Basic Mathematics for Economists - Rosser - Chapter 12 ppsx

• Check the second-order conditions for optimization of relevant economic tions using the quotient rule for differentiation.. To differentiate y with respect to x in this type of functio

12 Further topics in calculus

Learning objectives

After completing this chapter students should be able to:

• Use the chain, product and quotient rules for differentiation

• Choose the most appropriate method for differentiating different forms offunctions

• Check the second-order conditions for optimization of relevant economic tions using the quotient rule for differentiation

func-• Integrate simple functions

• Use integration to determine total cost and total revenue from marginal cost andmarginal revenue functions

• Understand how a definite integral relates to the area under a function and applythis concept to calculate consumer surplus

12.1 Overview

In this chapter, some techniques are introduced that can be used to differentiate functions thatare rather more complex than those encountered inChapters 8,9,10and11 These are thechain rule, the product rule and the quotient rule As you will see in the worked examples, it

is often necessary to combine several of these methods to differentiate some functions Theconcept of integration is also introduced

12.2 The chain rule

The chain rule is used to differentiate ‘functions within functions’ For example, if we havethe function

To differentiate y with respect to x in this type of function we use the chain rule which

to allow differentiation via the chain rule Assume, for example, that you wish to find anexpression for the slope of the non-linear demand function

p = (150 − 0.2q) 0.5

(1)The basic rules for differentiation explained in Chapter 8 cannot cope with this sort offunction However, if we define a new function

The marginal revenue productivity theory of the demand for labour

In the marginal revenue productivity theory of the demand for labour, the rule for profitmaximization is to employ additional units of labour as long as the extra revenue generated

by selling the extra output produced by an additional unit of labour exceeds the marginal cost

of employing this additional unit of labour This rule applies in the short run when inputsother than labour are assumed fixed

The optimal amount of labour is employed when

MRPL= MR × MPL

This is the rule for determining the profit-maximizing amount of labour which you shouldencounter in your microeconomics course

Example 12.3

A firm is a monopoly seller of good q and faces the demand schedule p = 200−2q, where p

is the price in pounds, and the short-run production function q = 4L 0.5 If it can buy labour

at a fixed wage of £8, how much L should be employed to maximize profit, assuming other

inputs are fixed?

Using the chain rule we need to derive a formula for MRPLin terms of L and then set it equal

to £8, given that MCLis fixed at this wage rate As

which is the optimal employment level

In the example above the idea of a ‘short-run production function’ was used to simplify

the analysis, where the input of capital (K) was implicitly assumed to be fixed Now that

you understand how an MRPLfunction can be derived we can work with full production

functions in the format Q = f(K, L) The effect of one input increasing while the other is held constant can now be shown by the relevant partial derivative.

MPL= ∂Q

∂L

The same chain rule can be used for partial derivatives, and full and partial derivatives can

be combined, as in the following examples

Example 12.4

A firm operates with the production function q = 45K 0.7 L 0.4and faces the demand function

p = 6,980 − 6q Derive its MRPLfunction

As we wish to derive MRPLas a function of L, we substitute the production function given

in the question into (3) for q Thus

MRPL= [6,980 − 12(45K 0.7 L 0.4 ) ]18K 0.7 L −0.6

= 125,640K 0.7 L −0.6 − 9,720K 1.4 L −0.2

Note that the value MRPL will depend on the amount that K is fixed at, as well as the value of L.

Point elasticity of demand

The chain rule can help the calculation of point elasticity of demand for some non-lineardemand functions

Sometimes it may be possible to simplify an expression in order to be able to differentiate

it, but one may instead use the chain rule if it is more convenient The same result will beobtained by both methods, of course

Example 12.6

Differentiate the function y = (6 + 4x)2

Test Yourself, Exercise 12.1

1 A firm operates in the short run with the production function q = 2L 0.5and faces

the demand schedule p = 60 − 4q where p is price in pounds If it can employ

labour at a wage rate of £4 per hour, how much should it employ to maximizeprofit?

2 If a supply schedule is given by p = (2 + 0.05q)2show (a) by multiplying out,

and (b) by using the chain rule, that its slope is 2.2 when q is 400.

3 The return R on a sum M invested at i per cent for 3 years is given by the formula

6 A firm operates with the production function q = 0.4K 0.5 L 0.5and sells its output

in a market where it is a monopoly with the demand schedule p = 60 − 2q If K

is fixed at 25 units and the wage rate is £7 per unit of L, derive the MRPLfunctionand work out how much L the firm should employ to maximize profit

7 A firm faces the demand schedule p = 650 − 3q and the production function

q = 4K 0.5 L 0.5 and has to pay £8 per unit to buy L If K is fixed at 4 units how

much L should the firm use if it wishes to maximize profits?

8 If a firm operates with the total cost function TC= 4 + 10(9 + q2) 0.5, what is its

marginal cost when q is 4?

9 Given the production function q = (6K 0.5 + 0.5L 0.5 ) 0.3, find MPLwhen K is 16 and L is 576.

12.3 The product rule

The product rule allows one to differentiate two functions which are multiplied together

If y = uv where u and v are both functions of x, then according to the product rule dy

by the two methods They should, of course, be the same

We are given the function

When it is not possible to multiply out the different components of a function then onemust use the product rule to differentiate One may also need to use the chain rule to helpdifferentiate the different sub-functions.

We now need to check which of these values of q satisfies the second-order condition

for a maximum (You should immediately be able to see why it will not be 2,600 byobserving what happens when this quantity is substituted into the demand function.) To

derive d TR/dq we need to use the product rule again to differentiate dTR/dq From (1)

double check that the other stationary point will not maximize TR by substituting the value

As we have already left this example in mid-solution once already, it will not do any harm

to leave it once again Although the second-order condition could be worked out using theproduct rule it is more convenient to use the quotient rule in this case and so we shall continuethis problem later, in Example 12.13

Example 12.10

In a perfectly competitive market the demand schedule is p = 120 − 0.5q2and the supply

schedule is p = 20 + 2q2 If the government imposes a per-unit tax t on the good sold in this market, what level of t will maximize the government’s tax yield?

The government’s tax yield (TY) is tq Substituting (1) for q, this gives

Test Yourself, Exercise 12.2

4 If Q = 120K 0.5 (250− 0.5K) 0.3 at what value of K will dQ/dK = 0? (That is,

find the first-order condition for maximization of Q.)

5 In a perfectly competitive market the demand schedule is p = 600 − 4q 0.5 and

the supply schedule is p = 30 + 6q 0.5 What level of a per-unit tax levied on thegood sold in this market will maximize the government’s tax yield?

6 Make up your own function involving the product of two sub-functions and thendifferentiate it using the product rule

7 For the demand schedule p = (60 − 0.1q) 0.5:

(a) derive an expression for the slope of the demand schedule;

(b) demonstrate that this slope gets flatter as q increases from 0 to 600;

(c) find the output at which total revenue is a maximum

12.4 The quotient rule

The quotient rule allows one to differentiate two functions where one function is divided bythe other function

If y = u/v where u and v are functions of x, then according to the quotient rule

Therefore, according to the quotient rule,

This solution could also have been found using the product rule, since any function in the

form y = u/v can be written as y = uv−1 We can check this by reworking Example 12.11

and differentiating the function y = 4x2(8+ 0.2x)−1.

Defining relevant sub-functions and differentiating them

The answers (1) and (2) are identical, as expected

Whether one chooses to use the quotient rule or the product rule depends on the functions

to be differentiated Only practice will give you an idea of which will be the easier to use forspecific examples

Example 12.12

Derive a function for marginal revenue (in terms of q) if a monopoly faces the non-linear demand schedule p= 252

(4+ q) 0.5

which holds when K = 10.

To derive d2Q/ dK2let u = 768 − 76.8K and v = (160 − 8K) 0.6 K 0.6 Therefore,

du

and, using the product rule,

Therefore, the second-order condition for a maximum is satisfied when K = 10

Minimum average cost

In your introductory economics course you were probably given an intuitive geometricalexplanation of why a marginal cost schedule cuts a U-shaped average cost curve at itsminimum point The quotient rule can now be used to prove this rule

In the short run, with only one variable input, assume that total cost (TC) is a function

of q Thus, MC = dTC/dq (as explained inChapter 8) and, by definition, AC= TC/q.

To differentiate AC using the quotient rule let

To check second-order conditions we need to find d AC/dq From (1) above we knowthat

Therefore, the second-order condition for a minimum is satisfied when MC= AC and MC

is rising Thus, although MC may cut AC at another point when MC is falling, when MC isrising it cuts AC at its minimum point

12.5Individual labour supply

Not all of you will have encountered the theory of individual labour supply Neverthelessyou should now be able to understand the following example which shows how the utility-maximizing combination of work and leisure hours can be found when an individual’s utilityfunction, wage rate and maximum working day are specified

Example 12.14

In the theory of individual labour supply it is assumed that an individual derives utility from

both leisure (L) and income (I ) Income is determined by hours of work (H ) multiplied by the hourly wage rate (w), i.e I = wH.

Assume that each day a total of 12 hours is available for an individual to split betweenleisure and work, the wage rate is given as £4 an hour and that the individual’s utility function

is U = L 0.5 I 0.75 How will this individual balance leisure and income so as to maximizeutility?

Given a maximum working day of 12 hours, then hours of work H = 12 − L.

Therefore, given an hourly wage of £4, income earned will be

3 Using your answer from Test Yourself, Exercise 12.2.4, show that the

second-order condition for a maximum value of the function Q= 1200.5 (250− 0.5K) 0.3

is satisfied when K is 312.5 and evaluate d2Q/ dK2

4 For the demand schedule p = (800−0.4q) 0.5 find which value of q will maximize

total revenue, using the quotient rule to check the second-order condition

5 Assume that an individual can choose the number of hours per day that they work

up to a maximum of 12 hours This individual attempts to maximize the utility

function U = L 0.4 I 0.6 where L is defined as hours not worked out of the 12-hour maximum working day, and I is income, equal to hours worked (H ) times the

hourly wage rate of £15 What mix of leisure and work will be chosen?

6 Show that when a firm faces a U-shaped short-run average variable cost (AVC)schedule, its marginal cost schedule will always cut the AVC schedule at itsminimum point when MC is rising

12.6 Integration

Integrating a function means finding another function which, when it is differentiated, givesthe first function It is basically differentiation in reverse, and the rules for integration arethe reverse of those for differentiation Unlike differentiation, which we have seen to be veryuseful in optimization problems, the mathematical technique of integration is not as widelyused in economics and so we shall only look at some of the basic ideas involved

Assume that you wish to integrate the function

However, although this is one solution, the same derivative can be obtained from other

functions For example, if y = 35 + 6x2+ 8x3then we also get

dy

dx = 12x + 24x2

In fact, whatever constant term starts the function the same derivative will be obtained.Because constant numbers disappear when a function is differentiated, we cannot know whatconstant should appear in an integrated function unless further information is available We

therefore simply include a ‘constant of integration’ (C) in the integral.

The notation used for integration is

y =



f(x) dx

This means that y is the integral of the function f(x) The sign∫ is known as the integration

sign The ‘dx’ signifies that if y is differentiated with respect x the result will equal f(x).

We can therefore write the integral of the above example as

where a and n are given parameters and n

the rule for differentiation you should have no problems in seeing how the answers beloware derived

Total cost functions can usually be split into fixed and variable components The integral

of marginal cost will give total variable costs plus a constant of integration which shouldequal Total Fixed Cost (TFC) For example, if we are given the information that total variablecost is

Once the TR function corresponding to a given MR function has been derived then one simply

has to divide this through by q to arrive at the demand function.

If a firm faces the marginal cost schedule MC= 180 + 0.3q2

and the marginal revenue schedule MR= 540 − 0.6q2

and total fixed costs are £65, what is the maximum profit it can make? (Assume that thesecond-order condition for a maximum is met.)

Test Yourself, Exercise 12.4

1 Find the integrals for the following functions:

(c) 120x4− 60x3

(d) 42− 18x−2(e) 90x 0.5 − 44x −1.2

2 Find the total variable cost functions corresponding to the following marginal costfunctions:

(c) MC= 35 + 0.9q2

(d) MC= 62 − 16q + 1.5q2(e) MC= 185 − 24q + 1.2q3

12.7 Definite integrals

The integrals we have looked at so far are called ‘indefinite integrals’ Another form ofintegral is the ‘definite integral’ This is specified with two values of the independent variable

(placed at the top and bottom of the integration sign) and is defined as the value of the integral

at one value minus its value at the other value

For example, the definite integral8

3 6x2dx is the value of this integral when xis 8 minus its value when x is 3 Thus, given that

y = 20 + 4x

C

D 0

A

Figure 12.1

function y = 20 + 4x which is illustrated in Figure 12.1, i.e the area BFDA This would be

equal to the definite integral

Total area BFDA= ABCD + BFC = 56

Definite integrals of marginal cost functions

This concept of the definite integral has several applications in economics To evaluate TVCfrom an MC function for a given value of output one simply substitutes the given quantityinto the TVC function This is the same as evaluating the definite integral between zero andthe given quantity For example, assume that you wished to find the value of TVC when

q = 8 and you are given the function MC = 7.5 + 0.3q2 This value would be