Basic Mathematics for Economists phần 8 docx

Substituting (6), (7) and (8) into (5) 30,000 − 50(0.72L) − 30L −25(0.96L) −20(1.5L) = 0 30,000 − 36L − 30L − 24L − 30L = 0 30,000 = 120L 250 = L Substituting this value for L into (6), (7) and (8) K = 0.72(250) = 180 R = 0.96(250) = 240 M = 1.5(250) = 375 Using these optimal values of L, K, M and R gives the maximum output level Q = 160K 0.3 L 0.25 R 0.2 M 0.25 = 160(180 0.3 )(250 0.25 )(240 0.2 )(375 0.25 ) = 39,786.6 units Example 11.14 A firm operates with the production function Q = 20K 0.5 L 0.25 R 0.4 . The input prices per unit are £20 for K, £10 for L and £5 for R. What is the cheapest way of producing 1,200 units of output? Solution This time output is the constraint such that 20K 0.5 L 0.25 R 0.4 = 1,200 and the objective function to be minimized is the cost function TC = 20K +10L +5R The corresponding Lagrange function is therefore G = 20K +10L +5R + λ(1,200 − 20K 0.5 L 0.25 R 0.4 ) Differentiating to get stationary points ∂G ∂K = 20 − λ10K −0.5 L 0.25 R 0.4 = 0 λ = 2K 0.5 L 0.25 R 0.4 (1) ∂G ∂L = 10 − λ5K 0.5 L −0.75 R 0.4 = 0 λ = 2L 0.75 K 0.5 R 0.4 (2) ∂G ∂R = 5 − λ8K 0.5 L 0.25 R −0.6 = 0 λ = 5R 0.6 8K 0.5 L 0.25 (3) ∂G ∂λ = 1,200 − 20K 0.5 L 0.25 R 0.4 = 0(4) © 1993, 2003 Mike Rosser Equating (1) and (2) 2K 0.5 L 0.25 R 0.4 = 2L 0.75 K 0.5 R 0.4 K = L (5) Equating (2) and (3) 2L 0.75 K 0.5 R 0.4 = 5R 0.6 8K 0.5 L 0.25 16L = 5R 3.2L = R (6) Substituting (5) and (6) into (4) to eliminate R and K 1,200 −20(L) 0.5 L 0.25 (3.2L) 0.4 = 0 1,200 −20(3.2) 0.4 L 1.15 = 0 60 = 1.5924287L 1.15 37.678296 = L 1.15 23.47 = L Substituting this value for L into (5) and (6) gives K = 23.47 R = 3.2(23.47) = 75.1 Checking that these values do give the required 1,200 units of output: Q = 20K 0.5 L 0.25 R 0.4 = 20(23.47) 0.5 (23.47) 0.25 (75.1) 0.4 = 1,200 The cheapest cost level for producing this output will therefore be 20K + 10L + 5R = 20(23.4) + 10(23.47) +5(75.1) = £1,079.60 Example 11.15 A firm operates with the production function Q = 45K 0.4 L 0.3 R 0.3 and can buy input K at £80 a unit, L at £35 and R at £50. What is the cheapest way it can produce an output of 75,000 units? Solution The output constraint is 45K 0.4 L 0.3 R 0.3 = 75,000 andthe objective function to be minimized is TC = 80K +35L + 50R. The corresponding Lagrange function is thus G = 80K +35L + 50R + λ(75,000 − 45K 0.4 L 0.3 R 0.3 ) © 1993, 2003 Mike Rosser Differentiating to get first-order conditions for a minimum ∂G ∂K = 80 − λ18K −0.6 L 0.3 R 0.3 = 0 λ = 80K 0.6 18L 0.3 R 0.3 (1) ∂G ∂L = 35 − λ13.5K 0.4 L −0.7 R 0.3 = 0 λ = 35L 0.7 13.5K 0.4 R 0.3 (2) ∂G ∂R = 50 − λ13.5K 0.4 L 0.3 R −0.7 = 0 λ = 50R 0.7 13.5K 0.4 L 0.3 (3) ∂G ∂λ = 75,000 − 45K 0.4 L 0.3 R 0.3 = 0(4) Equating (1) and (2) 80K 0.6 18L 0.3 R 0.3 = 35L 0.7 13.5K 0.4 R 0.3 1,080K = 630L 12K 7 = L (5) As we have L in terms of K we now need to use (1) and (3) to get R in terms of K. Thus equating (1) and (3) 80K 0.6 18L 0.3 R 0.3 = 50R 0.7 13.5K 0.4 L 0.3 1,080K = 900R 1.2K = R (6) Substituting (5) and (6) into (4) 75,000 −45K 0.4  12K 7  0.3 (1.2K) 0.3 = 0 75,000 −45  12 7  0.3 (1.2) 0.3 K 0.4 K 0.3 K 0.3 = 0 75,000 = 55.871697K 1,342.3612 = K Substituting this value into (5) L = 12 7 (1,342.3612) = 2,301.1907 Substituting into (6) R = 1.2(1,342.3612) = 1,610.8334 Thus, optimum values are K = 1,342.4 L = 2,301.2 R = 1,610.8 (to 1 dp) © 1993, 2003 Mike Rosser Total expenditure on inputs will then be 80K + 35L + 50R = 80(1,342.4) + 35(2,301.2) + 50(1,610.8) = £268,474 Test Yourself, Exercise 11.4 1. A firm has a budget of £570 to spend on the three inputs x, y and z whose prices per unit are respectively £4, £6 and £3. What combination of x, y and z will maximize output given the production function Q = 2x 0.2 y 0.3 z 0.45 ? 2. A firm uses inputs K, L and R to manufacture good Q. It has a budget of £828 and its production function (for positive values of Q)is Q = 20K +16L + 12R − 0.2K 2 − 0.1L 2 − 0.3R 2 If P K = £20, P L = £10 and P R = £6, what is the maximum output it can produce? Assume that second-order conditions for a maximum are satisfied for the relevant Lagrangian. 3. What amounts of the inputs x, y and z should a firm use to maximize output if it faces the production function Q = 2x 0.4 y 0.2 z 0.6 and it has a budget of £600, given that the prices of x, y and z are respectively £4, £1 and £2 per unit? 4. A firm buys the inputs x, y and z for £5, £10 and £2 respectively per unit. If its production function is Q = 60x 0.2 y 0.4 z 0.5 how much can it produce for an outlay of £8,250? 5. Inputs K, L, R and M cost £10, £6, £15 and £3 respectively per unit. What is the cheapest way of producing an output of 900 units if a firm operates with the production function Q = 20K 0.4 L 0.3 R 0.2 M 0.25 ? 6. Make up your own constrained optimization problem for an objective function with three variables and solve it. 7. A firm faces the production function Q = 50K 0.5 L 0.2 R 0.25 and is required to produce an output level of 1,913 units. What is the cheapest way of doing this if the per-unit costs of inputs K, L and R are £80, £24 and £45 respectively? © 1993, 2003 Mike Rosser 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 [...]... the product rule, let u = 12K 0.4 v = (160 − 8K)0.4 and and so du = 4.8K −0.6 dK and dv = 0.4(160 − 8K)−0.6 ( 8) dK = −3.2(160 − 8K)−0.6 Therefore, dQ = 12K 0.4 (−3.2)(160 − 8K)−0.6 + (160 − 8K)0.4 4.8K −0.6 dK − 38. 4K + (160 − 8K)4 .8 = (160 − 8K)0.6 K 0.6 7 68 − 76.8K = (160 − 8K)0.6 K 0.6 (1) Setting (1) equal to zero for a stationary point must mean 7 68 − 76.8K = 0 K = 10 As we have already left this... (160 − 8K)0.4 (2) Therefore, using the quotient rule and substituting (1) and (2) d2 Q dK 2 = (160 − 8K)0.6 K 0.6 (−76 .8) − (7 68 − 76.8K) 96 − 9.6K − 8K)0.4 K 0.4 (160 (160 − 8K)1.2 K 1.2 (160 − 8K)K(−76 .8) − 76 .8( 10 − K)9.6(10 − K) = (160 − 8K)1.6 K 1.6 At the stationary point when K = 10 several terms become zero, giving d2 Q −76 .8( 800) = . −8K) −0.6 ( 8) =−3.2(160 − 8K) −0.6 Therefore, dQ dK = 12K 0.4 (−3.2)(160 − 8K) −0.6 + (160 − 8K) 0.4 4.8K −0.6 = − 38. 4K + (160 − 8K)4 .8 (160 −8K) 0.6 K 0.6 = 7 68 −76.8K (160 −8K) 0.6 K 0.6 (1) Setting. (8 + 0.2x) −1 du dx = 8x dv dx =−0.2 (8 + 0.2x) −2 Thus, using the product rule dy dx = u dv dx + v du dx = 4x 2 [−0.2 (8 + 0.2x) −2 ]+ (8 +0.2x) −1 8x = −0.8x 2 + (8 +0.2x)8x (8 +0.2x) 2 = −0.8x 2 +. the quotient rule, dy dx = v du dx − u dv dx v 2 = (8 +0.2x)8x − 4x 2 (0.2) (8 +0.2x) 2 = 64x + 1.6x 2 − 0.8x 2 (8 +0.2x) 2 = 64x + 0.8x 2 (8 +0.2x) 2 (1) This solution could also have been found