Basic Mathematics for Economists phần 7 potx

where C is consumer expenditure, Y d is disposable income, Y is national income, I is investment, t is the tax rate and G is government expenditure. What is the marginal propensity to consume out of Y? What is the value of the govern- ment expenditure multiplier? How much does government expenditure need to be increased to achieve a national income of 700? 10.3 Second-order partial derivatives Second-order partial derivatives are found by differentiating the first-order partial derivatives of a function. When a function has two independent variables there will be four second-order partial derivatives. Take, for example, the production function Q = 25K 0.4 L 0.3 There are two first-order partial derivatives ∂Q ∂K = 10K −0.6 L 0.3 ∂Q ∂L = 7.5K 0.4 L −0.7 These represent the marginal product functions for K and L. Differentiating these functions a second time we get ∂ 2 Q ∂K 2 =−6K −1.6 L 0.3 ∂ 2 Q ∂L 2 =−5.25K 0.4 L −1.7 These second-order partial derivatives represent the rate of change of the marginal product functions. In this example we can see that the slope of MP L (i.e. ∂ 2 Q/∂L 2 ) will always be negative (assuming positive values of K and L) and as L increases, ceteris paribus, the absolute value of this slope diminishes. We can also find the rate of change of ∂Q/∂K with respect to changes in L and the rate of change of ∂Q/∂L with respect to K. These will be ∂ 2 Q ∂K∂L = 3K −0.6 L −0.7 ∂ 2 Q ∂L∂K = 3K −0.6 L −0.7 and are known as ‘cross partial derivatives’. They show how the rate of change of Q with respect to one input alters when the other input changes. In this example, the cross partial derivative ∂ 2 Q/∂L∂K tells us that the rate of change of MP L with respect to changes in K will be positive and will fall in value as K increases. You will also have noted in this example that ∂ 2 Q ∂K∂L = ∂ 2 Q ∂L∂K In fact, matched pairs of cross partial derivatives will always be equal to each other. Thus, for any continuous two-variable function y = f(x, z), there will be four second-order partial derivatives: (i) ∂ 2 y ∂x 2 (ii) ∂ 2 y ∂z 2 (iii) ∂ 2 y ∂x∂z (iv) ∂ 2 y ∂z∂x © 1993, 2003 Mike Rosser with the cross partial derivatives (iii) and (iv) always being equal, i.e. ∂ 2 y ∂x∂z = ∂ 2 y ∂z∂x Example 10.12 Derive the four second-order partial derivatives for the production function Q = 6K + 0.3K 2 L + 1.2L 2 and interpret their meaning. Solution The two first-order partial derivatives are ∂Q ∂K = 6 + 0.6KL ∂Q ∂L = 0.3K 2 + 2.4L and these represent the marginal product functions MP K and MP L . The four second-order partial derivatives are as follows: (i) ∂ 2 Q ∂K 2 = 0.6L This represents the slope of the MP K function. It tells us that the MP K function will have a constant slope along its length (i.e. it is linear) for any given value of L, but an increase in L will cause an increase in this slope (ii) ∂ 2 Q ∂L 2 = 2.4 This represents the slope of the MP L function and tells us that MP L is a straight line with slope 2.4. This slope does not depend on the value of K. (iii) ∂ 2 Q ∂K∂L = 0.6K This tells us that MP K increases if L is increased. The rate at which MP K rises as L is increased will depend on the value of K. (iv) ∂ 2 Q ∂L∂K = 0.6K This tells us that MP L will increase if K is increased and that the rate of this increase will depend on the value of K. Thus, although the slope of the MP L schedule will always be 2.4, from (ii) above, its actual position will depend on the amount of K used. Some other applications of second-order partial derivatives are given below. © 1993, 2003 Mike Rosser Example 10.13 A firm sells two competing products whose demand schedules are q 1 = 120 − 0.8p 1 + 0.5p 2 q 2 = 160 + 0.4p 1 − 12p 2 How will the price of good 2 affect the marginal revenue of good 1? Solution To find the total revenue function for good 1 (TR 1 ) in terms of q 1 we first need to derive the inverse demand function p 1 = f(q 1 ). Thus, given q 1 = 120 − 0.8p 1 + 0.5p 2 0.8p 1 = 120 + 0.5p 2 − q 1 p 1 = 150 + 0.625p 2 − 1.25q 1 TR 1 = p 1 q 1 = (150 + 0.625p 2 − 1.25q 1 )q 1 = 150q 1 + 0.625p 2 q 1 − 1.25q 2 1 Thus MR 1 = ∂TR 1 ∂q 1 = 150 + 0.625p 2 − 2.5q 1 This marginal revenue function will have a constant slope of −2.5 regardless of the value of p 2 or the amount of q 1 sold. The effect of a change in p 2 on MR 1 is shown by the cross partial derivative ∂TR 1 ∂q 1 ∂p 2 = 0.625 Thus an increase in p 2 of one unit will cause an increase in the marginal revenue from good 1 of 0.625, i.e. although the slope of the MR 1 schedule remains constant at −2.5, its position shifts upward if p 2 rises. (Note that in order to answer this question, we have formulated the total revenue for good 1 as a function of one price and one quantity, i.e. TR 1 = f(q 1 ,p 2 ).) Example 10.14 A firm operates with the production function Q = 820K 0.3 L 0.2 and can buy inputs K and L at £65 and £40 respectively per unit. If it can sell its output at a fixed price of £12 per unit, what is the relationship between increases in L and total profit? Will a change in K affect the extra profit derived from marginal increases in L? © 1993, 2003 Mike Rosser Solution TR = PQ = 12(820K 0.3 L 0.2 ) TC = P K K + P L L = 65K + 40L Therefore profit will be π = TR − TC = 12(820K 0.3 L 0.2 ) − (65K + 40L) = 9,840K 0.3 L 0.2 − 65K − 40L The effect of an increase in L on profit is shown by the first-order partial derivative: ∂π ∂L = 1,968K 0.3 L −0.8 − 40 (1) This effect will be positive as long as 1,968K 0.3 L −0.8 > 40 However, if L is continually increased while K is held constant, the value of the term 1,968K 0.3 L −0.8 will eventually fall below 40 and so ∂π/∂L will become negative. To determine the effect of a change in K on the marginal profit function with respect to L, we need to differentiate (1) with respect to K, giving ∂ 2 π ∂L∂K = 0.3(1,968K −0.7 L −0.8 ) = 590.4K −0.7 L −0.8 This cross partial derivative will be positive as long as K and L are positive. This is what we would expect and so an increase in K will have a positive effect on the extra profit generated by marginal increases in L. The magnitude of this impact will depend on the values of K and L. Second-order and cross partial derivatives can also be derived for functions with three or more independent variables. For a function with three independent variables, such as y = f(w,x,z)there will be the three second-order partial derivatives ∂ 2 y ∂w 2 ∂ 2 y ∂x 2 ∂ 2 y ∂z 2 plus the six cross partial derivatives ∂ 2 y ∂w∂x = ∂ 2 y ∂x∂w ∂ 2 y ∂x∂z = ∂ 2 y ∂z∂x ∂ 2 y ∂w∂z = ∂ 2 y ∂z∂w These are arranged in pairs because, as with the two-variable case, cross partial derivatives will be equal if the two stages of differentiation involve the same two variables. © 1993, 2003 Mike Rosser Example 10.15 For the production function Q = 32K 0.5 L 0.25 R 0.4 derive all the second-order and cross partial derivatives and show that the cross partial derivatives with respect to each possible pair of independent variables will be equal to each other. Solution The three first-order partial derivatives will be ∂Q ∂K = 16K −0.5 L 0.25 R 0.4 ∂Q ∂L = 8K 0.5 L −0.75 R 0.4 ∂Q ∂R = 12.8K 0.5 L 0.25 R −0.6 The second-order partial derivatives will be ∂ 2 Q ∂K 2 =−8K −1.5 L 0.25 R 0.4 ∂ 2 Q ∂L 2 =−6K 0.5 L −1.75 R 0.4 ∂ 2 Q ∂R 2 =−7.68K 0.5 L 0.25 R −1.6 plus the six cross partial derivatives: ∂ 2 Q ∂K∂L = 4K −0.5 L −0.75 R 0.4 = ∂ 2 Q ∂L∂K ∂ 2 Q ∂L∂R = 3.2K 0.5 L −0.75 R −0.6 = ∂ 2 Q ∂R∂L ∂ 2 Q ∂R∂K = 6.4K −0.5 L 0.25 R −0.6 = ∂ 2 Q ∂K∂R Second-order derivatives for multi-variable functions are needed to check second-order conditions for optimization, as explained in the next section. Test Yourself, Exercise 10.3 1. For the production function Q = 8K 0.6 L 0.5 derive a function for the slope of the marginal product of L. What effect will a marginal increase in K have upon this MP L function? 2. Derive all the second-order and cross partial derivatives for the production function Q = 35KL + 1.4LK 2 + 3.2L 2 and interpret their meaning. © 1993, 2003 Mike Rosser 3. A firm operates three plants with the joint total cost function TC = 58 +18q 1 + 9q 2 q 3 + 0.004q 2 1 q 2 3 + 1.2q 1 q 2 q 3 Find all the second-order partial derivatives for TC and demonstrate that the cross partial derivatives can be arranged in three equal pairs. 10.4 Unconstrained optimization: functions with two variables For the two variable function y = f(x, z) to be at a maximum or at a minimum, the first-order conditions which must be met are ∂y ∂x = 0 and ∂y ∂z = 0 These are similar to the first-order conditions for optimization of a single variable function thatwereexplainedinChapter9.Tobeatamaximumorminimum,thefunctionmustbeat a stationary point with respect to changes in both variables. The second-order conditions and the reasons for them were relatively easy to explain in the case of a function of one independent variable. However, when two or more indepen- dent variables are involved the rationale for all the second-order conditions is not quite so straightforward. We shall therefore just state these second-order conditions here and give a brief intuitive explanation for the two-variable case before looking at some applications. The second-order conditions for the optimization of multi-variable functions with more than two variablesareexplainedinChapter15usingmatrixalgebra. For the optimization of two variable functions there are two sets of second-order conditions. For any function y = f(x, z). (1) ∂ 2 y ∂x 2 < 0 and ∂ 2 y ∂z 2 < 0 for a maximum ∂ 2 y ∂x 2 > 0 and ∂ 2 y ∂z 2 > 0 for a minimum These are similar to the second-order conditions for the optimization of a single variable function. The rate of change of a function (i.e. its slope) must be decreasing at a stationary point for that point to be a maximum and it must be increasing for a stationary point to be a minimum. The difference here is that these conditions must hold with respect to changes in both independent variables. (2) The other second-order condition is  ∂ 2 y ∂x 2  ∂ 2 y ∂z 2  >  ∂ 2 y ∂x∂z  2 This must hold at both maximum and minimum stationary points. To get an idea of the reason for this condition, imagine a three-dimensional model with x and z being measured on the two axes of a graph and y being measured by the height above the flat surface on which the x and z axes are drawn. For a point to be the peak of the y ‘hill’ then, as well as the slope being zero at this point, one needs to ensure that, whichever © 1993, 2003 Mike Rosser direction one moves, the height will fall and the slope will become steeper. Similarly, for a point to be the minimum of a y ‘trough’ then, as well as the slope being zero, one needs to ensure that the height will rise and the slope will become steeper whichever direction one moves in. As moves can be made in directions other than those parallel to the two axes, it can be mathematically proved that the condition  ∂ 2 y ∂x 2  ∂ 2 y ∂z 2  >  ∂ 2 y ∂x∂z  2 satisfies these requirements as long as the other second-order conditions for a maximum or minimum also hold. Note also that all the above conditions refer to the requirements for local maximum or minimum values of a function, which may or may not be global maxima or minima. Refer backtoChapter9ifyoucannotrememberthedifferencebetweenthesetwoconcepts. Let us now look at some applications of these rules for the unconstrained optimization of a function with two independent variables. Example 10.16 A firm produces two products which are sold in two separate markets with the demand schedules p 1 = 600 − 0.3q 1 p 2 = 500 − 0.2q 2 Production costs are related and the firm faces the total cost function TC = 16 +1.2q 1 + 1.5q 2 + 0.2q 1 q 2 If the firm wishes to maximize total profits, how much of each product should it sell? What will the maximum profit level be? Solution The total revenue is TR = TR 1 + TR 2 = p 1 q 1 + p 2 q 2 = (600 − 0.3q 1 )q 1 + (500 − 0.2q 2 )q 2 = 600q 1 − 0.3q 2 1 + 500q 2 − 0.2q 2 2 Therefore profit is π = TR − TC = 600q 1 − 0.3q 2 1 + 500q 2 − 0.2q 2 2 − (16 + 1.2q 1 + 1.5q 2 + 0.2q 1 q 2 ) = 600q 1 − 0.3q 2 1 + 500q 2 − 0.2q 2 2 − 16 − 1.2q 1 − 1.5q 2 − 0.2q 1 q 2 =−16 + 598.8q 1 − 0.3q 2 1 + 498.5q 2 − 0.2q 2 2 − 0.2q 1 q 2 © 1993, 2003 Mike Rosser First-order conditions for maximization of this profit function are ∂π ∂q 1 = 598.8 − 0.6q 1 − 0.2q 2 = 0 (1) and ∂π ∂q 2 = 498.5 − 0.4q 2 − 0.2q 1 = 0(2) Simultaneous equations (1) and (2) can now be solved to find the optimal values of q 1 and q 2 . Multiplying (2) by 3 1,495.5 − 1.2q 2 − 0.6q 1 = 0 Rearranging (1) 598.8 − 0.2q 2 − 0.6q 1 = 0 Subtracting gives 896.7 − q 2 = 0 Giving the optimal value 896.7 = q 2 Substituting this value for q 2 into (1) 598.8 − 0.6q 1 − 0.2(896.7) = 0 598.8 − 179.34 = 0.6q 1 419.46 = 0.6q 1 699.1 = q 1 Checking second-order conditions by differentiating (1) and (2) again: ∂ 2 π ∂q 2 1 =−0.6 < 0 ∂ 2 π ∂q 2 2 =−0.4 < 0 This satisfies one set of second-order conditions for a maximum. The cross partial derivative will be ∂ 2 π ∂q 1 ∂q 2 =−0.2 Therefore  ∂ 2 π ∂q 2 1  ∂ 2 π ∂q 2 2  = (−0.6)(−0.4) = 0.24 > 0.04 = (−0.2) 2 =  ∂ 2 π ∂q 1 ∂q 2  2 and so the remaining second-order condition for a maximum is satisfied. The actual profit is found by substituting the optimum values q 1 = 699.1 and q 2 = 896.7. into the profit function. Thus π =−16 + 598.8q 1 − 0.3q 2 1 + 498.5q 2 − 0.2q 2 2 − 0.2q 1 q 2 =−16 + 598.8(699.1) − 0.3(699.1) 2 + 498.5(896.7) − 0.2(896.7) 2 − 0.2(699.1)(896.7) = £432, 797.02 © 1993, 2003 Mike Rosser Example 10.17 A firm sells two products which are partial substitutes for each other. If the price of one product increases then the demand for the other substitute product rises. The prices of the two products (in £) are p 1 and p 2 and their respective demand functions are q 1 = 517 − 3.5p 1 + 0.8p 2 q 2 = 770 − 4.4p 2 + 1.4p 1 What price should the firm charge for each product to maximize its total sales revenue? Solution For this problem it is more convenient to express total revenue as a function of price rather than quantity. Thus TR = TR 1 + TR 2 = p 1 q 1 + p 2 q 2 = p 1 (517 − 3.5p 1 + 0.8p 2 ) + p 2 (770 − 4.4p 2 + 1.4p 1 ) = 517p 1 − 3.5p 2 1 + 0.8p 1 p 2 + 770p 2 − 4.4p 2 2 + 1.4p 1 p 2 = 517p 1 − 3.5p 2 1 + 770p 2 − 4.4p 2 2 + 2.2p 1 p 2 First-order conditions for a maximum are ∂TR ∂p 1 = 517 − 7p 1 + 2.2p 2 = 0 (1) and ∂TR ∂p 2 = 770 − 8.8p 2 + 2.2p 1 = 0(2) Multiplying (1) by 4 2,068 − 28p 1 + 8.8p 2 = 0 Rearranging and adding (2) 770 + 2.2p 1 − 8.8p 2 = 0 2,838 − 25.8p 1 = 0 2,838 = 25.8p 1 110 = p 1 Substituting this value of p 1 into (1) 517 − 7(110) + 2.2p 2 = 0 2.2p 2 = 253 p 2 = 115 Checking second-order conditions: ∂ 2 TR ∂p 2 1 =−7 < 0 ∂ 2 TR ∂p 2 2 =−8.8 < 0 ∂ 2 TR ∂p 1 ∂p 2 = 2.2 © 1993, 2003 Mike Rosser  ∂ 2 TR ∂q 2 1  ∂ 2 TR ∂q 2 2  = (−7)(−8.8) = 61.6 > 4.84 = (2.2) 2 =  ∂ 2 TR ∂q 1 ∂q 2  2 Therefore all second-order conditions for a maximum value of total revenue are satisfied when p 1 = £110 and p 2 = £115. Example 10.18 A multiplant monopoly operates two plants whose total cost schedules are TC 1 = 8.5 + 0.03q 2 1 TC 2 = 5.2 + 0.04q 2 2 If it faces the demand schedule p = 60 − 0.04q where q = q 1 +q 2 , how much should it produce in each plant in order to maximize profits? Solution The total revenue is TR = pq = (60 − 0.04q)q = 60q − 0.04q 2 Substituting (q 1 + q 2 ) for q gives TR = 60(q 1 + q 2 ) − 0.04(q 1 + q 2 ) 2 = 60q 1 + 60q 2 − 0.04q 2 1 − 0.08q 1 q 2 − 0.04q 2 2 Thus, subtracting the two total cost schedules, profit is π = TR − TC 1 − TC 2 = 60q 1 + 60q 2 − 0.04q 2 1 − 0.08q 1 q 2 − 0.04q 2 2 − 8.5 − 0.03q 2 1 − 5.2 − 0.04q 2 2 =−13.7 + 60q 1 + 60q 2 − 0.07q 2 1 − 0.08q 2 2 − 0.08q 1 q 2 First-order conditions for a maximum value of π require ∂π ∂q 1 = 60 − 0.14q 1 − 0.08q 2 = 0 (1) and ∂π ∂q 2 = 60 − 0.16q 2 − 0.08q 1 = 0(2) Multiplying (1) by 2 120 − 0.28q 1 − 0.16q 2 = 0 Rearranging and subtracting (2) 60 − 0.08q 1 − 0.16q 2 = 0 60 − 0.2q 1 = 0 60 = 0.2q 1 300 = q 1 © 1993, 2003 Mike Rosser [...]... = 80. 471 179 (4) Substituting (4) into (1) L0.25 = 42 (80. 471 179 )0 .7 = 4. 871 7455 186 L = (L0.25 )4 = (4. 871 7455)4 = 563.29822 Therefore, first-order conditions suggest that the optimum values are L = 563.3 and K = 80. 47 (to 2 dp) © 1993, 2003 Mike Rosser Checking second-order conditions: ∂ 2π = (−0 .7) 186K −1 .7 L0.25 ∂K 2 = −130.2(80. 47) −1 .7 (563.3)0.25 = −0.3653 576 < 0 ∂ 2π = (−0 .75 )155K 0.3 L−1 .75 ∂L2... (−0 .75 )155K 0.3 L−1 .75 ∂L2 = −116.25(80. 47) 0.3 (563.3)−1 .75 = −0.0066 572 < 0 ∂ 2π = (0.25)186K −0 .7 L−0 .75 ∂K∂L = 46.5(80. 47) −0 .7 (563.3)−0 .75 = 0.0186404 ∂ 2π ∂K 2 ∂ 2π ∂L2 ∂ 2π ∂K∂L 2 = (−0.3653 576 )(−0.0066 572 ) = 0.0024323 = (−0.0186404)2 = 0.0003 475 Therefore ∂ 2π ∂ 2π > ∂K 2 ∂L2 ∂ 2π ∂K∂L 2 and so all second-order conditions for maximum profit are satisfied when K = 80. 47 and L = 563.3 The actual profit will... gives TC = 40K + 15( 275 ,5 67. 6)K −1.5 © 1993, 2003 Mike Rosser (2) Differentiating and setting equal to zero to find a stationary point dTC = 40 − 22.5( 275 ,5 67. 6)K −2.5 = 0 dK 22.5( 275 ,5 67. 6) 40 = K 2.5 22.5( 275 ,5 67. 6) K 2.5 = = 155,006 .78 40 K = 119.16268 (3) Substituting this value into (1) gives L= 275 ,5 67. 6 = 211.84 478 (119.1628)1.5 This time we can check the second-order condition for minimization Differentiating... conditions for a maximum are ∂π = 114K −0 .7 L0.2 R 0.25 − 30 = 0 ∂K ∂π = 76 K 0.3 L−0.8 R 0.25 − 16 = 0 ∂L ∂π = 95K 0.3 L0.2 R −0 .75 − 12 = 0 ∂R (1) (2) (3) From (1) 114L0.2 R 0.25 = 30K 0 .7 R 0.25 = 30K 0 .7 114L0.2 (4) Substituting (4) into (2) 76 K 0.3 L−0.8 30K 0 .7 = 16 114L0.2 76 K 0.3 (30K 0 .7 ) = 16L0.8 (114L0.2 ) 2,280K = 1,824L K = 0.8L (5) Substituting (5) into (4) R 0.25 = 30(0.8L)0 .7 30(0.8)0 .7 L0.5... (1) into the budget constraint 5L + 8K = 70 gives 5L + 8(2 + 0.8L) = 70 5L + 16 + 6.4L = 70 11.4L = 54 L = 4 .74 © 1993, 2003 Mike Rosser (to 2 dp) Substituting this result into (1) K = 2 + 0.8(4 .74 ) = 5 .79 Therefore maximum output is Q = 120L + 200K − L2 − 2K 2 = 120(4 .74 ) + 200(5 .79 ) − (4 .74 )2 − 2(5 .79 )2 = 1,6 37. 28 This technique can also be applied to consumer theory, where utility is maximized subject... 620K 0.3 L0.25 TC = 42K + 5L Therefore the profit function the firm wishes to maximize is π = TR − TC = 620K 0.3 L0.25 − 42K − 5L First-order conditions for a maximum require ∂π = 186K −0 .7 L0.25 − 42 = 0 ∂K ∂π = 155K 0.3 L−0 .75 − 5 = 0 ∂L giving 186L0.25 = 42K 0 .7 L0.25 = and 42 0 .7 K 186 (1) 155K 0.3 = 5L0 .75 31K 0.3 = L0 .75 (2) Taking (1) to the power of 3 L0 .75 = 42 0 .7 K 186 3 = 423 2.1 K 1863 (3)... 0.018q1 = −0.2 − 0.018(169. 87) = −3.2 576 6 < 0 2 ∂q1 ∂ 2π = −0.2 − 0.03q2 = −0.2 − 0.03(131.58) = −4.1 474 < 0 2 ∂q2 ∂ 2π = −0.2 ∂q1 ∂q2 Thus, using the shorthand notation for the above second-order derivatives, (π11 )(π22 ) = (−3.2 576 6)(−4.1 474 ) = 13.51 > 0.04 = (−0.2)2 = (π12 )2 Therefore all second-order conditions for a maximum value of profit are satisfied when q1 = 169. 87 and q2 = 131.58 When a function... 30(0.8L)0 .7 30(0.8)0 .7 L0.5 = 114L0.2 114 © 1993, 2003 Mike Rosser (6) Taking each side of (6) to the power of 3 R 0 .75 = 27, 000(0.8)2.1 L1.5 1143 Inverting R −0 .75 = 1143 27, 000(0.8)2.1 L1.5 (7) Substituting (7) and (5) into (3) 95K 0.3 L0.2 R −0 .75 − 12 = 0 95(0.8L)0.3 L0.2 (114)3 = 12 27, 000(0.8)2.1 L1.5 95(0.8)0.3 L0.3 L0.2 (114)3 = 324,000 (0.8)2.1 L1.5 95(114)3 = 324,000 (0.8)1.8 L 95(114)3 =L 324,000(0.8)1.8... TC2 = [8.5 + 0.03(300)2 ] + [5.2 + 0.04(225)2 ] = 2 ,70 8.5 + 2,030.2 = 4 ,73 8 .7 π = TR − TC = 20, 475 − 4 ,73 8 .7 = £15 ,73 6.30 Note that this method could also be used to solve the multiplant monopoly problems in Chapter 5 that only involved linear functions The unconstrained optimization method used here is, however, a more general method that can be used for both linear and non-linear functions Example 10.19... 0.35492q1 − 320 (4) Using the quadratic formula to solve (4) √ −0.35492 ± −b ± b2 − 4ac = q1 = 2a (0.35492)2 − 4(0.009)(−320) 0.018 √ −0.35492 ± 11.64598 = 0.018 −0.35492 ± 3.412619 = 0.018 Disregarding the negative solution, this gives plant 1 output 3.0 576 99 0.018 = 169. 872 16 = 169. 87 q1 = (to 2 dp) Substituting this value for q1 into (3) q2 = 0 .77 46(169. 872 16) = 131.58 (to 2 dp) Checking second-order . (−0 .75 )155K 0.3 L −1 .75 =−116.25(80. 47) 0.3 (563.3) −1 .75 =−0.0066 572 < 0 ∂ 2 π ∂K∂L = (0.25)186K −0 .7 L −0 .75 = 46.5(80. 47) −0 .7 (563.3) −0 .75 = 0.0186404  ∂ 2 π ∂K 2  ∂ 2 π ∂L 2  = (−0.3653 576 )(−0.0066 572 ). 31K 0.3 K 1.8 = 31(186) 3 42 3 = 2, 692.481 K = 80. 471 179 (4) Substituting (4) into (1) L 0.25 = 42 186 (80. 471 179 ) 0 .7 = 4. 871 7455 L = (L 0.25 ) 4 = (4. 871 7455) 4 = 563.29822 Therefore, first-order conditions suggest. p 1 q 1 + p 2 q 2 = p 1 (5 17 − 3.5p 1 + 0.8p 2 ) + p 2 (77 0 − 4.4p 2 + 1.4p 1 ) = 517p 1 − 3.5p 2 1 + 0.8p 1 p 2 + 77 0p 2 − 4.4p 2 2 + 1.4p 1 p 2 = 517p 1 − 3.5p 2 1 + 77 0p 2 − 4.4p 2 2 + 2.2p 1 p 2 First-order