Jacques I. Mathematics for Economics and Business

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Partial Differentiation

410

Isocost curve A line showing all combinations of two factors which can be bought for a

ﬁxed cost.

Method of substitution The method of solving constrained optimization problems

whereby the constraint is used to eliminate one of the variables in the objective function.

Objective function A function that one seeks to optimize (usually) subject to constraints.

Key Terms

Practice Problems

4 Find the maximum value of

z = 6x − 3x

+ 2y

subject to the constraint

y − x

= 2

5 Find the maximum value of

z = 80x − 0.1x

+ 100y − 0.2y

subject to the constraint

x + y = 500

6 A firm’s production function is given by

Q = 10K

1/2

1/4

Unit capital and labour costs are $4 and $5 respectively and the firm spends a total of $60 on these

inputs. Find the values of K and L which maximize output.

7 A firm’s production function is given by

Q = 50KL

Unit capital and labour costs are $2 and $3 respectively. Find the values of K and L which minimize

total input costs if the production quota is 1200.

8 A firm’s production function is given by

Q = 2L

1/2

+ 3K

1/2

where Q, L and K denote the number of units of output, labour and capital respectively. Labour costs

are $2 per unit, capital costs are $1 per unit and output sells at $8 per unit. If the firm is prepared to

spend $99 on input costs, find the maximum profit and the values of K and L at which it is achieved.

[You might like to compare your answer with the corresponding unconstrained problem that you

solved in Practice Problem 7 of Section 5.4.]

9 A consumer’s utility function is

U = ln x

+ 2 ln x

Find the values of x

and x

which maximize U subject to the budgetary constraint

+ 3x

= 18

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section 5.6

Lagrange multipliers

We now describe the method of Lagrange multipliers for solving constrained optimization

problems. This is the preferred method, since it handles non-linear constraints and problems

involving more than two variables with ease. It also provides some additional information that

is useful when solving economic problems.

To optimize an objective function

f(x, y)

subject to a constraint

ϕ(x, y) = M

we work as follows.

Step 1

Deﬁne a new function

g(x, y, λ) = f (x, y) +λ[M −ϕ(x, y)]

Objectives

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

 Use the method of Lagrange multipliers to solve constrained optimization

problems.

 Give an economic interpretation of Lagrange multipliers.

 Use Lagrange multipliers to maximize a Cobb–Douglas production function

subject to a cost constraint.

 Use Lagrange multipliers to 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.

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Step 2

Solve the simultaneous equations

= 0

for the three unknowns, x, y and λ.

The basic steps of the method are straightforward. In step 1 we combine the objective func-

tion and constraint into a single function. To do this we ﬁrst rearrange the constraint as

M −ϕ(x, y)

and multiply by the scalar (i.e. number) λ (the Greek letter ‘lambda’). This scalar is called the

Lagrange multiplier. Finally, we add on the objective function to produce the new function

g(x, y, λ) = f(x, y) +λ[M −ϕ(x, y)]

This is called the Lagrangian function. The right-hand side involves the three letters x, y and λ,

so g is a function of three variables.

In step 2 we work out the three ﬁrst-order partial derivatives

and equate these to zero to produce a system of three simultaneous equations for the three

unknowns x, y and λ. The point (x, y) is then the optimal solution of the constrained problem.

The number λ can also be given a meaning and we consider this later. For the moment we con-

sider a simple example to get us started.

∂g

∂λ

∂g

∂y

∂g

∂x

∂g

∂λ

∂g

∂y

∂g

∂x

Partial Differentiation

412

Example

Use Lagrange multipliers to ﬁnd the optimal value of

− 3xy + 12x

subject to the constraint

2x + 3y = 6

Solution

Step 1

In this example

f(x, y) = x

− 3xy + 12x

ϕ(x, y) = 2x + 3y

M = 6

so the Lagrangian function is given by

g(x, y, λ) = x

− 3xy + 12x +λ(6 − 2x − 3y)

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Step 2

Working out the three partial derivatives of g gives

= 2x − 3y + 12 − 2λ

=−3x − 3λ

= 6 − 2x − 3y

so we need to solve the simultaneous equations

2x − 3y + 12 − 2λ=0

−3x − 3λ=0

6 − 2x − 3y = 0

that is,

2x − 3y − 2λ=−12 (1)

−3x − 3λ=0 (2)

2x + 3y = 6 (3)

We can eliminate x from equation (2) by multiplying equation (1) by 3, multiplying equation (2) by 2 and

adding. Similarly, x can be eliminated from equation (3) by subtracting equation (3) from (1). These opera-

tions give

−9y − 12λ=−36 (4)

−6y − 2λ=−18 (5)

The variable y can be eliminated by multiplying equation (4) by 6 and equation (5) by 9, and subtracting to

get

−54λ=−54 (6)

so λ=1. Substituting this into equations (5) and (2) gives y = 8/3 and x =−1 respectively.

The optimal solution is therefore (−1, 8/3) and the corresponding value of the objective function

− 3xy + 12x

(−1)

− 3(−1)(

/3) + 12(−1) =−3

∂g

∂λ

∂g

∂y

∂g

∂x

5.6 • Lagrange multipliers

413

Practice Problem

1 Use Lagrange multipliers to optimize

− xy

subject to

x + y = 12

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Looking back at the worked example and your own solution to Practice Problem 1, notice

that the third equation in step 2 is just a restatement of the original constraint. It is easy to see

that this is always the case because if

g(x, y, λ) = f(x, y) +λ[M −ϕ(x, y)]

then

= M −ϕ(x,y)

The equation

= 0

then implies the constraint

ϕ(x, y) = M

It is possible to make use of second-order partial derivatives to classify the optimal point.

Unfortunately, these conditions are quite complicated and are considered in Appendix 3. In all

problems that we consider there is only a single optimum and it is usually obvious on economic

grounds whether it is a maximum or a minimum.

∂g

∂λ

∂g

∂λ

Partial Differentiation

414

Example

A monopolistic producer of two goods, G1 and G2, has a joint total cost function

TC = 10Q

+ Q

+ 10Q

where Q

and Q

denote the quantities of G1 and G2 respectively. If P

and P

denote the corresponding

prices then the demand equations are

= 50 − Q

+ Q

= 30 + 2Q

− Q

Find the maximum proﬁt if the ﬁrm is contracted to produce a total of 15 goods of either type. Estimate the

new optimal proﬁt if the production quota rises by 1 unit.

Solution

The ﬁrst thing that we need to do is to write down expressions for the objective function and constraint. The

objective function is proﬁt and is given by

π=TR − TC

The total cost function is given to be

TC = 10Q

+ Q

+ 10Q

However, we need to use the demand equations to obtain an expression for TR. Total revenue from the sale

of G1 is

= P

= (50 − Q

+ Q

= 50Q

− Q

+ Q

and total revenue from the sale of G2 is

= P

= (30 + 2Q

− Q

= 30Q

+ 2Q

− Q

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TR = TR

+ TR

= 50Q

− Q

+ Q

+ 30Q

+ 2Q

− Q

= 50Q

− Q

+ 3Q

+ 30Q

− Q

Hence

π=TR − TC

= (50Q

− Q

+ 3Q

+ 30Q

− Q

) − (10Q

+ Q

+ 10Q

)

= 40Q

− Q

+ 2Q

+ 20Q

− Q

The constraint is more easily determined. We are told that the ﬁrm produces 15 goods in total, so

+ Q

= 15

The mathematical problem is to maximize the objective function

π=40Q

− Q

+ 2Q

+ 20Q

− Q

subject to the constraint

+ Q

= 15

Step 1

The Lagrangian function is

g(Q

, Q

, λ) = 40Q

− Q

+ 2Q

+ 20Q

− Q

+λ(15 − Q

− Q

)

Step 2

The simultaneous equations

= 0, = 0, = 0

are

40 − 2Q

+ 2Q

−λ=0

+ 20 − 2Q

−λ=0

15 − Q

− Q

= 0

that is,

−2Q

+ 2Q

−λ=−40 (1)

− 2Q

−λ=−20 (2)

+ Q

= 15 (3)

The obvious way of solving this system is to add equations (1) and (2) to get

−2λ=−60

so λ=30. Putting this into equation (1) gives

−2Q

+ 2Q

=−10 (4)

Equations (3) and (4) constitute a system of two equations for the two unknowns Q

and Q

. We can

eliminate Q

by multiplying equation (3) by 2 and adding equation (4) to get

= 20

∂g

∂λ

∂g

∂Q

∂g

∂Q

5.6 • Lagrange multipliers

415



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so Q

= 5. Substituting this into equation (3) gives

= 15 − 5 = 10

The maximum proﬁt is found by substituting Q

= 10 and Q

= 5 into the formula for π to get

π=40(10) − (10)

+ 2(10)(5) + 20(5) − 5

= 475

The ﬁnal part of this example wants us to ﬁnd the new optimal proﬁt when the production quota rises

by 1 unit. One way of doing this is just to repeat the calculations replacing the previous quota of 15 by 16,

although this is extremely tedious and not strictly necessary. There is a convenient shortcut based on the

value of the Lagrange multiplier λ. To understand this, let us replace the production quota, 15, by the vari-

able M, so that the Lagrangian function is

g(Q

, Q

, λ, M ) = 40Q

− Q

+ 2Q

+ 20Q

− Q

+λ(M − Q

− Q

)

The expression on the right-hand side involves Q

, Q

, λ and M, so g is now a function of four variables. If

we partially differentiate with respect to M then

=λ

We see that λ is a multiplier not only in the mathematical but also in the economic sense. It represents the

(approximate) change in g due to a 1 unit increase in M. Moreover, if the constraint is satisﬁed, then

+ Q

= M

and the expression for g reduces to

40Q

− Q

+ 2Q

+ 20Q

− Q

which is equal to proﬁt. The value of the Lagrange multiplier represents the change in optimal proﬁt

brought about by a 1 unit increase in the production quota. In this case, λ=30, so proﬁt rises by 30 to

become 505.

∂g

∂M

Partial Differentiation

416

The interpretation placed on the value of λ in this example applies quite generally. Given an

objective function

f(x, y)

and constraint

ϕ(x, y) = M

the value of λ gives the approximate change in the optimal value of f due to a 1 unit increase

in M.

Practice Problem

2 A consumer’s utility function is given by

U(x

, x

) = 2x

+ 3x

where x

and x

denote the number of items of two goods G1 and G2 that are bought. Each item costs

$1 for G1 and $2 for G2. Use Lagrange multipliers to find the maximum value of U if the consumer’s

income is $83. Estimate the new optimal utility if the consumer’s income rises by $1.

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5.6 • Lagrange multipliers

417

Example

Use Lagrange multipliers to ﬁnd expressions for K and L which maximize output given by a Cobb–Douglas

production function

Q = AK

(A, α and β are positive constants)

subject to a cost constraint

K + P

L = M

Solution

This example appears very hard at ﬁrst sight because it does not involve speciﬁc numbers. However, it is easy

to handle such generalized problems provided that we do not panic.

Step 1

The Lagrangian function is

g(K, L, λ) = AK

+λ(M − P

K − P

Step 2

The simultaneous equations

= 0, = 0, = 0

are

AαK

α−1

−λP

= 0 (1)

AβK

β−1

−λP

= 0 (2)

M − P

K − P

L = 0 (3)

These equations look rather forbidding. Before we begin to solve them it pays to simplify equations (1) and

(2) slightly by introducing Q = AK

. Notice that

AαK

α−1

AβK

β−1

so equations (1), (2) and (3) can be written

−λP

= 0 (4)

−λP

= 0 (5)

K + P

L = M (6)

Equations (4) and (5) can be rearranged to give

λ= and λ=

βQ

αQ

βQ

αQ

βQ

β(AK

)

αQ

α(AK

)

∂g

∂λ

∂g

∂L

∂g

∂K



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so that

and hence

= (divide both sides by Q and turn both sides upside down)

that is,

K = P

L (multiply through by α) (7)

Substituting this into equation (6) gives

L + P

L = M

αL +βL = (multiply through by β/P

)

(α+β)L = (factorize)

L = (divide through by α+β)

Finally, we can put this into equation (7) to get

K =

The values of K and L which optimize Q are therefore

and

βM

(α+β)P

αM

(α+β)P

αM

(α+β)P

αM

α+β

βM

(α+β)P

βM

βQ

αQ

Partial Differentiation

418

Practice Problem

3 Use Lagrange multipliers to find expressions for x

and x

which maximize the utility function

U = x

1/2

+ x

1/2

subject to the general budgetary constraint

+ P

= M

The previous example illustrates the power of mathematics when solving economics prob-

lems. The main advantage of using algebra and calculus rather than just graphs and tables of

numbers is their generality. In future, if we need to maximize any particular Cobb–Douglas

production function subject to any particular cost constraint, then all we have to do is to quote

the result of the previous example. By substituting speciﬁc values of M, α, β, P

and P

into the

general formulas for K and L, we can write down the solution in a matter of seconds. In fact,

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we can use mathematics to generalize still further. Rather than work with production functions

of a prescribed form such as

Q = AK

we can obtain results pertaining to any production function

Q = f(K, L)

For instance, we can use Lagrange multipliers to justify a result that we derived graphically in

Section 5.5. At the beginning of that section we showed that when output is maximized subject

to a cost constraint, the ratio of marginal product to price is the same for all inputs. To obtain

this result using Lagrange multipliers we simply write down the Lagrangian function

g(K, L, λ) = f(K, L) +λ(M − P

K − P

which corresponds to a production function

f(K, L)

and cost constraint

K + P

L = M

The simultaneous equations

= 0, = 0, = 0

are

−λP

= 0 (1)

−λP

= 0 (2)

M − P

K − P

L = 0 (3)

because

= MP

and = MP

Equations (1) and (2) can be rearranged to give

λ= and λ=

as required.

∂f

∂L

∂f

∂K

∂g

∂λ

∂g

∂L

∂g

∂K

5.6 • Lagrange multipliers

419

Lagrange multiplier The number

which is used in the Lagrangian function. In eco-

nomics this gives the change in the value of the objective function when the value of the con-

straint is increased by 1 unit.

Lagrangian The function f(x, y) +λ[M −ϕ(x, y)], where f(x, y) is the objective function

and φ(x, y) = M is the constraint. The stationary point of this function is the solution of the

associated constrained optimization problem.

Key Terms

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