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 of functions. • Check the second-order conditions for optimization of relevant economic func- tions using the quotient rule for differentiation. • Integrate simple functions. • Use integration to determine total cost and total revenue from marginal cost and marginal revenue functions. • Understand how a definite integral relates to the area under a function and apply this concept to calculate consumer surplus. 12.1 Overview In this chapter, some techniques are introduced that can be used to differentiate functions that arerathermorecomplexthanthoseencounteredinChapters8,9,10and11.Thesearethe chain 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. The concept of integration is also introduced. 12.2 The chain rule The chain rule is used to differentiate ‘functions within functions’. For example, if we have the function y = f(z) and we also know that there is a second functional relationship z = g(x) then we can write y as a function of x in the form y = f[g(x)] © 1993, 2003 Mike Rosser To differentiate y with respect to x in this type of function we use the chain rule which states that dy dx = dy dz dz dx One economics example of a function within a function occurs in the marginal revenue productivity theory of the demand for labour, where a firm’s total revenue depends on output which, in turn, depends on the amount of labour employed. An applied example is explained later. However, we shall first look at what is perhaps the most frequent use of the chain rule, which is to break down an awkward function artificially into two components in order to allow differentiation via the chain rule. Assume, for example, that you wish to find an expression for the slope of the non-linear demand function p = (150 − 0.2q) 0.5 (1) ThebasicrulesfordifferentiationexplainedinChapter8cannotcopewiththissortof function. However, if we define a new function z = 150 − 0.2q (2) then (1) above can be rewritten as p = z 0.5 (3) (Note that in both (1) and (3) the functions are assumed to hold for p ≥ 0 only, i.e. negative roots are ignored.) Differentiating (2) and (3) we get dz dq =−0.2 dp dz = 0.5z −0.5 Thus, using the chain rule and then substituting equation (2) back in for z,weget dp dq = dp dz dz dq = 0.5z −0.5 (−0.2) = −0.1 z 0.5 = −0.1 (150 − 0.2q) 0.5 Some more examples of the use of the chain rule are set out below. Example 12.1 The present value of a payment of £1 due in 8 years’ time is given by the formula PV = 1 (1 + i) 8 where i is the given interest rate. What is the rate of change of PV with respect to i? © 1993, 2003 Mike Rosser Solution If we let z = 1 + i (1) then we can write PV = 1 z 8 = z −8 (2) Differentiating (1) and (2) gives dz di = 1 dPV dz =−8z −9 Therefore, using the chain rule, the rate of change of PV with respect to i will be dPV di = dPV dz dz di =−8z −9 = −8 (1 + i) 9 Example 12.2 If y = (48 + 20x −1 + 4x +0.3x 2 ) 4 , what is dy/dx? Solution Let z = 48 + 20x −1 + 4x +0.3x 2 (1) and so dz dx =−20x −2 + 4 +0.6x (2) Substituting (1) into the function given in the question y = z 4 and so dy dz = 4z 3 (3) Therefore, using the chain rule and substituting (2) and (3) dy dx = dy dz dz dx = 4z 3 (−20x −2 + 4 +0.6x) = 4(48 + 20x −1 + 4x +0.3x 2 ) 3 (−20x −2 + 4 +0.6x) © 1993, 2003 Mike Rosser The marginal revenue productivity theory of the demand for labour In the marginal revenue productivity theory of the demand for labour, the rule for profit maximization 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 inputs other than labour are assumed fixed. The optimal amount of labour is employed when MRP L = MC L where MRP L is the marginal revenue product of labour, defined as the additional revenue generated by an additional unit of labour, and MC L is the marginal cost of an additional unit of labour. The MC L is normally equal to the wage rate unless the firm is a monopsonist (sole buyer) in the labour market. If all relevant functions are assumed to be continuous then the above definitions can be rewritten as MRP L = dTR dL MC L = dTC L dL where TR is total sales revenue (i.e. pq) and TC L is the total cost of labour. If a firm is a monopoly seller of a good, then we effectively have to deal with two functions in order to derive its MRP L function since total revenue will depend on output, i.e. TR = f(q), and output will depend on labour input, i.e. q = f(L). Therefore, using the chain rule, MRP L = dTR dL = dTR dq dq dL (1) We already know that dTR dq = MR dq dL = MP L Therefore, substituting these into (1), MRP L = MR × MP L This is the rule for determining the profit-maximizing amount of labour which you should encounter 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? © 1993, 2003 Mike Rosser Solution Using the chain rule we need to derive a formula for MRP L in terms of L and then set it equal to £8, given that MC L is fixed at this wage rate. As MRP L = dTR dL = dTR dq dq dL (1) we need to find dTR/dq and dq/dL. Given p = 200 − 2q, then TR = pq = (200 −2q)q = 200q − 2q 2 Therefore dTR dq = 200 − 4q (2) Given q = 4L 0.5 , then the marginal product of labour will be dq dL = 2L −0.5 (3) Thus, substituting (2) and (3) into (1) MRP L = (200 − 4q)2L −0.5 = (400 − 8q)L −0.5 As all units of L cost £8, setting this function for MRP L equal to the wage rate we get 400 − 8q L 0.5 = 8 400 − 8q = 8L 0.5 (4) Substituting the production function q = 4L 0.5 into (4), as we are trying to derive a formula in terms of L,gives 400 − 8(4L 0.5 ) = 8L 0.5 400 − 32L 0.5 = 8L 0.5 400 = 40L 0.5 10 = L 0.5 100 = L 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 MRP L function 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. © 1993, 2003 Mike Rosser Thus MP L = ∂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 0.4 and faces the demand function p = 6,980 − 6q. Derive its MRP L function. Solution By definition MRP L = ∂TR/∂L, where K is assumed fixed. We know that TR = pq = (6,980 −6q)q = 6,980q −6q 2 Therefore dTR dq = 6,980 − 12q (1) From the production function q = 45K 0.7 L 0.4 we can derive MP L = ∂q ∂L = 18K 0.7 L −0.6 (2) Using the chain rule and substituting (1) and (2) MRP L = ∂TR ∂L = dTR dq ∂q ∂L = (6,980 − 12q)18K 0.7 L −0.6 (3) As we wish to derive MRP L as a function of L, we substitute the production function given in the question into (3) for q. Thus MRP L =[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 MRP L 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-linear demand functions. © 1993, 2003 Mike Rosser Example 12.5 Find point elasticity of demand when q = 10 if p = (120 − 2q) 0.5 . Solution Point elasticity is defined as e = (−1) p q 1 dp dq (1) Create a new variable z = 120 − 2q. Thus p = z 0.5 and so, by differentiating: dz dq =−2 dp dz = 0.5z −0.5 Therefore dp dq = dp dz dz dq = 0.5z −0.5 (−2) = 0.5(120 − 2q) −0.5 (−2) = −1 (120 − 2q) 0.5 and so, inverting this result, 1 dp/dq =−(120 − 2q) 0.5 When q = 10, then from the original demand function price can be calculated as p = (120 − 20) 0.5 = 100 0.5 = 10 Thus, substituting these results into formula (1), point elasticity will be e = (−1) 10 10 (−1)(120 − 2q) 0.5 = (120 − 20) 0.5 = 100 0.5 = 10 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 be obtained by both methods, of course. Example 12.6 Differentiate the function y = (6 + 4x) 2 . © 1993, 2003 Mike Rosser Solution (i) By multiplying out y = (6 + 4x) 2 = 36 + 48x + 16x 2 Therefore dy dx = 48 + 32x (ii) Using the chain rule, let z = 6 + 4x so that y = z 2 . Thus dy dx = dy dz dz dx = 2z × 4 = 2(6 + 4x)4 = 48 + 32x Test Yourself, Exercise 12.1 1. A firm operates in the short run with the production function q = 2L 0.5 and 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 maximize profit? 2. If a supply schedule is given by p = (2 + 0.05q) 2 show (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 R = M(1 + i) 3 What is the rate of change of R with respect to i? 4. If y = (3 + 0.6x 2 ) 0.5 what is dy/dx? 5. If a firm faces the total cost function TC = (6 + x) 0.5 , what is its marginal cost function? 6. A firm operates with the production function q = 0.4K 0.5 L 0.5 and sells its output in a market where it is a monopoly with the demand schedule p = 60 − 2q.IfK is fixed at 25 units and the wage rate is £7 per unit of L, derive the MRP L function and 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 +q 2 ) 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 MP L when K is 16 and L is 576. © 1993, 2003 Mike Rosser 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 dx = u dv dx + v du dx As with the chain rule, one may find it convenient to split a single awkward function into two artificial functions even if these functions do not have any particular economic meaning. The following examples show how this rule can be used. Example 12.7 If y = (7.5 + 0.2x 2 )(4 + 8x −1 ), what is dy/dx? Solution This function could in fact be multiplied out and differentiated without using the product rule. However, let us first use the product rule and then we can compare the answers obtained by the two methods. They should, of course, be the same. We are given the function y = (7.5 + 0.2x 2 )(4 + 8x −1 ) so let u = 7.5 +0.2x 2 v = 4 +8x −1 Therefore du dx = 0.4x dv dx =−8x −2 Thus, using the product rule and substituting these results in, we get dy dx = u dv dx + v du dx = (7.5 + 0.2x 2 )(−8x −2 ) + (4 + 8x −1 )0.4x =−60x −2 − 1.6 +1.6x + 3.2 = 1.6 + 1.6x − 60x −2 (1) The alternative method of differentiation is to multiply out the original function. Thus y = (7.5 + 0.2x 2 )(4 + 8x −1 ) = 30 + 60x −1 + 0.8x 2 + 1.6x and so dy dx =−60x −2 + 1.6x +1.6(2) The answers (1) and (2) are the same, as we expected. © 1993, 2003 Mike Rosser When it is not possible to multiply out the different components of a function then one must use the product rule to differentiate. One may also need to use the chain rule to help differentiate the different sub-functions. Example 12.8 A firm faces the non-linear demand schedule p = (650 − 0.25q) 1.5 . What output should it sell to maximize total revenue? Solution When the demand function in the question is substituted for p then TR = pq = (650 −0.25q) 1.5 q To differentiate TR using the product rule, first let u = (650 − 0.25q) 1.5 v = q Thus, employing the chain rule du dq = 1.5(650 − 0.25q) 0.5 (−0.25) =−0.375(650 −0.25q) 0.5 and also dv dq = 1 Therefore, using the product rule dTR dq = u dv dq + v du dq = (650 − 0.25q) 1.5 + q(−0.375)(650 − 0.25q) 0.5 = (650 − 0.25q) 0.5 (650 − 0.25q − 0.375q) = (650 − 0.25q) 0.5 (650 − 0.625q) (1) For a stationary point dTR dq = (650 − 0.25q) 0.5 (650 − 0.625q) = 0 Therefore, either 650 − 0.25q = 0 or 650 − 0.625q = 0 2,600 = q or 1,040 = q 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 by observing what happens when this quantity is substituted into the demand function.) To © 1993, 2003 Mike Rosser [...]... giving 0 + (−975)(−0 .125 ) d2 TR → +∞ = 2 dq 0 Therefore this second value for q obviously does not satisfy second-order conditions for a maximum Example 12. 9 At what level of K is the function Q = 12K 0.4 (160 − 8K)0.4 at a maximum? (This is Example 11.1 (reworked) which was not completed in the last chapter. ) © 1993, 2003 Mike Rosser Solution We need to differentiate the function Q = 12K 0.4 (160 − 8K)0.4... −0.1644 < 0 dt 2 Therefore, the second-order condition for a maximum is satisfied when t = 66.67 Maximum tax revenue is raised when the per-unit tax is £66.67 © 1993, 2003 Mike Rosser Test Yourself, Exercise 12. 2 1 2 3 4 5 6 7 If y = (6x + 7)0.5 (2.6x 2 − 1.9), what is dy/dx? What output will maximize total revenue given the non-linear demand schedule p = (60 − 2q)1.5 ? Derive a function for the marginal... = −0 .125 (650 − 0.25q)−0.5 dq using the chain rule and dv = −0.625 dq Therefore, employing the product rule du dv d2 TR +v =u 2 dq dq dq = (650 − 0.25q)0.5 (−0.625) + (650 − 0.625q)(−0 .125 )(650 − 0.25q)−0.5 = (650 − 0.25q)(−0.625) + (650 − 0.625q)(−0 .125 ) (650 − 0.25q)0.5 (2) Substituting the value q = 1,040 into (2) gives (390)(−0.625) + 0 d2 TR = = 12. 34 < 0 dq 2 3900.5 Therefore, the second-order... give you an idea of which will be the easier to use for specific examples Example 12. 12 Derive a function for marginal revenue (in terms of q) if a monopoly faces the non-linear 252 demand schedule p = (4 + q)0.5 © 1993, 2003 Mike Rosser Solution TR = pq = 252q (4 + q)0.5 Defining u = 252q gives and v = (4 + q)0.5 dv = 0.5(4 + q)−0.5 dq du = 252 dq Therefore, using the quotient rule dTR = MR = dq v dv... optimization problem and then use the quotient rule to check the second-order condition If we return to the unfinished Example 12. 9 we can now see how the quotient rule can be used to check the second-order condition Example 12. 13 The objective is to find the value of K which maximizes Q = 12K 0.4 (160 − 8K)0.4 In Example 12. 9, first-order conditions were satisfied when 768 − 76.8K dQ = dK (160 − 8K)0.6... in Chapter 8) and, by definition, AC = TC/q To differentiate AC using the quotient rule let u = TC v=q and giving du dTC = = MC dq dq dv =1 dq Therefore, using the quotient rule, first-order conditions for a minimum are dAC qMC − TC = =0 dq q2 qMC − TC = 0 MC = (1) (or q → ∞, which we disregard) TC = AC q Therefore, MC = AC when AC is at a stationary point © 1993, 2003 Mike Rosser (2) To check second-order... the second-order condition for maximization of utility is satisfied when 7.2 hours are worked and 4.8 hours are taken as leisure Test Yourself, Exercise 12. 3 1 2 dy (3x + 0.4x 2 ) what is ? dx (8 − 6x 1.5 )0.5 Derive a function for marginal revenue for the demand schedule If y = p= 3 4 5 6 720 (25 + q)0.5 Using your answer from Test Yourself, Exercise 12. 2.4, show that the secondorder condition for a maximum... Substituting (1) for q, this gives TY = t (40 − 0.4t)0.5 (2) We need to set dTY/dt = 0 for the first-order condition for maximization of TY From (2) let u=t v = (40 − 0.4t)0.5 and giving du =1 dt dv = 0.5(40 − 0.4t)−0.5 (−0.4) dt = −0.2(40 − 0.4t)−0.5 Therefore, using the product rule dTY = t (−0.2)(40 − 0.4t)−0.5 + (40 − 0.4t)0.5 dt −0.2t + 40 − 0.4t = (40 − 0.4t)0.5 40 − 0.6t = =0 (40 − 0.4t)0.5 (3) For finite... MC − MC = q dq dq dq and dv = 2q dq Therefore d2 AC = dq 2 q2 q dMC dq − (qMC − TC)2q (3) q4 The first-order condition for a minimum is satisfied when qMC = TC, from (2) above Substituting this result into (3) the second term in the numerator disappears and we get d2 AC dq 2 dMC dq = q4 1 dMC = >0 q dq q2 q when dMC >0 dq Therefore, the second-order condition for a minimum is satisfied when MC = AC and... each day a total of 12 hours is available for an individual to split between leisure and work, the wage rate is given as £4 an hour and that the individual’s utility function is U = L0.5 I 0.75 How will this individual balance leisure and income so as to maximize utility? © 1993, 2003 Mike Rosser Solution Given a maximum working day of 12 hours, then hours of work H = 12 − L Therefore, given an hourly . giving d 2 TR dq 2 = 0 + (−975)(−0 .125 ) 0 →+∞ Therefore this second value for q obviously does not satisfy second-order conditions for a maximum. Example 12. 9 AtwhatlevelofKisthefunctionQ=12K 0.4 (160−8K) 0.4 atamaximum?(Thisis Example11.1(reworked)whichwasnotcompletedinthelastchapter.) ©. rule. Assume, for example, that you wish to find an expression for the slope of the non-linear demand function p = (150 − 0.2q) 0.5 (1) ThebasicrulesfordifferentiationexplainedinChapter8cannotcopewiththissortof function appropriate method for differentiating different forms of functions. • Check the second-order conditions for optimization of relevant economic func- tions using the quotient rule for differentiation. •