Basic Mathematics for Economists - Rosser - Chapter 11 pps

The substitution method is mainly suitable for problems where a function withonly two variables is maximized or minimized subject to one constraint.. Solution The problem is to maximize

11 Constrained optimization

Learning objectives

After completing this chapter students should be able to:

• Solve constrained optimization problems by the substitution method

• Use the Lagrange method to set up and solve constrained maximization andconstrained minimization problems

• Apply the Lagrange method to resource allocation problems in economics

11.1 Constrained optimization and resource allocation

Chapters 9and10dealt with the optimization of functions without any constraints imposed.However, in economics we often come across resource allocation problems that involvethe optimization of some variable subject to certain limitations For example, a firm maytry to maximize output subject to a budget constraint for expenditure on inputs, or it maywish to minimize costs subject to a specified output being produced We have already seen

inChapter 5how constrained optimization problems with linear constraints and objectivefunctions can be tackled using linear programming This chapter now explains how problemsinvolving the constrained optimization of non-linear functions can be tackled, using partialdifferentiation

We shall consider two methods:

(i) constrained optimization by substitution, and

(ii) the Lagrange multiplier method

The Lagrange multiplier method can be used for most types of constrained optimizationproblems The substitution method is mainly suitable for problems where a function withonly two variables is maximized or minimized subject to one constraint We shall considerthis simpler substitution method first

11.2 Constrained optimization by substitution

Consider the example of a firm that wishes to maximize output Q = f(K, L), with a fixed budget M for purchasing inputs K and L at set prices PK and PL This problem is illustrated

inFigure 11.1 The firm needs to find the combination of K and L that will allow it to reach

the optimum point X which is on the highest possible isoquant within the budget constraint

with intercepts M/PK and M/PL.

To determine a solution for this type of economic resource allocation problem we have toreformulate it as a mathematical constrained optimization problem The following examplessuggest ways in which this can be done

Example 11.1

A firm faces the production function Q = 12K 0.4 L 0.4 and can buy the inputs K and L atprices per unit of £40 and £5 respectively If it has a budget of £800 what combination of Kand L should it use in order to produce the maximum possible output?

Solution

The problem is to maximize the function Q = 12K 0.4 L 0.4subject to the budget constraint

(In all problems in this chapter, it is assumed that each constraint ‘bites’; e.g all the budget

is used in this example.)

The theory of the firm tells us that a firm is optimally allocating a fixed budget if the last £1spent on each input adds the same amount to output, i.e marginal product over price should

be equal for all inputs This optimization condition can be written as

MPK

PK = MPL

PL

(2)

The marginal products can be determined by partial differentiation:

Note that although this method allows us to derive optimum values of K and L that satisfy

condition (2) above, it does not provide a check on whether this is a unique solution, i.e there

is no second-order condition check However, it may be assumed that in all the problems inthis section the objective function is maximized (or minimized depending on the question)when the basic economic rules for an optimum are satisfied

The above method is not the only way of tackling this problem by substitution An native approach, explained below, is to encapsulate the constraint within the function to bemaximized, and then maximize this new objective function

Substituting (2) into the objective function Q = 12K L gives

Q = 12K 0.4 (160− 8K) 0.4

(3)

We are now faced with the unconstrained optimization problem of finding the value of K that

maximizes the function (3) which has the budget constraint (1) ‘built in’ to it by substitution

This requires us to set d Q/dK = 0 However, it is not straightforward to differentiate thefunction in (3), and we must wait until further topics in calculus have been covered beforeproceeding with this solution (seeChapter 12, Example 12.9)

To make sure that you understand the basic substitution method, we shall use it to tackleanother constrained maximization problem

L = 3K = 18

The examples of constrained optimization considered so far have only involved outputmaximization when a firm faces a Cobb–Douglas production function, but the same techniquecan also be applied to other forms of production functions

Example 11.3

A firm faces the production function

Q = 120L + 200K − L2− 2K2

for positive values of Q It can buy L at £5 a unit and K at £8 a unit and has a budget of £70.

What is the maximum output it can produce?

Substituting this result into (1)

The utility a consumer derives from consuming the two goods A and B can be assumed to be

determined by the utility function U = 40A 0.25 B 0.5 If A costs £4 a unit and B costs £10 aunit and the consumer’s income is £600, what combination of A and B will maximize utility?

B = 0.8(50) = 40

The substitution method can also be used for constrained minimization problems If

output is given and a firm is required to minimize the cost of this output, then one variablecan be eliminated from the production function before it is substituted into the cost functionwhich is to be minimized

Differentiating and setting equal to zero to find a stationary point

Thus cost minimization is achieved when K = 119.16 and L = 212.84 (to 2 dp) and so total

production costs will be

TC= 40(119.16) + 15(211.84) = £7,944

Test Yourself, Exercise 11.1

1 If a firm has a budget of £378 what combination of K and L will maximize output given the production function Q = 40K 0.6 L 0.3and prices for K and L of £20 perunit and £6 per unit respectively?

2 A firm faces the production function Q = 6K 0.4 L 0.5 If it can buy input K at £32

a unit and input L at £8 a unit, what combination of L and K should it use to maximize production if it is constrained by a fixed budget of £36,000?

3 A consumer spends all her income of £120 on the two goods A and B Good A

costs £10 a unit and good B costs £15 What combination of A and B will she purchase if her utility function is U = 4A 0.5 B 0.5?

4 If a firm faces the production function Q = 4K 0.5 L 0.5, what is the maximum

output it can produce for a budget of £200? The prices of K and L are given as £4

per unit and £2 per unit respectively

5 Make up your own constrained optimization problem for an objective functionwith two independent variables and solve it using the substitution method

6 A firm faces the production function Q = 2K L and can buy L at £240 a unitand K at £4 a unit.

(a) If it has a budget of £16,000 what combination of K and L should it use to

maximize output?

(b) If it is given a target output of 40 units of Q what combination of K and L

should it use to minimize the cost of this output?

7 A firm has a budget of £1,140 and can buy inputs K and L at £3 and £8 respectively

a unit Its output is determined by the production function

Q = 6K + 20L − 0.025K2− 0.05L2

for positive values of Q What is the maximum output it can produce?

8 A firm operates with the production function Q = 30K 0.4 L 0.2and buys inputs Kand L at £12 per unit and £5 per unit respectively What is the cheapest way ofproducing 750 units of output? (Work to nearest whole units of K and L.)

11.3 The Lagrange multiplier: constrained maximization

with two variables

The best way to explain how to use the Lagrange multiplier is with an example and so weshall work through the problem in Example 11.1 from the last section using the Lagrangemultiplier method

The firm is trying to maximize output Q = 12K 0.4 L 0.4 subject to the budget constraint

40K + 5L = 800 The first step is to rearrange the budget constraint so that zero appears on

one side of the equality sign Therefore

We then write the ‘Lagrange equation’ or ‘Lagrangian’ in the form

G = (function to be optimized) + λ(constraint)

where G is just the value of the Lagrangian function and λ is known as the ‘Lagrange

multiplier’ (Do not worry about where these terms come from or what their actual values

are They are just introduced to help the analysis Note also that in other texts a ‘curly L’ is

often used to represent the Lagrange function This can confuse students because economicsproblems frequently involve labour, represented by L, as one of the variables in the function

to be optimized This text therefore uses the notation ‘G’ to avoid this confusion However,

if you are already accustomed to using the ‘curly L’ you can, of course, continue to use it

when answering problems yourself What matters is whether you understand the analysis,not what symbols you use.)

In this problem the Lagrange function is thus

G = 12K 0.4 L 0.4 + λ(800 − 40K − 5L) (2)

Next, derive the partial derivatives of G with respect to K, L and λ and set them equal to zero, i.e find the stationary points of G that satisfy the first-order conditions for a maximum.

You will note that (5) is the same as the budget constraint (1) We now have a set of three linear

simultaneous equations in three unknowns to solve for K and L The Lagrange multiplier λ

can be eliminated as, from (3),

constraint We shall just accept this result without going into the proof of why this is so.Strictly speaking we should now check the second-order conditions in the above problem

to be sure that we actually have a maximum rather than a minimum These, however, arerather complex, involving an examination of the function at and near the stationary pointsfound, and are discussed in the next section For the time being you can assume that oncethe stationary points of a Lagrangian function have been found the second-order conditionsfor a maximum will automatically be met Some more examples are worked through so thatyou can become familiar with this method

Example 11.6

A firm can buy two inputs K and L at £18 per unit and £8 per unit respectively and faces

the production function Q = 24K 0.6 L 0.3 What is the maximum output it can produce for a

budget of £50,000? (Work to nearest whole units of K, L and Q.)

Solution

The budget constraint is 50,000 − 18K − 8L = 0 and the function to be maximized is

Q = 24K 0.6 L 0.3 The Lagrangian for this problem is therefore

We can check that when these whole values of K and L are used the total cost will be

Solution

To help solve this problem the relevant budget schedules and indifference curves are illustrated

in Figure 11.2, although the indifference curves are not accurately drawn to scale The originaloptimum is at X The price fall for A causes the budget line to become flatter and swing round,giving a new equilibrium at Y

Hicks’s method for splitting the total change in A into its income and substitution effectsrequires one to draw a ‘ghost’ budget line parallel to the new budget line (reflecting the new

Figure 11.2

relative prices) but tangential to the original indifference curve This is shown by the brokenline tangential to indifference curve I at H From X to H is the substitution effect and from H to

Y is the income effect of the price change This problem requires us to find the corresponding

values of A and B for the three tangency points X, Y and H and then to comment on the

direction of these changes

The original equilibrium is the combination of A and B that maximizes the utility function

U = 40A 0.5 B 0.5subject to the budget constraint 600= 20A + 5B These values of A and

Bcan be found by deriving the stationary points of the Lagrange function

Thus, A = 15 and B = 60 at the original equilibrium at X.

When the price of A falls to 10, the budget constraint becomes

There are several ways of finding the values of A and B that correspond to point H We

know that H is on the same indifference curve as point X, and therefore the utility function

will take the same value at both points We can find the value of utility at X where A= 15

and B = 60 This will be

Thus the substitution effect of A’s price fall, from X to H, increases consumption of A from

15 to 21.2 units and decreases consumption of B from 60 to 42.4 units This effect is negative(i.e quantity rises when price falls) in line with standard consumer theory

The income effect, from H to Y, increases consumption of A from 21.2 to 30 units andalso increases consumption of B from 42.4 back to its original 60 unit level As both incomeeffects are positive, both A and B must be normal goods

Test Yourself, Exercise 11.2

Use the Lagrange method to answer questions 1, 2, 3, 4, 6(a) and 7 from TestYourself, Exercise 11.1

11.4 The Lagrange multiplier: second-order conditions

Inasmuch as it involves setting the first derivatives of the objective function equal to zero,the Lagrange method of solving constrained optimization problems is similar to the method

of solving unconstrained optimization problems involving functions of several variables thatwas explained inChapter 10 However, one cannot simply apply the same set of second-orderconditions to check for a maximum or minimum because of the special role that the Lagrangemultiplier takes The mathematics required to prove why this is so, and to explain whatadditional second-order conditions are necessary for a Lagrangian function to be a maximum

or minimum, becomes rather complex We shall therefore just look at an intuitive explanation

of what these conditions involve here The use of matrix algebra to check second-orderconditions in constrained optimization problems will then be explained later inChapter 15.First, we shall consider the conditions for a maximum If we assume that a function has two

independent variables then both the function to be maximized, f(A, B), and the constraint

could take on several possible forms, as illustrated in Figure 11.3 These diagrams are allconstructed on the same basis as isoquant maps Thus the lines I, II and III represent differentlevels of the objective function with its value increasing as one moves away from the origin

In Figure 11.3(a), the objective function is convex to the origin and the constraint CD islinear Maximization of the objective function occurs at the tangency point T

T

C R

S

III II I

Figure 11.3

InFigure 11.3(b), the objective function is concave to the origin and so it is maximized

subject to the linear constraint CD at the corner point C Thus the tangency point T does not

determine the maximum value

In Figure 11.3(c), the objective function is convex to the origin but the constraint CD isnon-linear and more sharply curved than the objective function Thus the tangency point T

is not the maximum value Higher values of the objective function can be found at points R

and S, for example

From the above examples we can see that, for the two-variable case, a linear constraintand an objective function that is convex to the origin will ensure maximization at the point

of tangency In other cases, tangency may not ensure maximization If a problem involvesthe maximization of production subject to a linear budget constraint this means that output

is maximized where the slope of the isoquant is equal to the slope of the budget constraint

We have already seen inChapter 10how a Cobb–Douglas production function in the form

Q = AK α L β will correspond to a set of isoquants which continually decline and become

flatter as L is increased, i.e are convex to the origin Thus in any constrained optimization

problem where one is attempting to maximize a production function in the Cobb–Douglasformat subject to a linear budget constraint, the input combination that satisfies the first-orderconditions will be a maximum

If the Lagrangian represents other concepts with similar shaped functions and constraints,such as utility, the same conditions apply In all such cases, one assumes that the independentvariables in the objective function must take positive values and so any negative mathematicalsolutions can be disregarded

Although we cannot illustrate functions with more than two variables diagrammatically, thesame basic principles apply when one is attempting to maximize a function with three or morevariables subject to a linear constraint Thus any Cobb–Douglas production function withmore than two inputs will be at a maximum, subject to a specified linear budget constraint,when the first-order conditions for optimization of the relevant Lagrange equation are met Forthe purpose of answering the problems in this Chapter, and for most constrained maximizationproblems that you will encounter in a first-year economics course, it can be assumed thatthe stationary points of the Lagrange function will satisfy the second-order conditions for amaximum

The properties of the objective function and constraint that guarantee that a Lagrangefunction is minimized when first-order conditions are met are the reverse of the propertiesrequired for a maximum, i.e the objective function must be linear and the constraint must

be convex to the origin Thus if one is required to find values of K and L that minimize a

budget function in the form