chapter 2 in managerial economic

Chapter 2 Economic Optimisation

29

This chapter introduces a number of fundamental principles of economic

analysis.

In chapter one, we defined managerial economics and discussed the various

objectives that managers aim to achieve. The definition specifically states the

application of decision science tools in analysing and evaluating decision

alternatives.

As managers we encounter problems daily. Be it minor or major in nature, each

problem requires serious attention. Of utmost importance is selecting the optimal

course of action in light of available options and objectives. Effective managers

must be able to collect, organise and process these information.

Economists build models to better understand and portray the essential link in

terms of appropriate decision variables, costs and benefits. For very complex

scenario, models are used to breakdown and subdivide aspects of the problem

where necessary. It is easy to build models, but to build a good model requires in

depth knowledge of economic concepts and methodology, which means that

managers must have prior knowledge of basic economics and mathematics.

2 BASIC TRAINING - ECONOMIC OPTIMISATION


30

Even though solutions are easily arrived at via various means, economists place

greater importance on number crunching and precision techniques which offer a

realistic means in dealing with the complexities of goal-oriented managerial

activities. Economists find calculus with specific reference to derivative or

marginal analysis a vital tool.

This chapter places great emphasis on helping students understand marginal

concepts and the rules of differentiation. To reinforce students’ understanding,

chapter texts explicitly illustrate applications of marginal concept (be it

unconstrained or constrained) in optimisation process.

Key terms for review:

Functions

Total, average and

marginal value

Lagrangian multiplier

Aggregate Approach

Marginal Approach

Differentiation


31

CHAPTER OVERVIEW


32

Learning Objectives

After reading this chapter, the students should be able to:

1. Write functions to represent your economic problems.

2. Build a model for the economic problem at hand.

3. Relate the relationships that exist between total, average, and

marginal concepts irrespective of whether it is revenue, cost or

product.

4. Perform optimisation analysis graphically via both aggregate and

marginal approach.

5. Conduct first- and second-order derivations.

6. Apply marginal analysis in decision making

7. Distinguish between maximum and minimum values in optimisation

problems.

8. Perform optimisation analysis for multivariate functions.

9. Solve constrained optimisation problems by the Lagrangian

technique.

10. Find an optimal solution to economic problems via

the mathematical format.


33

INTRODUCTION Mathematics scares a lot of people. But we need it in many things we do. This section

intends to show you the practical applications of economic theory because in your

capacity as managers, you will resort to and rely on many concepts, graphs and simple

numerical examples to assist you in your decision making. Moreover, explanations of

economic term, concepts and methods of analysis rely primarily on verbal definitions,

numerical tables, and graphs. Appropriate discussions will centre on the same material

using both algebra and calculus. In addition, exercises and problems will be slotted in to

give students ample opportunity to reinforce their understanding.

Many students have already learnt the mathematics employed in this text, however,

some would have studied this material some time ago and may benefit from a review.

This chapter introduces a number of fundamental principles of economic analysis, basic

economic relations, the tools and techniques of optimisation. First, we will examine the

ways of presenting relationships. Subsequently, we will be examining the relationships

between total, average, and marginal concepts. Then, we will move on to examine

optimisation analysis. To find an optimal solution to complex problems and to facilitate

the above and forth coming discussions, calculus with specific emphasis on rules of

differentiation will be discussed. Finally, we will apply the rules of differentiation to

unconstrained and constrained optimisation problems.

2.1 FUNCTIONAL RELATIONSHIP & ECONOMIC MODELS

Mathematics is an important instructional vehicle in managerial economics. Economists

study variables such as price, output, revenue, cost, and profit. Economists try to

understand how and why the values of these variables change and what conditions will

lead to optimal values.


34

NOTES

Optimal may refer to maximum value (as in the case of profit),or it may

refer to minimum value (as in the case of cost.)

FUNCTION

• Single variable

In mathematics, the relationship of one variable’s value to the value of other variables

is expressed in terms of function. For example;

y = f(x) (Eqn.2.1a)

y is said to be a function of x, where y is the dependent variable, f is the function and

x is the independent variable. The above function (Eqn.2.1a) has only one

independent variable (i.e. x).

• Multivariable

But in most cases, functional relationship involves more than one independent

variable (multivariate function) for example, equation 2.1.1 below:

y = f(x, z,...n) ( Eqn.2.1b)

NOTES

Multivariate function will be discussed further under the partial derivative

topic.


35

FUNCTIONAL FORMS Economic data are best presented in the forms of:

1) algebraic equations, 2.1a and 2.1b ;

2) tables ( table 1 – next page ); or

3) graphs (diagram 1- next page)

• Algebraic equation

An equation is an expression of the functional relationships or connection among

economic variables. Five key functions are used in this chapter; demand, total

revenue, total cost, and profit. To illustrate the different ways of expressing a

function, we will try to use as many functions as possible. In economics, the

general functional relationship for total revenue is that it is dependent on the

number of units sold, i.e. TR = f(Q).

RECALL

Total revenue is defined as the unit price of a product (P) multiplied by the

units sold (Q),i.e. TR = PQ

This equation is read, “Total revenue is a function of output.” It merely states some

relation exist between output and total revenue but does not indicate the specific relation

between them. The value of the dependent variable (on the left hand side of the equal

sign) depends on the size of the independent variable (on the right hand side of the

equal sign). As such, a more precise expression would be:

TR = PQ

or TR = 7Q - 0.1Q2 …(Eqn.2.1c)

where P is the price at which each unit of Q is sold.


36

Similarly, the algebraic equations of price, cost and profit are respectively

presented as such:

P = 7 - 0.1Q …(Eqn.2.1d)

TC = 10 + 8Q - 0.3Q2+ 0.01Q3 ...(Eqn.2.1e)

π = -10 - Q + 0.2Q2 - 0.01Q3 …(Eqn 2.1f)

Equation 2.1g below is an example of an algebraic equation for a multivariate

function where there are more than 1 independent variable.

Q = 2K + 3L + LK … (Eqn.2.1g)

NOTES

For the purpose of discussion in this chapter, we will adhere consistently to

equations 2.1c to 2.1g (wherever applicable).

* Table

Table is the simplest and most direct form of presenting data. The total revenue

function (2.2) is depicted in table 1, which shows the relationship between total

revenue and quantity over a selected range of output.


37

Table 1 Total revenue schedule of a firm

Quantity(Q) Price(P) Total revenue(TR)=P.Q

0 7 0

10 6 60

20 5 100

30 4 120

40 3 120

50 2 100

60 1 60

70 0 0

• Graphs

The following diagrams show algebraic and graphical expressions for demand,

total revenue, cost, and profit.

π


38

LINEAR, QUADRATIC, AND CUBIC FUNCTIONS

The demand function in equation 2.1d is in linear form. As a result we will derive a

straight line to relate the relationship. The total revenue is expressed as a quadratic.

This is a result of a second power in the independent variable. This is easily recognised

by its parabolic shape. In addition, the independent variable can be raised to a third

power, and the corresponding relationship is called a cubic function e.g. cost and profit

functions (with 2 loops).

ECONOMIC MODELS - TYPE AND USES In the above discussions, you have encounter jargon, graphs and algebraic equations

and you will keep on seeing and using them later. They are used to model relationship

that exists between and among variables and to facilitate your analysis and decision-

making.

• Type- symbolic model

Models are defined as simplified representation of a complex situation. For example,

symbolic modes use jargons and symbols to represent reality. Managerial economics

are liberal with jargon words (for example, cost, revenue, profit) which are only verbal

models of things and phenomenon. Diagrams and mathematical expressions

similarly model a situation by means of lines or stating the relationship between and

among variables. Where diagram (1.2) are economic models depicted graphically

whilst equation (2.1c - 2.1g) are algebraic models that describe exactly the same

economic phenomenon.

• Uses

All models serve three main purposes:

1) pedagogical purpose - as a device to teach about the operation of a complex

system ,

2) explanatory purpose - as a device to explain the relationship between and among

events in a logical fashion,

3) predictive purpose - to predict future behaviour.


39

2.2 TOTAL, AVERAGE AND MARGINAL RELATIONSHIP

In the theory of demand, cost, production and market structures specific functional

relationships called total, average and marginal functions are used. In optimisation

analysis, the relationship between total, average and marginal is important I

understanding the principle of managerial economics. Table 2 shows the relationship

between total, average and marginal revenue of a firm with TR = 7Q - 0.1Q2. The first

and second columns display the relationship between output and total revenue. Column

3 and 4 is the derived average revenue and marginal revenue respectively. Here, we

assume the firm is in imperfect competition.

RECALL

An average relation is the dependent variable divided by the independent

variable. A marginal relation is the change in the dependent variable

caused by a one unit change in the independent variable.

Table 2 Total, average and marginal revenue of a firm

Q TR AR MR

PxQ TR÷Q ∆TR÷∆Q

0 0 - -

10 60 6 6

20 100 5 4

30 120 4 2

40 120 3 0

50 100 2 - 2

60 60 1 - 4

70 0 0 - 6

Remember TR = 7Q-0.1Q2


40

AVERAGE REVENUE

Column 3 depicts average revenue. The average revenue of the tenth output is $6,

which is simply a division of total revenue with output. Each subsequent row in the

column is similarly derived.

MARGINAL REVENUE

In column 4, the marginal revenue earned from the tenth unit of output is $6. This is the

change from $0 when there is no output, to $60 earned, when 10 units were sold, thus a

one unit change is equivalent to $6. thus, the change in total revenue as a result of a

change in output is called marginal revenue. Each marginal revenue derived in

subsequent rows is calculated on the same basis.

Diagram 2 Total

revenue function,

Demand function,

and Revenue

maximising price

and quantity


41

SUMMARISING

Let us summarise what we have discussed so far.

The key relationships between total, average and marginal are: 1) The value of the average function at any point is the slope of a ray

drawn from that point to the total function.

2) The value of the marginal function at any point slope of a line drawn

tangent to the total function at that point.

3) When total function increases, both average function and marginal

function is positive

4) When total function is decreasing , average function is still positive but

marginal function is negative

5) When total function is at its maximum, marginal function equals zero.

Example

A firm’s demand function is defined as Q =14 - 2P. Calculate total revenue when price is

equal to 3 and when price is equal to 4. What is marginal revenue equal to between P=3

and P=4?

Explain your answer.

SUGGESTED SOLUTIONS

When P=4, Q = 14 - 2(3) = 6 ∴TR = PQ = $24

P=3, Q = 14 - 2(4) = 8 ∴TR = PQ = $24


42

There is no change in TR from P=4 to P=3, thus MR is equal to 0. Thus this firm will

maximise total revenue when output is between 6 units and 8 units.

IMPORTANCE OF CONCEPTS

The concepts of average, marginal and total relations should be thoroughly understood

as they are widely used in short-run optimisation problems (as shown in 2.3, 2.6, 2.7 and

2.8).

NOTES

In diagram 2b, we relate price to quantity in a linear fashion. In a linear

equation, the coefficient (-0.1) is the change in the dependent variable (P)

over the change in the independent variable (Q), (i.e., ∆P/∆Q). In other

words, it represents the slope of the equation, which measures its

steepness.

The slope of a function is critical to economic analysis because it is the essence of marginal analysis. Decisions made by managers involve some sort of change in one

variable relative to a change in other variables. For example, whether or not to raise

price would depend on the resulting change in revenue or profit. Many other economic

decisions rely on marginal analysis, including purchasing additional equipment and hiring

extra staff because of importance to you is the change in cost, productivity, or profit

associated with a change in your resource allocation.

We can use concepts of calculus to clarify the relationship between average, marginal,

and total. Therefore, elementary calculus will be discussed in 2.4.


43

2.3 OPTIMISATION ANALYSIS

In economic analysis, optimisation generally means finding either the maximum or

minimum value of a variable of a function, thus a firm determines the output level at

which it maximises total profit or minimises cost.

APPROACH FOR FINDING OPTIMAL VALUE

We can use either the aggregate approach or the marginal approach. Likewise, it can be

illustrated graphically or calculated using calculus. Thus, we will begin by showing

graphically how optimal output is obtained by the aggregate approach followed by the

marginal approach. (In 2.6, we will use calculus to ascertain that the outcome is the

same as shown graphically).

OPTIMISATION BY TOTAL REVENUE AND TOTAL COST APPROACH ( AGGREGATE APPROACH)

The aggregate approach looks at aggregate, for example, total profit is the difference

between total revenue and total cost i.e.:

π = TR - TC

If given TR = 7Q - 0.1Q2 and TC = 10 + 8Q - 0.3Q2 + 0.01Q 3

. π = 7Q - 0.1Q2 - 10 - 8Q + 0.3Q2 - 0.01Q 3

= - 10 - Q + 0.2Q2 - 0.01Q 3


44

Diagram 3 Total revenue, total cost, marginal revenue, marginal cost and profit

maximisation.

We can show profit is maximised via graph, by plotting the total revenue and total cost

functions for a range of output, Q. Diagram 3a shows the plotted total revenue and total

cost whilst 3b shows profit function. The graphical representation shows that profit is

maximised at Q2* =10 that is when the positive difference between TR and TC is the

greatest (3a) or alternatively when the profit function is at its maximum (3b).

π

π


45

OPTIMISATION BY MARGINAL APPROACH (MR=MC)

RECALL

According to the marginal analysis, profit is maximised when MR=MC.

As long as marginal benefit exceed marginal cost, it pays for the firm

to increase output and the firm will continue doing so till marginal

revenue equals marginal cost (at this point the firm would be

maximising profit). It stems from the fact that the distance between

revenue and cost function is maximised at the point where their

slopes are the same. Because the slopes of total revenue and total

cost measures marginal revenue and marginal cost, hence when

these slopes are equal, MR = MC.

The maximum can also be located by finding the derivative or marginal of the function

then determining the value of Q at which the derivative (Marginal Mπ ) is equal to 0. For

our hypothetical example, MR=MC occurs at Q2* = 10 as illustrated in 3a. The relevance

of marginal revenue and marginal cost relations to profit maximisation can be

demonstrated by considering the general expression

π = TR – TC

d π dTR dTC

Marginal profit is M π = ------ = ------ - ------

dQ dQ dQ

given that dTR /dQ is MR and dTC/dQ is MC it follows that

M π = MR – MC


46

Because maximization of any function requires that the first derivative equal zero, profit

maximisation occurs where

M π = MR – MC = 0

or where MR = MC. at output Q = 10.

2.4 METHODS OF DIFFERENTIATION CALCULUS - DEFINITION AND PURPOSE

Even though tables and graphs are useful for explaining concepts, equations are more

suitable for problem solving. One reason being that the technique of differential calculus

can be employed to locate maximum and minimum point more precisely.

Calculus is a mathematical technique that enables you to find instantaneous rate of

change of a continuous function. Moreover, it is especially useful in constrained

optimisation problems that often characterise managerial decision making.

CONCEPT OF A DERIVATIVE

A derivative is a precise specification of the marginal relation. As mentioned in earlier

pages, the slope is a measure of the change in y relative to a infinitesimally small change

in x. To find the magnitude of the slope, we need to employ derivatives. Differentiation is

the process of determining the derivatives of a function. Consider the general function y

= f(x).

NOTES

Differentiation means finding the change in y (∆y) for a change in x ( ∆ x)

as the change of x approaches 0.


47

Expressed formally:

dy ∆y

----- = lim ----

dx ∆x→0 ∆x

This notation is read: “The derivative of y with respect to x equals the limit of the change

in y relative to the change in x as the change in x approaches zero.” Mathematicians use

d to represent very small changes in a variable. The ratio ∆y/∆x is a general

specification of the marginal concept.

2.5 RULES OF DIFFERENTIATION Let us spend a few minutes reviewing basic differentiation. If you need additional

material, refer to the MAT 153 (Business Calculus) module or any Calculus text

FIRST AND SECOND DERIVATIVES OF FUNCTIONS

Rule for finding derivative

Take for example a function expressed in the general form of

y = bxn

The rule for finding derivative is

dy --- = nbxn-1 dx


48

NOTES

Derivative (or first derivative ) is denoted by dy/dx.

Let us try to use the rule..

Thus, suppose we have the equation

y = 5x2

The derivative of equation (2.7) according to this rule is

dy

--- = 2 . 5x 2-1 = 10x

dx

meaning that if x=3, the instantaneous rate of change of y with respect to x is 10(3) or

30.

In economic analysis, there are several other rules for differentiating a function, such as

differentiating a logarithmic function or ‘function of a function’. (For a detailed summary of

rules of differentiation, which will be used extensively in this text, please refer to the table

3 that follows.)


49

Table 3 RULES FOR DIFFERENTIATING FUNCTIONS

Function Derivative

1. Constant Function y = a dy/dx = 0

2. Power function y = axb dy/dx = b.a.xb-1

3. Sums & differences function y = u + v dy/dx = du/dx + dv/dx

y = u - v dy/dx = du/dx - dv/dx

4. Product of two functions y = u . v dy/dx = u. dv/dx + v.du/dx

5. Quotient of two functions y = u/v v(du/dx) - u(dv/dx)

dy/dx = -------------------------

v2

6. Function of a function y = f(u)

where u = g(x) dy/dx = (dy/du) . (du/dx)

Do not worry; do not let the mathematical symbols bother you. The process of

differentiating is rather simple, actually

Examples:

1. y = 3 dy/dx = 0

2. y = 3x3 dy/dx = 3.3 x3-1 = 9x2

The following examples comprise of more than one function. Can you identify and write

down the functions U and V in the box provided?

3. y = 3x3 + (x2 + 2) dy/dx = (3.3x3-1) +(2.x2-1) = 9x2 + 2x

y = 3x3 - (x2 + 2) dy/dx = (3.3x3-1) - (2.x2-1) = 9x2- 2x

Examples 4,5 and 6 take into account the product of two functions

4. y = (3x3 ) ( x2 + 2) dy/dx = (3x3)(2x) + (x2 + 2)(3.3x3-1)


50

= (6x4) +(x2 +2)9x2 = 6x4 +9x4 +18x2

=15x4 +18x2

5. y = (3x3 )/( x2 + 2) (x2 + 2)(3.3x3-1) - (3x3)(2.x2-1)

dy/dx = --------------------------------------

( x2 + 2)2

(x2 + 2) 9x2 - (3x3 )( 2x)

= -------------------------------

(x2 + 2)2

(x2 + 2) 9x2 - 6x4

= --------------------

(x2 + 2)2

3x4 + 18x2

= --------------

(x2 + 2)2

6. y = u2 + 5 and u = 3x2 dy/dx = (2u) ( 6x) = 2.3x2(6x)

= 36x3

But, almost all of the mathematical examples and problems involving calculus only

require the rules for constant, powers, sum and differences.

Take this example that involves all the above.

π = - 10 - Q + 0.2Q2 - 0.01Q3

dπ/dQ = -1 + 0.4Q - 0.03Q2


51

Example Determine the derivative of this function:

y = 2000 - 200x2 + 3x3

SUGGESTED SOLUTIONS

dy/dx = - 400x + 9x2

FIRST DERIVATIVE

As mentioned earlier, the first derivative of a non-linear function is also called the

marginal function.

Turning now to the total revenue function;

TR = 7Q - 0.1Q2

When we take the first derivative of the above total function;

d(TR)

------ = 7 - 0.2Q = MR

dQ

the outcome is also known as the marginal revenue (MR) function.

As opposed to the above power function, take the case of a linear function such as the

demand equation P = 7 - 0.1Q.


52

The derivative is

dP

----- = - 0.1 dQ

NOTES

(dP/dQ is equal to a constant value - 0.1). Here, we see that the first derivative of

a linear function is simply the value of the b coefficient, - 0.1 (i.e. the slope of the

linear function itself).

SUMMARISING

First derivative of 1) a linear function is the slope and is a constant

2) a quadratic function is the marginal function

SECOND DERIVATIVE- DEFINITION

Sometimes, mathematical calculations require the use of second derivative. Generally,

the second derivative is the derivative of its first derivative. Confusing? Not to worry. It

simply means that the second derivative of a function is a measure of the rate of change

of the first derivative.

PROCEDURE FOR FINDING SECOND DERIVATIVE

The procedure for finding the second derivative is quite simple. All the rules for finding

the first derivative also apply to obtaining the second derivative.

For example, using the same TR function


53

TR = 7Q - 0.1Q2

d(TR)

first derivative: --------- = 7 - 0.2Q

dQ

When we derive it for the second time

d2 (TR)

------- = - 0.2

dQ2

NOTES

Second derivative is denoted with a superscript 2.

Second order derivative is very important in optimisation problems. (We will deal with this

in more detail in 2.6 .)

Example

Determine the first- and second- order derivatives of this function:

C = 2000 - 200x2 + 3x 3

Solution

First derivative: dC/dx = - 400x +9x2

Second derivative: d2C/dx = - 400 + 18X


54

2.6 OPTIMISATION BY CALCULUS

MARGINAL ANALYSIS IN DECISION MAKING

Managerial decision making requires one to find minimum or maximum value of a

function. When a function is at a minimum or maximum, its slope or marginal value is

equal to zero, thus the derivative must be equated to zero. To illustrate we will discuss

both aggregate and marginal approaches.

AGGREGATE APPROACH

A primary objective of managerial economics is finding optimal values of key variables.

Beside being done graphically, the optimisation analysis is conducted much more

expediently with the use of calculus - using marginal analysis and derivative concepts.

For our hypothetical example, the profit function is obtained as follows:

π = TR - TC

= 7Q - 0.1Q2 - 10 - 8Q + 0.3Q2 - 0.01Q 3

= -10 - Q + 0.2Q2 - 0.01Q 3

Let us now employ calculus to determine the point of profit maximisation. We begin by

i) finding the first derivative of the profit function,

ii) setting it equal to zero, and

iii) solving for the value of Q that satisfies this condition.

d π/dQ = - 1 + 0.4Q - 0.03Q 2 = 0

(-10 +Q)(10 - 3Q) = 0

Q2* = 10 or Q1* =10/3

MARGINAL APPROACH

Alternatively, you can use the marginal approach. First,

i) find the derivative of total revenue and total cost, then

ii) set them equal to each other and finally

As expected, Q2*

coincides with the

point shown in

diagram 3a.


55

iii) solve for Q*

MR = dTR/dQ = 7 - 0.2Q and MC = dTC/dQ = 8 - 0.6Q + 0.03Q2

By equating MR = MC = Mπ = 0

7 - 0.2Q = 8 - 0.6Q + 0.03Q2

-1 + 0.4Q - 0.03Q2 = 0

Q2* = 10 or Q1* =10/3

Marginal profit, the derivative of total profit is Mπ = dTR/dπ

Mπ = dTR/dQ - dTC/dQ .

Given that dTR/dQ by definition is marginal revenue, MR and dTC/dQ is marginal cost,

MC, it follows that Mπ = MR = MC. Since maximisation of any function requires that the

first derivative = 0, profit maximisation occurs at Mπ = MR = MC = 0 or where MR = MC =

0.

Example

A firm’s demand function is Q = 16-P and its total cost function is defined as

TC = 3 + Q + 0.25Q2.

Form the firm’s profit function and then determine the level of output that yields the

maximum profit. What is the level of profit at the optimum?


56

SUGGESTED SOLUTIONS

TR = PxQ

given Q = 16 - P

∴ P = 16 - Q

hence TR = 16Q - Q2 and with TC = 3 + Q + 0.25Q2

π = 16Q - Q2 - 3 - Q - 0.25Q2

= - 3 + 15Q - 1.25Q2

Using calculus:

first derivative: dπ/dQ = 15 - 2.5Q = 0

implies Q = 6

and the second derivative d2π/ dQ2 = - 2.5, which implies that Q = 6 is a maximum.(see

page 15 for second-order condition)

Alternatively set MR = MC

hence MR = 16 -2Q

MC = 1 + 0.5Q

16 - 2Q = 1 + 0.5Q

2.5Q = 15

Q = 6.

To find profit, substitute Q = 6 into the profit function, hence

profit = - 3 + 15(6) - 1.25 (36)

= 42


57

DISTINGUISHING MAXIMUM AND MINIMUM VALUES IN THE OPTIMISATION PROBLEMS

In economic analysis, finding optimum values generally means finding the maximum or

minimum value of a variable, depending on the type of function discussed. For example,

when a profit or total revenue function is analysed, the maximum value is the focus,

whilst the minimum value would be the main concern if the total cost function is analysed

SECOND DERIVATIVE- REVISITED

However, there are instances when the function has both a maximum and a minimum

value, (diagram1d - when a function is in cubic form). Since marginal value equals zero

for both maximum and minimum values of a function, that means, the method described

previously cannot tell us whether the optimum is a maximum or minimum. Hence further

analysis is required. A formal mathematical procedure to distinguish between maximum

or minimum value, requires the use of second derivative, which is, differentiation of the

first derivative. As mentioned in our earlier discussions, rules for finding first derivative

also apply to obtaining the second derivative.

The rule is that if the second derivative is positive,

we have a minimum, and if the second derivative is

negative, we have a maximum.

Using mathematical notation, we can then state that the first-and second-order

conditions for determining maximum and minimum values of a function are as follows.

First order Second order

condition condition

Maximum value dy/dx = 0 d2 /dx 2 < 0

Minimum value dy/dx = 0 d2 /dx2 > 0

−.

RULES


58

USE OF MARGINALS TO MAXIMISE THE DIFFERENCE BETWEEN TWO FUNCTIONS

Now to illustrate,

Suppose the firm has the following revenue and cost function (note: we use a cubic

function)

TR = 7Q - 0.1Q2

TC = 10 + 8Q - 0.3Q2 + 0.01Q3

Based on these equations, the firm’s total profit function is

π = - 10 - Q + 0.2Q2 - 0.01Q3

Let us now employ calculus along with the first- and second- order conditions to

determine the point at which the firm maximises its profit.

d π

--- = -1 + 0.4Q - 0.03Q2 = marginal profit

d Q

= (-10 + Q) (10-3Q) = 0

Q1* = 10/3 Q*2 = 10

Although both Q1 * and Q2 * fulfil the first-order condition, only one satisfies the second-

order condition. To find out which one, let us find the second derivative of the function by

taking the derivative of the marginal profit function or second derivative of the profit

function:

d2π

---- = 0.4 - 0.06Q

dQ2


59

By substitution, we see that when Q= 10/3, the value of the second derivative is a

positive number:

d2π/ dQ2 = 0.4 - 0.06(10/3) = 0.2

On the other hand, when Q=10, the value of the second derivative is a negative number:

d2π/ dQ2 = 0.4 - 0.06(10) = - 0.2

Thus, we see that only Q2* enables us to adhere to the second-order condition of

d2π /dQ2 < 0.

As shown by graph (3a, 3b) and calculus above, only Q2* is THE optimal output. Our

example also shows that MR=MC at profit maximising output level, the converse does

not hold true. Profit is not necessarily maximised at any point where MR=MC (for

example Q2*.=10/3)

NOTES

Optimal output is derived when total profit function reaches a maximum. This

corresponds to the point when Mπ = 0 at Q2*. To be the optimal output, it must

satisfy both the F.O.C. and S.O.C that is

dπ/dQ = 0 and d2π/ dQ2 < 0.


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Example

Given the demand function Q = 1000 - 40P, where Q is quantity and P is price,

determine the quantity that results in maximum total revenue.

SUGGESTED SOLUTIONS

Given Q = 1000 - 40P

∴ P = 25 - 0.025Q

and TR = PQ

= 25Q - 0.025Q2

To determine the quantity that maximises TR, take the first derivative, set it equal to

zero, and solve for Q.

d(TR)

------ = MR = 25 - 0.05Q = 0

dQ

Q = 500

That Q=500 is a maximum is clear because the second derivative is negative, that is,

d2(TR)

------- = - 0.05 < 0

dQ2


61

CHECKLIST

Take a few minutes to check if you are now able to:

Write functions to represent your economic problems.

Build a model for the economic problem at hand.

Relate the relationships that exist between total, average, and

marginal concepts irrespective of whether it is revenue, cost or

product.

Perform optimisation analysis, graphically via both aggregate and

marginal approach.

Conduct first- and second-order derivations.

Apply marginal analysis in decision making

Distinguish between maximum and minimum values in

optimisation problems.

2.7 MULTIVARIATE OPTIMISATION

In this section, we determine maximum and minimum values of a function with more than

one variable. First, we introduce the concept of partial derivative and then use it to

examine the maximisation process of a multivariate function

PARTIAL DERIVATIVE

The functions discussed so far involve only one variable. Many functional relationships

which you will encounter later involve more than one independent variable. For a

multivariate function, for example y = f(x,z), the concept of partial derivative is employed.


62

SOME EXAMPLES OF MULTIVARIATE FUNCTIONS

Most economic relationship involves more than one variable, for example;

demand of a product (Qdx) is influenced by the price of the product itself (Px ),

price of a related product (Py, where y could be a substitute or complement),

income (I), taste(T) and others, that is {Qdx = f(Px, Py, I,T,…)}; total revenue (TR)

depends on output (Q), advertisement (A), and taste factors (T), that is {TR=

f(Q,A,T)} and output (Q)is a function of labour (L) and capital inputs(K). Hence, it

is important to determine the marginal effect of each independent variable

separately. To do so, we use partial derivative.

For example, given y = f (x, z)

y = 3x2 -xz + z2

When analysing multivariate function, we first take the partial of y with respect to

x, i.e. ∂y/∂x. This indicates the slope relationship between y and x when z is held

constant. Similarly, the partial derivative of y with respect to z (∂y/∂z) is derived by

assuming x to be constant, and taking the first derivative of y with respect to z.

Thus,

∂y/∂x = 6x - z

and ∂y/∂z = 2z - x

The partial derivative ∂y/∂x means that a small change in x is associated with y

changing at the rate of (6x - z) when z is held constant. If z =2, the slope

associated with y and x is (6x - 2). Similarly, ∂y/∂z means a small change in z

is associated with y changing at the rate of (2z - x), when x is held constant.

NOTES

Partial derivative is denoted by ∂y/∂x.


63

MAXIMISING A MULTIVARIATE FUNCTION

To maximise or minimise a multivariate function, we have to set each partial derivative

equal to zero and solve simultaneously for the optimal value of the independent

variable.

For example, to maximise total profit function

π = xz -3x2 - z2 + 11x

we find partial derivative, set it equal to 0 and solve for x and z.

∂ π/∂ x = z - 6x + 11 = 0 …( i )

∂ π/ ∂z = x - 2z = 0 …( ii )

Multiplying ( i ) by 2 and add to ( ii ), we get

2z-12z + 22 = 0

x - 2z = 0

-------------------------------------------

11x = 22

x = 2

∴ z = 1

Substituting x = 2 into ( i ), we get z = 1 and substituting these values into the

profit function , we get total profit:

π = (2)(1)-3 (2)2 - (1)2 + 11(2) = 11

Thus when this firm maximises its

profit amounting to $11 when it

sells 2 units of x with 1 unit of y.


64

Example

Given a firm’s total profit depends on the sale of product x and product z is π = f(x,z) =

100x - x2 - xz - 2z2 + 190z,

Find partial derivative of π with respect to x and z. Find the combined output of x and z

that maximises profit and profit, at that combination.

SUGGESTED SOLUTIONS

∂π/∂x = 100 - 2x - z …( i )

and ∂π/∂z = -x - 4z + 190 …( ii )

To solve,

( i ) x 4 400 - 8x - 4z = 0 …( iii )

( iii) - (ii) 190 - x - 4z = 0

---------------------

210 - 7x = 0

x = 30

∴ z = 40

Substituting x =30 and z = 40 into the profit function gives profit amounting to $5300

π = 100(30) - (30)(30) - 30(40) - 2(40)(40)

= $5300


65

2.8 CONSTRAINED OPTIMISATION

All the above discussions are illustrations of unconstrained optimisation, but that is not a

realistic scenario.

SOLVING BY SUBSTITUTION METHOD

In chapter 1, we mentioned that in attempting to achieve goals, managers face

constrains. This would limit their options, hence the maximisation or minimisation of an

objective function is subjected to constrain. One way to solve a constrained optimisation

problem is the substitution method.

Suppose, a firm operates with this total cost function:

TC = 3x2 + z2 - xz

where x and z represent two products to be produced.

The manager is requested to determine the least-cost combination of x and z, subject to

the constraint that total output of both products is 10 units. Therefore, the constrained

optimisation problem is:

Minimise TC = 3x2 + z2 - xz,

subject to x + z = 10.

Solving the constraint for z and substituting this value into the objective function results

in

z = 10 - x,

and TC = 3x2 + (10-x)2 - x(10-x)

= 3x2 + 100 - 20x + x2 - 10x + x2

= 5x2 - 30x + 100

Now, it is possible to treat the above equation as an unconstrained minimisation

problem. To solve, we find the first derivative, set it equal to zero. And solve for value of

x:


66

dTC

------ = 10x -30 = 0

dx

10x = 30

x = 3

A check of the sign of the second derivative at that point has to be made to ensure a

minimum:

d2TC

----- = 10

dx2

Since the second derivative is positive, x = 3 is indeed a minimum.

When we substitute x = 3 into the constraint equation, we can determine the optimal

output for z.

x + z = 10

z = 10 -3

= 7

Thus a production of 3 units of x and 7 units of z is the least-cost combination for

producing 10 units. The total cost of this combination is

TC = 3(9) - (3)(7) + 49

= $55.

SOLVING BY LAGRANGIAN MULTIPLIER TECHNIQUE

Fallacy of the substitution technique

Unfortunately, the substitution technique discussed in the preceding section is not

always feasible. Some decision problems involve numerous and complex constraint

conditions. To overcome this, we have to resort to the Lagrangian multiplier technique.


67

LAGRANGIAN TECHNIQUE

This method optimises a function which incorporates the original objective function and

the constraint function. This combined equation called the Lagragian function is created

in such a way that when it is maximised or minimised, the original objective function is

also maximised or minimised and all constraints are satisfied.

Lagrangian function

The first step in this method is to form a Lagrangian function. This is simply done by

writing the objective function that the firm wishes to seek; (either maximises or

minimises) plus λ (lambda - denoting the Lagragian multiplier ) times the constraint

function and set equal to zero−.

For our hypothetical example;

Min.x, z TC subject to total output

(Objective function) (Constraint function)

Thus L TC = 3x2 + z2 - xz + λ( x + z - 10)

L TC is defined as the Lagrangian function for the constrained optimisation under

consideration.

The partial derivative of the above functions with respect to the three unknown, x,

z, λ and are as follows:

∂LTC

----- = 6x - z + λ ,

∂x


68

∂LTC

------ = 2z - x + λ

∂z

∂LTC

and ------ = x + z - 20

∂λ

At a minimum point on a multivariate function, all partial derivatives are equal to zero.

Setting these partial derivatives equal to zero, gives three equations and three unknown.

6x - z + λ = 0 … (i)

2z - x + λ = 0 … (ii)

x + z - 10 = 0 …(Iii)

The above can be solved simultaneously.

First we get rid of one unknown, ( λ ) by subtracting ( ii ) from ( i )

6x - z + λ - 2z + x - λ

gives 7x - 3z = 0 … ( iv )

Next, to get rid of z, multiply ( iii) by 3 and add to ( iv ) to give the solution for x:

3 x + 3z - 30 = 0

7x - 3z = 0

---------------------------------------

10x - 30 = 0

10x = 30

x = 3

Substituting x = 3 into equation ( iii ) yields z = 7;


69

3 + z = 10

z = 7

and substituting x = 3 and z = 7 into equation ( i ), gives the λ value:

6(3) - 7 +λ = 0

λ = -11

Here, λ = -11 is interpreted as the marginal cost of producing 10 units; meaning that if

the firm reduces output from 10 to 9 units, total cost will fall by approximately $11.Thus,

λ is the marginal effect on the objective function associated with per unit change

(increase or decrease) in the constraint. The marginal relation described by the multiplier

provides managers with economic data to evaluate the potential benefits or costs of

relaxing constraints.

NOTES

The use of Lagrangian multiplier method is further examined in chapter 4 to find

consumer’s maximum satisfaction from consumption. Again, in chapter 5, the

Lagrangian technique for constrained optimisation is used to develop the optimal

input proportion rule.

SUMMARISING

Steps to follow:

1. Form the Lagrangian function, (L ( ) )

2. Find partial derivative (L ( ) ) with respect to the first variable,

3. Find partial derivative (L ( ) ) with respect to the second variable,

4. Find partial derivate (L ( ) ) with respect to λ,

5. Finally, solve simultaneously to get values of the first and second variable

and λ.


70

The exercise below provides another perspective of the usefulness of the Lagrangian

method.

Example

Given the utility function as follows

u = 2x 0.75 - y 0.25

and the constraint function

1000 = 2x + 4y

a) Formulate a Lagrangian function

b) What are the values of x and y that will maximise utility?

c) Find the value of λ and interpret it.

SUGGESTED SOLUTIONS

a) L( U*) = 2x0.75 y0.25 - λ (1000 - 2x - 4y)

b) F.O.C.

∂ U*

---- = 0.75(2x -0.25 y0.25 ) + 2 λ = 0 …(1)

∂ x

∂ U*

---- = 0.25(2x0.75 y -0..75) + 4λ = 0 …(2)

∂ y

It would be great if you attempt the

question before looking at the

answer. Just follow the steps above


71

∂ U*

------ = - 1000 + 2x + 4y = 0 …(3)

∂λ

Divide equation (1) by (2)

1.5x -0.25 y0.25 - 2λ

---------------- = -----

0.5x0.75 y -0.75 - 4λ

3 y 2

---- = ----

x 4

2x = 12y

x = 6y …(4)

Substitute equation (4) into (3)

2 (6y) + 4y = 1000

y = 62.5

x = 6(62.5)

= 375

A combination of x = 375 and y = 62.5 will maximises utility.

c) Substitute x = 375 and y = 62.5 into equation (1)

5(375)-0.25 (62.5)0.25 + 2 λ = 0

λ = - 0.479

λ is the marginal effect of the objective function associated with the per unit change in

the constraint function. A λ of - 0.479 means that total utility will fall by 0.479 when the

income falls by one unit.


72

QUESTIONS

Q1. Assume a firm produces its product in a market described with the following

production function and price data.

Q = 2x + 3y + 6xy

Px = $4, Py = $6

where x and y are two variable input factors in the production of output Q, and Px

and Py are prices of x and y respectively.

a) What is the optimal input combination for x and y in this production if the firm

is operating with a $1000 budget constraint? (use the Lagrangian multiplier

method)

b) What is the increase in output that could be obtained from an additional

expenditure of $1?

Q2. A firm’s production function is given by

Q = L2 + 5LK + 4K2

The price of a unit of labour service and a unit of capital is $5 and $10

respectively. The firm has a cost limitation of $1000 per time period.

a) Formulate the Lagrangian function.

b) What should be the combination of K and L that will maximise output Q?

c) How much is the maximum output?


73

d) Interpret the meaning of the Lagrangian multiplier.

Q3. A. Einstein invented a new alarm system for which he receives royalty of 20% of

total revenue from the sales of the alarm system. If the total cost and demand

function is given by

TC = 200 000 + 30Q + 0.002Q

Q = 12 500 - 50P

a) Calculate the total revenue maximising price and output. What is the amount

of royalty Einstein would receive?

b) Calculate the profit maximising output and price. What is the amount of

royalty Einstein would receive at this output?

Q4. A company produces according to the following production function:

Q = 500L - 0.5L2 + 1000K -0.25K2 + 100E - 0.1E2

where Q is output per month,

L is number of workers employed per month,

K is machines used per month

and E is electricity used per month.

All output is sold at $10 per unit. L costs $2000 per month, machines cost $9000,

electricity costs $0.10 per kilowatt/hour.

a) Formulate the profit function.

b) What is the profit maximising level of labour, machine and electricity per

month?

c) What is the profit maximising level of output?


74

d) What is total profit per month?

Q5. The demand function faced by john associate is

P = 45 – 0.5Q

where Q represents output and p represents price

Its total cost is given by the equation

TC = Q 3- 8Q 2 + 57Q + 2

a) derive the equation of the total profit function

b) determine the output level that maximises total profit

c) calculate the maximum total profit

Q6. Given Q = 1500 –50P

a) determine the TR function

b) determine the rate of output that maximise TR

c) calculate maximum TR


75

SUGGESTED SOLUTIONS

Q1 Steps to follow:

1 ) Form the Lagrangian function, (L ( ) )

2 ) Partial derivate (L ( ) ) with respect to x,

3 ) Partial derivate (L ( ) ) with respect to y,

4 ) Partial derivate (L ( ) ) with respect to λ,

5 ) Finally, solve simultaneously to get values of x, y, and λ.

Objective function ⇒ Q = 2x + 3y + 6xy

Constraint function ⇒ 4x + 6y = 1000

a) LQ = 2x + 3y + 6xy + λ (4x + 6y - 1000)

∂LQ/ ∂x = 2 + 6y + 4λ = 0 …(1)

∂LQ/ ∂y = 3 + 6x + 6λ = 0 …(2)

∂LQ/ ∂λ = 4x + 6y - 1000 = 0 …(3)

(1) x 3 ⇒ 6 + 18y + 12λ = 0 …(4)

(2) x 2 ⇒ 6 + 12x + 12λ = 0 …(5)

(3) - (5) ⇒ 18y -12x = 0

y = 2x /3

Substitute into (3)

4x + 6 (2/3)x = 1000

4x + 4x = 1000

x = 125

y = 125(2/3)

= 83.33


76

Therefore optimal combination is x = 125 and y = 83.33 units

b) Substitute x = 125 and y =83.33 into (1)

2 + 6(83.33) = 4λ

λ = 502/4

= 125.5

When expenditures increase by $1, output will increase by 125.5 units.

Q2. See Q1 for steps to follow.

a) Objective function ⇒ Q = L2 + 5LK + 4K2

Constraint function ⇒ 5L + 10K = 1000

∴ LQ = L2 + 5LK + 4K2 + λ (5L +10K -1000)

b) ∂LQ / ∂L = 2L + 5K + 5λ = 0 …(1)

∂LQ/ ∂K = 5L + 8K + 10λ = 0 …(2)

∂LQ/ ∂λ = 5L + 10K - 1000 = 0 …(3)

(1) x 2 ⇒ 4L + 10K + 10λ = 0 …(4)

(4)- (2) ⇒ -L + 2K = 0 …(5)

From (5) ⇒ L = 2K …(6)

Substitute (6) into (3) ⇒ 5(2K) + 10K = 1000

20K = 1000

K = 50

∴ L = 100

Output is maximised when K=50 and L =100 .


77

c) Substitute K=50 and L =100 into the output function.

Q = L2 + 5LK + 4K2

= 100(100) + 5(100)(50) + 4(50)(50)

= 45 000

d) Substitute K =50, L=100 into equation (1)

2(100) + 5(50) + 5λ = 0

λ = -90

When production budget falls by a unit, the output falls by 90 units.

Q3.

Steps to follow:

1) Find TR

2) Conduct F.O.C on TR to get Q* and P*

3) Calculate 20% of TR

4) Conduct S.O.C. to reaffirm maximising.

a) TR = PxQ

given Q = 12500 - 50P

∴P = 250 - 0.02Q

and TR = 250Q - 0.02Q2

F.O.C. dTR/ dQ = 250 - 0.04Q = 0

Q* = 6250

Substitute Q*=6250 into P function: P* = 250 - 0.02(6250)

= $ 125

therefore TR = 125 x 6250

= $ 781 250

therefore the amount of royalty = 20%( 781 250)

= $ 156 250


78

b) Steps to follow:

1) Find profit

2) Conduct F.O.C on π function to get Q* and P*

3) Conduct S.O.C. to reaffirm maximising

4) Find TR by substitute Q* and P* into TR

5) Calculate 20% of royalty

π = TR - TC

= 250Q - 0.02Q2 - { 200 000 + 30Q + 0.0002Q2 }

= - 200 000 + 230Q - 0.022Q2

∂ π/∂Q = 230 - 0.044Q =0

Q* = 5 000

P* = 250 - 0.02(5000)

= $150

TR = PxQ

= 150(5 000)

= $ 750 000

Royalty = 20%(750 000) = $ 150 000

Q4. Steps to follow:

1) Find profit by subtracting TC from TR

2) Partial derivate π with respect to L, K, E

3) Solve for Q* for respective L, K,E

4) Conduct S.O.C. to reaffirm maximising

5) Find π by substituting into answer a)

a) π = TR - TC

TR = PxQ

given P = $10

and Q = 500L - 0.5L2 + 1000K -0.25K2 + 100E - 0.1E2


79

∴ TR = 10(500L - 0.5L2 + 1000K -0.25K2 + 100E - 0.1E2) given PL = 2000, PK = 9000, and PE = 0.1

∴ TC = 2000L + 9000K + 0.1E

∴ π = 10(500L - 0.5L2 + 1000K -0.25K2 + 100E - 0.1E2)

- 2000L + 9000K + 0.1E

b)To maximise π , input is employed to the point where the change in π

associated with hiring one additional unit of input is equal to zero.

Labour : ∂π/∂L = 10( 500 - L) - 2000 = 0

L* = 300

Machines: ∂π/∂K = 10( 1000 - 0.5K) - 9000 = 0

K* = 200

Electricity: ∂π/∂E = 10( 100 - 0.2E) - 0.1 = 0

E* = 499.95

c) Q = 500(300) - 0.5(300)2 + 1000(200) - 0.25(200)2 +100(499.95) –

0.1(499.95)2

= 320 000 units

d) π = TR - TC

TR ⇒ 10( 320 000) 3 200 000

TC ⇒ Labour : 2000(300) = 600 000

Machines: 9000(200) = 1 800 000

Electricity: 0.1(499.95) = 50

2 400 050

---------------

Total profit $799 950

---------------

Well, you have reached the end of the chapter. The math in this chapter has been

carefully explained step by step. Please make sure you have understood the steps used

in the calculation. It has also been a long chapter; if you find difficulties in understanding


80

the material, please go through the sections again carefully and note down questions

that you need to discuss with your lecturer or “classmates”

SUMMARY

The functional relationship y = f(x) means there is a systematic relationship

between the dependent variable y and the independent variable x.

Functional relationships can be expressed in algebraic equations, tables or

graphs.

Functions can be linear, quadratic or cubic.

Economic models are used to illustrate an economic principle, explain an economic

phenomenon, or to predict implications as a result of changes that affect the

functional relationship.

The key relationships between total, average, and marginal are:

⎩ The value of the average function at any point is the slope of a ray drawn from

that point to the total function.

⎩ The value of the marginal function at any point is the slope of a line drawn

tangent to the total function at that point.

⎩ When total function increases, both average function and marginal function is

positive.

⎩ When total function decreases, average function is still positive but marginal

function is negative.

⎩ When total function is at its maximum, marginal function equals zero.

Optimisation means finding either the maximum or minimum value of a variable of a

function.


81

The slope of a function y = f(x) is the change in y (∆y) divided by the corresponding

change in x (∆x). The derivative of this function dy/dx or f ’(x) is the slope of a straight

line drawn tangent to the function at that point.

The rules of differentiation contained in page will help you find derivatives of

functions encountered in managerial economics.

Higher order derivatives are found by taking the first derivative of each resultant

derivative, for example, second derivative is the derivative of the first derivative.

The maximum or minimum value of a function y = f(x) can be found by setting the

first derivative equal to zero (dy/dx = 0 ) and solving for the value of x.

A function is at a maximum if dy/dx = 0 and d2y/dx2 < 0. Conversely,

A function is at a minimum if dy/dx = 0 and d2y/dx2 > 0.

Profit is maximised when MR=MC or Mπ = 0 and dMπ /dQ< 0 (or alternatively

dπ/dQ = 0 and d2π/dQ2 < 0).

Revenue is maximised when MR = 0

Average cost is minimised when MC = AC and average cost is increasing as

output expands.

For multivariate function (function with more than one independent variable -

y = f (x z) ), the concept of partial derivative ( ∂y /∂x) has to be employed. When we

partial derivate y with respect to x, we hold z constant meaning that we isolate the

marginal effect on y from changes in x only.

To maximise or minimise a multivariate function, we set each partial derivative

equal to zero and solve the resulting set of simultaneous equations for the optimal

value of the independent variables.


82

For constrained optimisation problems, the Lagrangian technique has to be

adopted. The Lagrangian function incorporates the objective function and the

constraint conditions. The Lagrangian multiplier, λ, indicates the marginal effect on

the objective function as a result of a unit increase or decrease in the constraint

function.

STUDY NOTES

chapter 2 in managerial economic

Documents

economic problems

optimisation analysis

chapter overview chapter

marginal analysis

optimisation process

chapter texts

marginal concepts

chapter places great