
6
8
10
0 2 4
–4
F I G U R E A.2 Constant, Linear, and Quadratic Functions

RULES OF DIFFERENTIATION: PROFIT MAXIMIZATION
AT ILLINOIS POWER (CONTINUED)
As an example of the application of these rules, consider again the profit function for
Illinois Power, given by Equation A.6, that was discussed earlier:
␲ ⫽ ⫺40 ⫹ 140Q ⫺ 10Q2
In this example Q represents the X variable and ␲ represents the Y variable; that is, ␲
⫽ f(Q). The function f(Q) is the sum of three separate functions—a constant function,
f1(Q) ⫽ ⫺40, and two power functions, f2(Q) ⫽ 140Q and f3(Q) ⫽ ⫺10Q2
. Therefore,
applying the differentiation rules yields
⫽ 0 ⫹ 1 ⋅ 140 ⋅ Q1 ⫺ 1
⫹ 2 ⋅ (⫺10) ⋅ Q2 ⫺ 1
⫽ 140 ⫺ 20Q
This is the same result that was obtained earlier in Equation A.11 by the differentia-
tion process.
Product of Two Functions Suppose the variable Y is equal to the product of two sepa-
rate functions f1(X) and f2(X):
Y ⫽ f1(X) ⋅ f2(X) [A.18]
In this case the derivative of Y with respect to X is equal to the sum of the first function
times the derivative of the second, plus the second function times the derivative of the first.
[A.19]
For example, suppose we are interested in the derivative of the expression
Y ⫽ X2
(2X ⫺ 3)
Let f1(X) ⫽ X2
and f2(X) ⫽ (2X ⫺ 3). By the above rule (and the earlier rules for dif-
ferentiating constant and power functions), we obtain
⫽ X2
⋅ (2 ⫺ 0) ⫹ (2X ⫺ 3) ⋅ (2X)
⫽ 2X2
⫹ 4X2
⫺ 6X
⫽ 6X2
⫺ 6X
⫽ 6X(X ⫺ 1)
Quotient of Two Functions Suppose the variable Y is equal to the quotient of two sep-
arate functions f1(X) and f2(X):
[A.20]
For such a relationship the derivative of Y with respect to X is obtained as follows:
[A.21]
dY
dX
⫽
f2 冠X冡 ·
df1 冠X冡
dX
⫺ f1 冠X冡 ·
df2 冠X冡
dX
关f2 冠X冡兴2
Y ⫽
f1 冠X冡
f2 冠X冡
dY
dX
⫽ X2
·
d
dX
关冠2X ⫺ 3冡兴 ⫹ 冠2X ⫺ 3冡 ·
d
dX
关X2
兴
dY
dX
⫽ f1 冠X冡 ·
df2 冠X冡
dX
⫹ f2 冠X冡 ·
df1 冠X冡
dX
d␲
dQ
⫽
df1 冠Q冡
dQ
⫹
df2 冠Q冡
dQ
⫹
df3 冠Q冡
dQ
Example

As an example, consider the problem of finding the derivative of the expression
Letting f1(X) ⫽ 10X2
and f2(X) ⫽ 5X ⫺ 1, we have
Functions of a Function (Chain Rule) Suppose Y is a function of the variable Z, Y ⫽
f1(Z); and Z is in turn a function of the variable X, Z ⫽ f2(X). The derivative of Y with
respect to X can be determined by first finding dY/dZ and dZ/dX and then multiplying
the two expressions together:
[A.22]
To illustrate the application of this rule, suppose we are interested in finding the de-
rivative (with respect to X) of the function
Y ⫽ 10Z ⫺ 2Z2
⫺ 3
where Z is related to X in the following way:8
Z ⫽ 2X2
⫺ 1
First, we find (by the earlier differentiation rules)
and then
dY
dX
⫽ 冠10 ⫺ 4Z冡 · 4X
dZ
dX
⫽ 4X
dY
dZ
⫽ 10 ⫺ 4Z
⫽
df1 冠Z冡
dZ
·
df2 冠X冡
dX
dY
dX
⫽
dY
dZ
·
dZ
dX
⫽
10X冠5X ⫺ 2冡
冠5X ⫺ 1冡2
⫽
50X2
⫺ 20X
冠5X ⫺ 1冡2
⫽
100X2
⫺ 20X ⫺ 50X2
冠5X ⫺ 1冡2
dY
dX
⫽
冠5X ⫺ 1冡 · 20X ⫺ 10X2
· 5
冠5X ⫺ 1冡2
Y ⫽
10X2
5X ⫺ 1
8
Alternatively, one can substitute Z ⫽ 2X2
⫺ 1 into Y ⫽ 10Z ⫺ 2Z2
⫺ 3 and differentiate Y with respect to
X. The reader is asked to demonstrate in Exercise 24 that this approach yields the same answer as the
chain rule.

Substituting the expression for Z in terms of X into this equation yields
⫽ (10 ⫺ 8X2
⫹ 4) ⋅ 4X
⫽ 40X ⫺ 32X3
⫹ 16X
⫽ 56X ⫺ 32X3
⫽ 8X(7 ⫺ 4X2
)
These rules for differentiating functions are summarized in Table A.1.
APPLICATIONS OF DIFFERENTIAL CALCULUS
TO OPTIMIZATION PROBLEMS
The reason for studying the process of differentiation and the rules for differentiating
functions is that these methods can be used to find optimal solutions to many kinds of
maximization and minimization problems in managerial economics.
Maximization Problem
Recall from the discussion of marginal analysis, a necessary (but not sufficient) condition
for finding the maximum point on a curve (for example, maximum profits) is that the
marginal value or slope of the curve at this point must be equal to zero. We can now ex-
press this condition within the framework of differential calculus. Because the derivative
of a function measures the slope or marginal value at any given point, an equivalent nec-
essary condition for finding the maximum value of a function Y ⫽ f(X) is that the deriv-
ative dY/dX at this point must be equal to zero. This is known as the first-order condition
for locating one or more maximum or minimum points of an algebraic function.
dY
dX
⫽ 关10 ⫺ 4冠2X2
⫺ 1冡兴 · 4X
T A B L E A.1
Summary of Rules for
Differentiating
Functions
Function Derivative
1. Constant Function
Y ⫽ a
2. Power Function
Y ⫽ aXb
3. Sums of Functions
Y ⫽ f1(X) ⫹ f2(X)
4. Product of Two Functions
Y ⫽ f1(X) ⋅ f2(X)
5. Quotient of Two Functions
6. Functions of a Function
Y ⫽ f1(Z), where Z ⫽ f2(X)
dY
dX
⫽
dY
dZ
·
dZ
dX
dY
dX
⫽
f2 冠X冡 ·
df1 冠X冡
dX
⫺ f1 冠X冡 ·
df2 冠X冡
dX
关f2 冠X冡兴2
Y ⫽
f1 冠X冡
f2 冠X冡
dY
dX
⫽ f1 冠X冡 ·
df2 冠X冡
dX
⫹ f2 冠X冡 ·
df1 冠X冡
dX
dY
dX
⫽
df1 冠X冡
dX
⫹
df2 冠X冡
dX
dY
dX
⫽ b · a · Xb⫺1
dY
dX
⫽ 0
First-Order Condition
A test to locate one or more
maximum or minimum
points of an algebraic
function.

FIRST-ORDER CONDITION: PROFIT MAXIMIZATION
Using the profit function (Equation A.6)
␲ ⫽ ⫺40 ⫹ 140Q ⫺ 10Q2
discussed earlier, we can illustrate how to find the profit-maximizing output level Q by
means of this condition. Setting the first derivative of this function (which was com-
puted previously) to zero, we obtain
0 ⫽ 140 ⫺ 20Q
Solving this equation for Q yields Q* ⫽ 7 units as the profit-maximizing output level.
The profit and first derivative functions and optimal solution are shown in Figure A.3.
As we can see, profits are maximized at the point where the function is neither increas-
ing nor decreasing; in other words, where the slope (or first derivative) is equal to zero.
d␲
dQ
⫽ 140 ⫺ 20Q
Example
dπ
dQ
= 140 – 20Q
= 0
= Marginal profit at Q = 7
dπ
dQ
($/unit)
dπ
dQ
= 140 – 20Q
π($)
1
Q units
100
π = –40 + 140Q – 10Q2
Total profit (π)
200
300
400
500
0
–10
2 3 4 5 6 7 8 9 10 11 12 13 14
1
50
100
150
200
0
–50
2 3 4 5 6 7 9 10 11 12 13 14
–100
–150
–200
Q units
F I G U R E A.3
Profit and First
Derivative Functions

Second Derivatives and the Second-Order Condition
Setting the derivative of a function equal to zero and solving the resulting equation for
the value of the decision variable does not guarantee that the point will be obtained at
which the function takes on its maximum value. (Recall the Stealth bomber example at
the start of the chapter.) The slope of a U-shaped function will also be equal to zero at
its low point and the function will take on its minimum value at the given point. In other
words, setting the derivative to zero is only a necessary condition for finding the maxi-
mum value of a function; it is not a sufficient condition. Another condition, known as
the second-order condition, is required to determine whether a point that has been de-
termined from the first-order condition is either a maximum point or minimum point
of the algebraic function.
This situation is illustrated in Figure A.4. At both points A and B the slope of the func-
tion (first derivative, dY/dX) is zero; however, only at point B does the function take on
its maximum value. We note in Figure A.4 that the marginal value (slope) is continually
decreasing in the neighborhood of the maximum value (point B) of the Y ⫽ f(X) func-
tion. First the slope is positive up to the point where dY/dX ⫽ 0, and thereafter the slope
becomes negative. Thus we must determine whether the slope’s marginal value (slope of
Second-Order Condition
A test to determine
whether a point that has
been determined from the
first-order condition is
either a maximum point or
a minimum point of the
algebraic function.
dY
dX
= 0
dY
dX
dY
dX
dY
dX
= 0
A
X
Y = f(X)
B
Y
X
F I G U R E A.4
Maximum and
Minimum Values of
a Function

the slope) is declining. A test to see whether the marginal value is decreasing is to take
the derivative of the marginal value and check to see if it is negative at the given point
on the function. In effect, we need to find the derivative of the derivative—that is, the
second derivative of the function—and then test to see if it is less than zero. Formally, the
second derivative of the function Y ⫽ f(X) is written as d2
Y/dX2
and is found by apply-
ing the previously described differentiation rules to the first derivative. A maximum point
is obtained if the second derivative is negative; that is, d2
Y/dX2
⬍ 0.
SECOND-ORDER CONDITION: PROFIT MAXIMIZATION
Returning to the profit-maximization example, the second derivative is obtained from
the first derivative as follows:
⫽ ⫺20
Because d2
␲/dQ2
⬍ 0, we know that a maximum-profit point has been obtained.
An opposite condition holds for obtaining the point at which the function takes on a
minimum value. Note again in Figure A.4 that the marginal value (slope) is continually
increasing in the neighborhood of the minimum value (point A) of the Y ⫽ (X) function.
First the slope is negative up to the point where dY/dX ⫽ 0, and thereafter the slope be-
comes positive. Therefore, we test to see if d2
Y/dX2
⬎ 0 at the given point. A minimum
point is obtained if the second derivative is positive; that is, d2
Y/dX2
⬎ 0.
Minimization Problem
In some decision-making situations, cost minimization may be the objective. As in
profit-maximization problems, differential calculus can be used to locate the optimal
points.
COST MINIMIZATION: KEYSPAN ENERGY
Suppose we are interested in determining the output level that minimizes average total
costs for KeySpan Energy, where the average total cost function might be approximated
by the following relationship (Q represents output):
C ⫽ 15 ⫺ .040Q ⫹ .000080Q2
[A.23]
Differentiating C with respect to Q gives
Setting this derivative equal to zero and solving for Q yields
0 ⫽ ⫺.040 ⫹ .000160Q
Q* ⫽ 250
dC
dQ
⫽ ⫺ .040 ⫹ .000160Q
d2
␲
dQ2 ⫽ 0 ⫹ 1 · 冠 ⫺ 20冡 · Q1⫺1
d␲
dQ
⫽ 140 ⫺ 20Q
Example
Example
http://
KeySpan Energy is a natural
gas distribution company.
You can access financial
information on KeySpan
Energy at their Internet site:
http://www.keyspanenergy.com

Taking the second derivative, we obtain
Because the second derivative is positive, the output level of Q ⫽ 250 is indeed the
value that minimizes average total costs.
Summarizing, we see that two conditions are required for locating a maximum or
minimum value of a function using differential calculus. The first-order condition de-
termines the point(s) at which the first derivative dY/dX is equal to zero. Having ob-
tained one or more points, a second-order condition is used to determine whether the
function takes on a maximum or minimum value at the given point(s). The second de-
rivative d2
Y/dX2
indicates whether a given point is a maximum (d2
Y/dX2
⬍ 0) or a min-
imum (d2
Y/dX2
⬎ 0) value of the function.
PARTIAL DIFFERENTIATION AND MULTIVARIATE
OPTIMIZATION
Thus far in the chapter, the analysis has been limited to a criterion variable Y that can
be expressed as a function of one decision variable X. However, many commonly used
economic relationships contain two or more decision variables. For example, a produc-
tion function relates the output of a plant, firm, industry, or country to the inputs em-
ployed—such as capital, labor, and raw materials. Another example is a demand function,
which relates sales of a product or service to such variables as price, advertising, pro-
motion expenses, price of substitutes, and income.
Partial Derivatives
Consider a criterion variable Y that is a function of two decision variables X1 and X2:9
Y ⫽ f (X1,X2)
Let us now examine the change in Y that results from a given change in either X1
or X2. To isolate the marginal effect on Y from a given change in X1—that is,
⌬Y/⌬X1—we must hold X2 constant. Similarly, if we wish to isolate the marginal ef-
fect on Y from a given change in X2—that is, ⌬Y/⌬X2—the variable X1 must be held
constant. A measure of the marginal effect of a change in any one variable on the
change in Y, holding all other variables in the relationship constant, is obtained from
the partial derivative of the function. The partial derivative of Y with respect to X1 is
written as ∂Y/∂X1 and is found by applying the previously described differentiation
rules to the Y ⫽ f(X1,X2) function, where the variable X2 is treated as a constant. Sim-
ilarly, the partial derivative of Y with respect to X2 is written as ∂Y/∂X2 and is found
by applying the differentiation rules to the function, where the variable X1 is treated
as a constant.
d2
C
dQ2 ⫽ ⫹ .000160
9
The following analysis is not limited to two decision variables. Relationships containing any number of
variables can be analyzed within this framework.
Partial Derivative
Measures the marginal effect
of a change in one variable
on the value of a multivari-
ate function, while holding
constant all other variables.

PARTIAL DERIVATIVES: INDIANA PETROLEUM COMPANY
To illustrate the procedure for obtaining partial derivatives, let us consider the follow-
ing relationship in which the profit variable, ␲, is a function of the output level of two
products (heating oil and gasoline) Q1 and Q2:
[A.24]
Treating Q2 as a constant, the partial derivative of ␲ with respect to Q1 is obtained:
⫽ 140 ⫺ 20Q1 ⫺ 6Q2
[A.25]
Similarly, with Q1 treated as a constant, the partial derivative of ␲ with respect to Q2
is equal to
⫽ 100 ⫺ 16Q2 ⫺ 6Q1
[A.26]
PARTIAL DERIVATIVES: DEMAND FUNCTION FOR SHIELD TOOTHPASTE
As another example, consider the following (multiplicative) demand function, where Q
⫽ quantity sold, P ⫽ selling price, and A ⫽ advertising expenditures:
Q ⫽ 3.0P⫺.50
A.25
[A.27]
The partial derivative of Q with respect to P is
⫽ ⫺1.5P⫺1.50
A.25
Similarly, the partial derivative of Q with respect to A is
⫽ .75P⫺.50
A⫺.75
Maximization Problem
The partial derivatives can be used to obtain the optimal solution to a maximization or min-
imization problem containing two or more X variables. Analogous to the first-order condi-
tions discussed earlier for the one-variable case, we set each of the partial derivatives equal
to zero and solve the resulting set of simultaneous equations for the optimal X values.
PROFIT MAXIMIZATION: INDIANA PETROLEUM COMPANY (CONTINUED)
Suppose we are interested in determining the values of Q1 and Q2 that maximize the
company’s profits given in Equation A.24. In this case, each of the two partial derivative
functions (Equations A.25 and A.26) would be set equal to zero:
0 ⫽ 140 ⫺ 20Q1 ⫺ 6Q2 [A.28]
0 ⫽ 100 ⫺ 16Q2 ⫺ 6Q1 [A.29]
∂Q
∂A
⫽ 3.0P⫺.50
冠.25A.25⫺1
冡
∂Q
∂P
⫽ 3.0A.25
冠 ⫺ .50P⫺.50⫺1
冡
∂␲
∂Q2
⫽ 0 ⫹ 0 ⫹ 100 ⫺ 0 ⫹ 2 · 冠⫺8冡 · Q2 ⫺ 6Q1
∂␲
∂Q1
⫽ 0 ⫹ 140 ⫹ 0 ⫹ 2 · 冠 ⫺ 10冡 · Q1 ⫺ 0 ⫺ 6Q2
␲ ⫽ ⫺ 60 ⫹ 140Q1 ⫹ 100Q2 ⫺ 10Q2
1 ⫺ 8Q2
2 ⫺ 6Q1Q2
Example
Example
Example

This system of equations can be solved for the profit-maximizing values of Q1 and
Q2.10
The optimal values are ⫽ 5.77 units and ⫽ 4.08 units.11
The optimal
total profit is equal to
␲* ⫽ ⫺60 ⫹ 140(5.77) ⫹ 100(4.08) ⫺ 10(5.77)2
⫺ 8(4.08)2
⫺ 6(5.77)(4.08) ⫽ 548.45
DEALING WITH IMPORT RESTRAINTS: TOYOTA
During the 1992 U.S. presidential campaign, there was extensive rhetoric about the “prob-
lem” of the U.S. balance of trade deficit with Japan, particularly about the level of Japan-
ese auto imports into the United States. Some of the proposals that have been advanced
to reduce the magnitude of this “problem,” and thereby assist the U.S. auto industry, in-
clude the imposition of rigid car import quotas. Japanese manufacturers would be forced
to restrict the number of cars that are exported to the United States.
Had rigid import quotas been imposed, Japanese manufacturers would have had to take
this constraint into consideration when making production, distribution, and new car in-
troduction plans. Japanese manufacturers could no longer just seek to maximize profits.
Profits could only be maximized subject to an aggregate constrained level of exports that
could be sent to the U.S. market.
Shortly after, Japanese manufacturers responded to import constraints by raising
prices, thereby making Japanese cars less price competitive with U.S.-built autos. In-
creases in prices by Japanese firms would give U.S. manufacturers additional room for
raising their own prices. Indeed, in early 1992 in the face of threatened import quotas,
Toyota announced significant price increases on the cars it sold in the United States. Over
the longer term, Japanese firms have shifted their product mix to more profitable, larger
(luxury) cars, abandoning some of their market share in smaller, less profitable vehicles
to U.S. manufacturers.
As can be seen in this example, the imposition of additional constraints on the operat-
ing activities of a firm can have a substantial impact on its short- and long-term pricing
and output strategies.
• Within the area of decision-making under certainty are two broad classes of
problems—unconstrained optimization problems and constrained optimization
problems.
• Marginal analysis is useful in making decisions about the expansion or contraction
of an economic activity.
• Differential calculus, which bears a close relationship to marginal analysis, can be
applied whenever an algebraic relationship can be specified between the decision
variables and the objective or criterion variable.
• The first-derivative measures the slope or rate of change of a function at a given
point and is equal to the limiting value of the marginal function as the marginal
value is calculated over smaller and smaller intervals, that is, as the interval
approaches zero.
Q *
2
Q *
1
10
The second-order conditions for obtaining a maximum or minimum in the multiple-variable case are
somewhat complex. A discussion of these conditions can be found in most basic calculus texts.
11
Exercise 25 at the end of the chapter requires the determination of these optimal values.
International
Perspectives
http://
You can read more about
trade tariffs and quotas at the
Internet site maintained by
the National Center for Policy
Analysis:
http://www.public-policy.
org/⬃ncpa/studies/s171/
s171.html
SUMMARY

• Various rules are available (see Table A.1) for finding the derivative of specific types
of functions.
• A necessary, but not sufficient, condition for finding the maximum or minimum
points of a function is that the first derivative be equal to zero. This is known as the
first-order condition.
• A second-order condition is required to determine whether a given point is a
maximum or minimum. The second derivative indicates that a given point is a
maximum if the second derivative is less than zero or a minimum if the second
derivative is greater than zero.
• The partial derivative of a multivariate function measures the marginal effect of a change
in one variable on the value of the function, holding constant all other variables.
• In constrained optimization problems, Lagrangian multiplier techniques can be used
to find the optimal value of a function that is subject to equality constraints.
Through the introduction of additional (artificial) variables into the problem, the
Lagrangian multiplier method converts the constrained problem into an
unconstrained problem, which can then be solved using ordinary differential
calculus procedures. Lagrangian multiplier techniques are discussed and illustrated
in the Appendix to this chapter.
1. Explain how the first and second derivatives of a function are used to find the
maximum or minimum points of a function Y ⫽ f(X). Illustrate your discussion
with graphs.
2. Why is the first-order condition for finding a maximum (or minimum) of a func-
tion referred to as a necessary, but not sufficient, condition?
3. Defining Q to be the level of output produced and sold, suppose that the firm’s
total revenue (TR) and total cost (TC) functions can be represented in tabular
form as shown below.
EXERCISES
Total Total Total Total
Output Revenue Cost Output Revenue Cost
Q TR TC Q TR TC
0 0 20 11 264 196
1 34 26 12 276 224
2 66 34 13 286 254
3 96 44 14 294 286
4 124 56 15 300 320
5 150 70 16 304 356
6 174 86 17 306 394
7 196 104 18 306 434
8 216 124 19 304 476
9 234 146 20 300 520
10 250 170
a. Compute the marginal revenue and average revenue functions.
b. Compute the marginal cost and average cost functions.

c. On a single graph, plot the total revenue, total cost, marginal revenue, and
marginal cost functions.
d. Determine the output level in the graph that maximizes profits (that is, prof-
it ⫽ total revenue ⫺ total cost) by finding the point where marginal revenue
equals marginal cost.
e. Check your result in part (d) by finding the output level in the tables devel-
oped in parts (a) and (b) that likewise satisfies the condition that marginal
revenue equals marginal cost.
4. Consider again the total revenue and total cost functions shown in tabular form
in the previous problem.
a. Compute the total, marginal, and average profit functions.
b. On a single graph, plot the total profit and marginal profit functions.
c. Determine the output level in the graph and table where the total profit func-
tion takes on its maximum value.
d. How does the result in part (c) in this exercise compare with the result in part
(d) of the previous exercise?
e. Determine total profits at the profit-maximizing output level.
5. Differentiate the following functions:
a. TC ⫽ 50 ⫹ 100Q ⫺ 6Q2
⫹ .5Q3
b. ATC ⫽ 50/Q ⫹ 100 ⫺ 6Q ⫹ .5Q2
c. MC ⫽ 100 ⫺ 12Q ⫹ 1.5Q2
d. Q ⫽ 50 ⫺ .75P
e. Q ⫽ .40X1.50
a. Y ⫽ 2X3
/(4X2
⫺ 1)
b. Y ⫽ 2X3
(4X2
⫺ 1)
c. Y ⫽ 8Z2
⫺ 4Z ⫹ 1, where Z ⫽ 2X2
⫺ 1 (differentiate Y with respect to X)
7. Defining Q to be the level of output produced and sold, assume that the firm’s
cost function is given by the relationship
TC ⫽ 20 ⫹ 5Q ⫹ Q2
Furthermore, assume that the demand for the output of the firm is a function of
price P given by the relationship
Q ⫽ 25 ⫺ P
a. Defining total profit as the difference between total revenue and total cost,
express in terms of Q the total profit function for the firm. (Note: Total rev-
enue equals price per unit times the number of units sold.)
b. Determine the output level where total profits are maximized.
c. Calculate total profits and selling price at the profit-maximizing output level.
d. If fixed costs increase from $20 to $25 in the total cost relationship, deter-
mine the effects of such an increase on the profit-maximizing output level
and total profits.
8. Using the cost and demand functions in Exercise 7:
a. Determine the marginal revenue and marginal cost functions.
b. Show that, at the profit-maximizing output level determined in part (b) of the
previous exercise, marginal revenue equals marginal cost. This illustrates the
economic principle that profits are maximized at the output level where mar-
ginal revenue equals marginal cost.

9. Using the cost and demand functions in Exercise 7, suppose the government
imposes a 20 percent tax on the net profits (that is, a tax on the difference between
revenues and costs) of the firm.
a. Determine the new profit function for the firm.
b. Determine the output level at which total profits are maximized.
c. Calculate total profits (after taxes) and the selling price at the profit-
maximizing output level.
d. Compare the results in parts (b) and (c) with the results in Exercise 7 above.
10. Suppose the government imposes a 20 percent sales tax (that is, a tax on revenue)
on the output of the firm. Answer questions (a), (b), (c), and (d) of the previous
exercise according to this new condition.
11. The Bowden Corporation’s average variable cost function is given by the following
relationship (where Q is the number of units produced and sold):
AVC ⫽ 25,000 ⫺ 180Q ⫹ .50Q2
a. Determine the output level (Q) that minimizes average variable cost.
b. How does one know that the value of Q determined in part (a) minimizes
rather than maximizes AVC?
12. Determine the partial derivatives with respect to all of the variables in the fol-
lowing functions:
a. TC ⫽ 50 ⫹ 5Q1 ⫹ 10 Q2 ⫹ .5Q1Q2
b. Q ⫽ 1.5L.60
K.50
c. QA ⫽ 2.5PA
⫺1.30
Y.20
PB
·40
13. Bounds Inc. has determined through regression analysis that its sales (S) are a
function of the amount of advertising (measured in units) in two different media.
This is given by the following relationship (X ⫽ newspapers, Y ⫽ magazines):
S(X,Y) ⫽ 200X ⫹ 100Y ⫺ 10X2
⫺ 20Y2
⫹ 20XY
a. Find the level of newspaper and magazine advertising that maximizes the
firm’s sales.
b. Calculate the firm’s sales at the optimal values of newspaper and magazine
advertising determined in part (a).
14. The Santa Fe Cookie Factory is considering an expansion of its retail piñon cookie
business to other cities. The firm’s owners lack the funds needed to undertake the
expansion on their own. They are considering a franchise arrangement for the new
outlets. The company incurs variable costs of $6 for each pound of cookies sold.
The fixed costs of operating a typical retail outlet are estimated to be $300,000 per
year. The demand function facing each retail outlet is estimated to be
P ⫽ $50 ⫺ .001Q
where P is the price per pound of cookies and Q is the number of pounds of cook-
ies sold. [Note: Total revenue equals price (P) times quantity (Q) sold.]
a. What price, output, total revenue, total cost, and total profit level will each
profit-maximizing franchise experience?
b. Assuming that the parent company charges each franchisee a fee equal to 5
percent of total revenues, recompute the values in part (a) above.
c. The Santa Fe Cookie Factory is considering a combined fixed/variable fran-
chise fee structure. Under this arrangement each franchisee would pay the
parent company $25,000 plus 1 percent of total revenues. Recompute the val-
ues in part (a) above.

d. What franchise fee arrangement do you recommend that the Santa Fe Cookie
Factory adopt? What are the advantages and disadvantages of each plan?
15. Several fast-food chains, including McDonald’s, announced that they have shifted
from the use of beef tallow, which is high in saturated fat, to unsaturated veg-
etable fat. This change was made in response to increasing consumer awareness
of the relationship between food and health.
a. Why do you believe McDonald’s used beef tallow prior to the change?
b. What impact would you expect the change to have on McDonald’s profits (i)
in the near term, (ii) in the long term?
c. Structure this situation as a constrained-optimization problem. What is the
objective function? What are the constraints?
a. TR ⫽ 50Q ⫺ 4Q2
b. VC ⫽ 75Q ⫺ 5Q2
⫹ .25Q3
c. MC ⫽ 75 ⫺ 10Q ⫹ .75Q2
d. Q ⫽ 50 ⫺ 4P
e. Q ⫽ 2.0 L.75
17. Determine the marginal cost function by differentiating the following total cost
function with respect to Q (output):
TC ⫽ a ⫹ bQ ⫹ cQ2
⫹ dQ3
where a, b, c, and d, are constants.
a.
b.
c. Y ⫽ 2Z2
⫹ 2Z ⫹ 3, where Z ⫽ X2
⫺ 2 (differentiate Y with respect to X)
19. Determine the partial derivatives with respect to all the variables in the following
functions:
a. Y ⫽ ␣ ⫹ ␤1X1 ⫹ ␤2X2 ⫹ ␤3X3, where ␣, ␤1, ␤2, and ␤3 are constants
b. , where ␣, ␤1, and ␤2 are constants
c. QA ⫽ , where a, b, c, and d are constants
20. Given the following total revenue function (where Q ⫽ output):
TR ⫽ 100 Q ⫺ 2Q2
a. Determine the level of output that maximizes revenues.
b. Show that the value of Q determined in part (a) maximizes, rather then min-
imizes, revenues.
21. Given the following total profit function (where Q ⫽ output):
␲ ⫽ ⫺250,000 ⫹ 20,000 Q ⫺ 2Q2
a. Determine the level of output that maximizes profits.
b. Show that the value of Q determined in part (a) maximizes, rather then min-
imizes, profits.
22. Given the following average total cost function (where Q ⫽ output):
ATC ⫽ 5,000 ⫺ 100Q ⫹ 1.0Q2
a. Determine the level of output that minimizes average total costs.
b. Show that the value of Q determined in part (a) minimizes, rather then max-
imizes, average total costs.
aPb
A Yc
Pd
B
Q ⫽ ␣L␤1
K␤2
Y ⫽ 1
3
X3
冠1
2
X2
⫺ 1冡
Y ⫽ 1
3
X3
兾冠1
2
X2
⫺ 1冡

WEB CHAPTER A Optimization Techniques
23. Determine the value(s) of X that maximize or minimize the following functions
and indicate whether each value represents either a maximum or minimum point
of the function. (Hint: The roots of the quadratic equation: aX2
⫹ bX ⫹ c ⫽ 0 are
a.
b.
24. Show that substituting
Z ⫽ 2X2
⫺ 1
into
Y ⫽ 10Z ⫺ 2Z2
⫺ 3
and differentiating Y with respect to X yields the same result as application of the
chain rule [dY/dX ⫽ 8X(7 ⫺ 4X2
)].
25. Show that the optimal solution to the set of simultaneous equations in the Indiana
Petroleum example, Equations A.28 and A.29, are ⫽ 5.77 and ⫽ 4.08.
26. As suggested in the text, import restrictions such as tariffs and quotas serve as con-
straints on the profit-maximization objective of importers. While international free
trade agreements such as the North American Free Trade Agreement have gener-
ally reduced or eliminated tariffs and quotas, some still remain. One of the import
quotas still in force in the United States applies to textiles. You can access infor-
mation on textile quotas from several sites on the Internet, including the World
Trade Organization’s website at http://www.wto.org/english/tratop_e/tratop_e.htm
and the Cato Institute website at http://www. cato.org/pubs/pas/pa-140es.html and
write a two-paragraph executive summary of how this import quota modifies the
profit-maximizing choices of both domestic and foreign textile manufacturers.
Q*
2
Q*
1
Y ⫽ ⫺1
3
X3
⫹ 60X2
⫺ 2,000X ⫹ 50,000
Y ⫽ 1
3
X3
⫺ 60X2
⫹ 2,000X ⫹ 50,000
X ⫽
⫺ b⫾兹b2
⫺ 4ac
2a
.冡
23
http://
Import Quotas as an
Optimization Constraint

Constrained Optimization and
Lagrangian Multiplier Techniques
SIMPLE CONSTRAINED OPTIMIZATION
Web Chapter A discussed some of the techniques for solving unconstrained optimization
problems. In this appendix the Lagrangian multiplier technique is developed to deal with
some classes of constrained optimization problems. In Web Chapter B linear program-
ming, a more general technique for dealing with constrained optimization, is developed.
Most organizations have constraints on their decision variables. The most obvious
constraints, and the easiest to quantify and incorporate into the analysis, are the limita-
tions imposed by the quantities of resources (such as capital, personnel, facilities, and
raw materials) available to the organization. Other more subjective constraints include
legal, environmental, and behavioral limitations on the decisions of the organization.
When the constraints take the form of equality relationships, classical optimization
procedures can be used to solve the problem. One method, which can be employed
when the objective function is subject to only one constraint equation of a relatively sim-
ple form, is to solve the constraint equation for one of the decision variables and then
substitute this expression into the objective function. This procedure converts the orig-
inal problem into an unconstrained optimization problem, which can be solved using
the calculus procedures developed in this Web Chapter.
CONSTRAINED PROFIT MAXIMIZATION: INDIANA PETROLEUM COMPANY
Consider again the two-product profit-maximization problem (Equation A.24) from
Web Chapter A. Suppose that the raw material (crude oil) needed to make the products
is in short supply and that the firm has a contract with a supplier calling for the deliv-
ery of 200 units of the given raw material during the forthcoming period. No other
sources of the raw material are available. Also, assume that all the raw material must be
used during the period and that none can be carried over to the next period in inventory.
Furthermore, suppose that Product 1 requires 20 units of the raw material to produce
one unit of output and Product 2 requires 40 units of raw material to produce one unit
of output. The constrained optimization problem can be written as follows:
[AA.1]
subject to 20Q1 ⫹ 40Q2 ⫽ 200 [AA.2]
The raw material constraint line and the profit function (curve) that is tangent to the
constraint line are shown together in Figure AA.1. Note that the solution to the uncon-
strained problem obtained earlier—Q1 ⫽ 5.77 and Q2 ⫽ 4.08—is not a feasible solution
to the constrained problem because it requires 20(5.77) ⫹ 40(4.08) ⫽ 278.6 units of
raw material when in fact only 200 units are available. Following the procedure just de-
scribed, we solve the constraint for Q1:
Maximize ␲ ⫽ ⫺60 ⫹ 140Q1 ⫹ 100Q2 ⫺ 10Q2
1 ⫺ 8Q2
2 ⫺ 6Q1Q2
Web Chapter A
Appendix
Example

⫽ 10 ⫺ 2Q2
Substituting this expression for Q1 in the objective function, we obtain
␲ ⫽ ⫺60 ⫹ 140(10 ⫺ 2Q2) ⫹ 100Q2 ⫺ 10(10 ⫺ 2Q2)2
⫽ ⫺60 ⫹ 1400 ⫺ 280Q2 ⫹ 100Q2 ⫺ 1000 ⫹ 400Q2
Taking the derivative of this expression with respect to Q2 yields
Setting d␲/dQ2 equal to zero and solving for Q2, we obtain
0 ⫽ 160 ⫺ 72Q2
⫽ 2.22 units
Q*
2 ⫽
160
72
d␲
dQ2
⫽ 160 ⫺ 72Q2
⫽ 340 ⫹ 160Q2 ⫺ 36Q2
2
⫺40Q2
2 ⫺ 8Q2
2 ⫺ 60Q2 ⫹ 12Q2
2
⫺8Q2
2 ⫺ 6冠10 ⫺ 2Q2冡Q2
Q1 ⫽
200
20
⫺
40Q2
20
25
WEB CHAPTER A APPENDIX Constrained Optimization and Lagrangian MultiplierTechniques
1
2.22 Q2
Q2
(units)
Raw material constraint
20 Q1
+ 40 Q2
= 200
π = 548.45 (unconstrained optimum)
π = 517.78 (constrained optimum)
Profit function
– 60 + 140 Q1
+ 100 Q2
– 10 Q1
2
– 8 Q2
– 6 Q1
Q2
= 517.78
1
5.56 Q1
Q1
units
0
2
3
4
5
6
2 3 4 5 6 7 8 9 10
*
*
2
F I G U R E AA.1 Constrained Profit Maximization: Indiana Petroleum Company

26 WEB CHAPTER A APPENDIX Constrained Optimization and Lagrangian MultiplierTechniques
In turn solving for Q1, we obtain
⫽ 5.56 units
Thus ⫽ 5.56 and ⫽ 2.22 is the optimal solution to the constrained profit-
maximization problem.
Using the constraint to substitute for one of the variables in the objective function,
as in the preceding example, will yield an optimal solution only when there is one con-
straint equation and it is possible to solve this equation for one of the decision variables.
With more than one constraint equation and/or a complex constraint relationship, the
more powerful method of Lagrangian multipliers can be employed to solve the con-
strained optimization problem.
LAGRANGIAN MULTIPLIER TECHNIQUES
The Lagrangian multiplier technique creates an additional artificial variable for each
constraint. Using these artificial variables, the constraints are incorporated into the ob-
jective function in such a way as to leave the value of the function unchanged. This new
function, called the Lagrangian function, constitutes an unconstrained optimization
problem. The next step is to set the partial derivatives of the Lagrangian function for
each of the variables equal to zero and solve the resulting set of simultaneous equations
for the optimal values of the variables.
LAGRANGIAN MULTIPLIERS: INDIANA PETROLEUM COMPANY
(CONTINUED)
The Lagrangian multiplier method can be illustrated using the example discussed in the
previous section. First, the constraint equation, which is a function ␦ of the two vari-
ables Q1 and Q2, is rearranged to form an expression equal to zero:
␦(Q1, Q2) ⫽ 20Q1 ⫹ 40Q2 ⫺ 200 ⫽ 0
Next we define an artificial variable ␭ (lambda) and form the Lagrangian function.12
L␲ ⫽ ␲(Q1, Q2) ⫺ ␭␦(Q1, Q2)
⫺␭(20Q1 ⫹ 40Q2 ⫺ 200)
As long as ␦(Q1,Q2) is maintained equal to zero, the Lagrangian function L␲ will not
differ in value from the profit function ␲. Maximizing L␲ will also maximize ␲. L␲ is
seen to be a function of Q1, Q2, and ␭. Therefore, to maximize L␲ (and also ␲), we need
to partially differentiate L␲ with respect to each of the variables, set the partial deriva-
⫽ ⫺ 60 ⫹ 140Q1 ⫹ 100Q2 ⫺ 10Q2
1 ⫺ 8Q2
2 ⫺ 6Q1Q2
Q*
2
Q*
1
Q*
1 ⫽ 10 ⫺ 2冠2.22冡
Example
12
To assist in the interpretation of the results, it is often useful to adopt the arbitrary convention that in the
case of a maximization problem the lambda term should be subtracted in the Lagrangian function. In the case
of a minimization problem, the lambda term should be added in the Lagrangian function.

tives equal to zero, and solve the resulting set of equations for the optimal values of Q1,
Q2, and ␭. The partial derivatives are equal to
Setting the partial derivatives equal to zero yields the equations
20Q1 ⫹ 6Q2 ⫹ 20␭ ⫽ 140
6Q1 ⫹ 16Q2 ⫹ 40␭ ⫽ 100
20Q1 ⫹ 40Q2 ⫽ 200
After solving this set of simultaneous equations, we obtain ⫽ 5.56, ⫽ 2.22, and
␭* ⫽ ⫹.774. (Note: These are the same values of Q1 and Q2 that were obtained earlier
in this section by the substitution method.)
If a problem has two or more constraints, then a separate ␭ variable is defined for each
constraint and incorporated into the Lagrangian function. In general, ␭ measures the
marginal change in the value of the objective function resulting from a one-unit change
in the value on the righthand side of the equality sign in the constraint relationship. In
the example above, ␭* equals $.774 and indicates that profits could be increased by this
amount if one more unit of raw material was available; that is, an increase from 200 units
to 201 units. The ␭ values are analogous to the dual variables of linear programming,
which are discussed in Web Chapter B.
1. What purpose do the artificial variables (␭s) serve in the solution of a constrained
optimization problem by Lagrangian multiplier techniques?
2. What do the artificial variables (␭s) measure in the solution of a constrained opti-
mization problem using Lagrangian multiplier techniques?
3. Bounds, Inc. has determined through regression analysis that its sales (S) are a
function of the amount of advertising (measured in units) in two different media.
This is given by the following relationship (X ⫽ newspapers, Y ⫽ magazines):
S(X,Y) ⫽ 200X ⫹ 100Y ⫺ 10X2
⫺ 20Y2
⫹ 20XY
Assume the advertising budget is restricted to 20 units.
a. Determine (using Lagrangian multiplier techniques) the level of newspaper and
magazine advertising that maximizes sales subject to this budget constraint.
b. Calculate the firm’s sales at this constrained optimum level.
c. Give an economic interpretation for the value of the Lagrangian multiplier
(␭) obtained in part (a).
d. Compare the answer obtained in parts (a) and (b) above with the optimal
solution to the unconstrained problem in Exercise 13 of Web Chapter A.
Q*
2
Q*
1
∂L␲
∂␭
⫽ 20Q1 ⫺ 40Q2 ⫹ 200
∂L␲
∂Q2
⫽ 100 ⫺ 16Q2 ⫺ 6Q1 ⫺ 40␭
∂L␲
∂Q1
⫽ 140 ⫺ 20Q1 ⫺ 6Q2 ⫺ 20␭
27
WEB CHAPTER A APPENDIX Constrained Optimization and Lagrangian MultiplierTechniques
EXERCISES

A Optimization Techniques

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