Jacques I. Mathematics for Economics and Business

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section 5.5

Constrained optimization

In Section 5.4 we described how to ﬁnd the optimum (that is, maximum or minimum) of a

function of two variables

z = f(x, y)

where the variables x and y are free to take any values. As we pointed out at the beginning of

that section, this assumption is unrealistic in many economic situations. An individual wishing

to maximize utility is subject to an income constraint and a ﬁrm wishing to maximize output

is subject to a cost constraint.

Objectives

At the end of this section you should be able to:

 Give a graphical interpretation of constrained optimization.

 Show that when a ﬁrm maximizes output subject to a cost constraint, the ratio

of marginal product to price is the same for all inputs.

 Show that when a consumer maximizes utility subject to a budgetary constraint,

the ratio of marginal utility to price is the same for all goods.

 Use the method of substitution to solve constrained optimization problems in

economics.

Advice

In this section we begin by proving some theoretical results before describing the method

of substitution. You might prefer to skip the theory at a first reading, and begin with the

two worked examples.

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In general, we want to optimize a function,

z = f(x, y)

called the objective function subject to a constraint

(x, y) = M

Here

, the Greek letter phi, is a known function of two variables and M is a known constant.

The problem is to pick the pair of numbers (x, y) which maximizes or minimizes f(x, y) as

before. This time, however, we limit the choice of pairs to those which satisfy

(x, y) = M

A graphical interpretation should make this clear. To be speciﬁc, let us suppose that a ﬁrm

wants to maximize output and that the production function is of the form

Q = f(K, L)

Let the costs of each unit of capital and labour be P

and P

respectively. The cost to the ﬁrm

of using as input K units of capital and L units of labour is

K + P

so if the ﬁrm has a ﬁxed amount, M, to spend on these inputs then

K + P

L = M

The problem is one of trying to maximize the objective function

Q = f(K, L)

subject to the cost constraint

K + P

L = M

Sketched in Figure 5.14 (overleaf) is a typical isoquant map. As usual, points on any one isoquant

yield the same level of output and as output rises the isoquants themselves move further away

from the origin. Also sketched in Figure 5.14 is the cost constraint. This is called an isocost

curve because it gives all combinations of K and L which can be bought for a ﬁxed cost, M.

The fact that

K + P

L = M

is represented by a straight line should come as no surprise to you by now. We can even rewrite

it in the more familiar ‘y = ax + b’ form and so identify its slope and intercept. In Figure 5.14,

L is plotted on the horizontal axis and K is plotted on the vertical axis, so we need to rearrange

K + P

L = M

to express K in terms of L. Subtracting P

L from both sides and dividing through by P

gives

K =− L +

The isocost curve is therefore a straight line with slope −P

and intercept M/P

. Graphically,

our constrained problem is to choose that point on the isocost line which maximizes output.

This is given by the point labelled A in Figure 5.14. Point A certainly lies on the isocost line and

it maximizes output because it also lies on the highest isoquant. Other points, such as B and C,

also satisfy the constraint but they lie on lower isoquants and so yield smaller levels of output

than A. Point A is characterized by the fact that the isocost line is tangential to an isoquant. In

other words, the slope of the isocost line is the same as that of the isoquant at A.

5.5 • Constrained optimization

401

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Partial Differentiation

402

Figure 5.14

Now we have already shown that the isocost line has slope −P

. In Section 5.2 we deﬁned

the marginal rate of technical substitution, MRTS, to be minus the slope of an isoquant, so at

point A we must have

= MRTS

We also showed that

MRTS =

the ratio of the input prices is equal to the ratio of their marginal products

This relationship can be rearranged as

so when output is maximized subject to a cost constraint

the ratio of marginal product to price is the same for all inputs

The marginal product determines the change in output due to a 1 unit increase in input. This

optimization condition therefore states that the last dollar spent on labour yields the same

addition to output as the last dollar spent on capital.

The above discussion has concentrated on production functions. An analogous situation

arises when we maximize utility functions

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5.5 • Constrained optimization

403

U = U(x

, x

)

where x

, x

denote the number of items of goods G1, G2 that an individual buys. If the prices

of these goods are denoted by P

and P

and the individual has a ﬁxed budget, M, to spend on

these goods then the corresponding constraint is

+ P

= M

This budgetary constraint plays the role of the cost constraint, and indifference curves are ana-

logous to isoquants. Consequently, we analyse the problem by superimposing the budget line

on an indifference map. The corresponding diagram is virtually indistinguishable from that of

Figure 5.14. The only change is that the axes would be labelled x

and x

rather than L and K.

Once again, the maximum point of the constrained problem occurs at the point of tangency,

so that at this point the slope of the budget line is that of an indifference curve. Hence

= MRCS

In Section 5.2 we derived the result

MRCS =

Writing the partial derivatives ∂U/∂x

more concisely as U

we can deduce that

that is,

the ratio of the prices of the goods is equal to the ratio of their marginal utilities

Again, this relationship can be rearranged into the more familiar form

so when utility is maximized subject to a budgetary constraint,

the ratio of marginal utility to price is the same for all goods consumed

If individuals allocate their budgets between goods in this way then utility is maximized when

the last dollar spent on each good yields the same addition to total utility. Under these cir-

cumstances, the consumer has achieved maximum satisfaction within the constraint of a ﬁxed

budget, so there is no tendency to reallocate income between these goods. Obviously, the con-

sumer’s equilibrium will be affected if there is a change in external conditions such as income

or the price of any good. For example, suppose that P

suddenly increases, while P

and M

remain ﬁxed. If this happens then the equation

turns into an inequality

∂U/∂x

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so equilibrium no longer holds. Given that P

has increased, consumers ﬁnd that the last dol-

lar spent no longer buys as many items of G1, so utility can be increased by purchasing more

of G2 and less of G1. By the law of diminishing marginal utility, the effect is to increase U

and

to decrease U

. The process of reallocation continues until the ratio of marginal utilities to

prices is again equal and equilibrium is again established.

The graphical approach provides a useful interpretation of constrained optimization. It has

also enabled us to justify some familiar results in microeconomics. However, it does not give

us a practical way of actually solving such problems. It is very difﬁcult to produce an accurate

isoquant or indifference map from any given production or utility function. We now describe

an alternative approach, known as the method of substitution. To illustrate the method we

begin with an easy example.

Partial Differentiation

404

Example

Find the minimum value of the objective function

z =−2x

+ y

subject to the constraint y = 2x − 1.

Solution

In this example we need to optimize the function

z =−2x

+ y

given that x and y are related by

y = 2x − 1

The obvious thing to do is to substitute the expression for y given by the constraint directly into the func-

tion that we are trying to optimize to get

z =−2x

+ (2x − 1)

=−2x

+ 4x

− 4x + 1

= 2x

− 4x + 1

Note the wonderful effect that this has on z. Instead of z being a function of two variables, x and y, it is now

just a function of the one variable, x. Consequently, the minimum value of z can be found using the theory

of stationary points discussed in Chapter 4.

At a stationary point

= 0

that is,

4x − 4 = 0

which has solution x = 1. Differentiating a second time we see that

= 4 > 0

conﬁrming that the stationary point is a minimum. The value of z can be found by substituting x = 1 into

z = 2x

− 4x + 1

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to get

z = 2(1)

− 4(1) + 1 =−1

It is also possible to ﬁnd the value of y at the minimum. To do this we substitute x = 1 into the constraint

y = 2x − 1

to get

y = 2(1) − 1 = 1

The constrained function therefore has a minimum value of −1 at the point (1, 1).

5.5 • Constrained optimization

405

Practice Problem

1 Find the maximum value of the objective function

z = 2x

− 3xy + 2y + 10

subject to the constraint y = x.

The method of substitution for optimizing

z = f(x, y)

subject to

ϕ(x, y) = M

may be summarized as follows.

Step 1

Use the constraint

ϕ(x, y) = M

to express y in terms of x.

Step 2

Substitute this expression for y into the objective function

z = f(x, y)

to write z as a function of x only.

Step 3

Use the theory of stationary points of functions of one variable to optimize z.

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To illustrate this we now use the method of substitution to solve two economic problems

that both involve production functions. In the ﬁrst example output is maximized subject to

cost constraint and in the second example cost is minimized subject to an output constraint.

Partial Differentiation

406

Advice

The most difficult part of the three-step strategy is step 1, where we rearrange the given

constraint to write y in terms of x. In the previous example and in Practice Problem 1 this step

was exceptionally easy because the constraint was linear. In both cases the constraint was

even presented in the appropriate form to begin with, so no extra work was required. In

general, if the constraint is non-linear, it may be difficult or impossible to perform the initial

rearrangement. If this happens then you could try working the other way round and

expressing x in terms of y, although there is no guarantee that this will be possible either.

However, when step 1 can be tackled successfully, the method does provide a really quick

way of solving constrained optimization problems.

Example

A ﬁrm’s unit capital and labour costs are $1 and $2 respectively. If the production function is given by

Q = 4LK + L

ﬁnd the maximum output and the levels of K and L at which it is achieved when the total input costs are

ﬁxed at $105. Verify that the ratio of marginal product to price is the same for both inputs at the optimum.

Solution

We are told that 1 unit of capital costs $1 and that 1 unit of labour costs $2. If the ﬁrm uses K units of

capital and L units of labour then the total input costs are

K + 2L

This is ﬁxed at $105, so

K + 2L = 105

The mathematical problem is to maximize the objective function

Q = 4LK + L

subject to the constraint

K + 2L = 105

The three-step strategy is as follows:

Step 1

Rearranging the constraint to express K in terms of L gives

K = 105 − 2L

Step 2

Substituting this into the objective function

Q = 4LK + L

gives

Q = 4L(105 − 2L) + L

= 420L − 7L

and so output is now a function of the one variable, L.

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Step 3

At a stationary point

= 0

that is,

420 − 14L = 0

which has solution L = 30. Differentiating a second time gives

=−14 < 0

conﬁrming that the stationary point is a maximum.

The maximum output is found by substituting L = 30 into the objective function

Q = 420L − 7L

to get

Q = 420(30) − 7(30)

= 6300

The corresponding level of capital is found by substituting L = 30 into the constraint

K = 105 − 2L

to get

K = 105 − 2(30) = 45

The ﬁrm should therefore use 30 units of labour and 45 units of capital to produce a maximum output

of 6300.

Finally, we are asked to check that the ratio of marginal product to price is the same for both inputs. From

the formula

Q = 4LK + L

we see that the marginal products are given by

==4K + 2L and MP

==4L

so at the optimum

= 4(45) + 2(30) = 240

and

= 4(30) = 120

The ratios of marginal products to prices are then

==120

and

==120

which are seen to be the same.

120

240

∂Q

∂K

∂Q

∂L

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407

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Partial Differentiation

408

Practice Problem

2 An individual’s utility function is given by

U = x

where x

and x

denote the number of items of two goods, G1 and G2. The prices of the goods are $2

and $10 respectively. Assuming that the individual has $400 available to spend on these goods, find

the utility-maximizing values of x

and x

. Verify that the ratio of marginal utility to price is the same

for both goods at the optimum.

Example

A ﬁrm’s production function is given by

Q = 2K

1/2

Unit capital and labour costs are $4 and $3 respectively. Find the values of K and L which minimize total

input costs if the ﬁrm is contracted to provide 160 units of output.

Solution

Given that capital and labour costs are $4 and $3 per unit, the total cost of using K units of capital and L

units of labour is

TC = 4K + 3L

The ﬁrm’s production quota is 160, so

1/2

= 160

The mathematical problem is to minimize the objective function

TC = 4K + 3L

subject to the constraint

1/2

= 160

Step 1

Rearranging the constraint to express L in terms of K gives

1/2

= (divide both sides by 2K

1/2

)

L = (square both sides)

Step 2

Substituting this into the objective function

TC = 4K + 3L

gives

6400

1/2

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TC = 4K +

and so total cost is now a function of the one variable, K.

Step 3

At a stationary point

= 0

that is,

4 − = 0

This can be written as

4 =

so that

= = 4800

Hence

K =√4800 = 69.28

Differentiating a second time gives

=>0 because K > 0

conﬁrming that the stationary point is a minimum.

Finally, the value of L can be found by substituting K = 69.28 into the constraint

L =

to get

L ==92.38

We are not asked for the minimum cost, although this could easily be found by substituting the values of K

and L into the objective function.

6400

69.28

6400

38 400

(TC)

19 200

d(TC)

19 200

5.5 • Constrained optimization

409

Practice Problem

3 A firm’s total cost function is given by

TC = 3x

+ 2x

+ 7x

where x

and x

denote the number of items of goods G1 and G2, respectively, that are produced. Find

the values of x

and x

which minimize costs if the firm is committed to providing 40 goods of either

type in total.

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