ECON Answers Homework #4. = 0 6q q 2 = 0 q 3 9q = 0 q = 12. = 9q 2 108q AC(12) = 3(12) = 500

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1 ECON Answers Homework #4 Exercise 1: (a)(i) The average cost function AC is: AC = T C q = 3q 2 54q q (ii) In order to nd the point where the average cost is minimum, I solve the rst-order condition and check with the second-order condition that I have a minimum. - FOC: dac(q) dq - SOC for maximization (of AC): For any q > 0 = 0 6q q 2 = 0 q 3 9q = 0 q = 12 d2 ( AC) dq 2 0 for any q > q 3 0 which is true for any positive q. Hence I conclude that q = 12 is the global minimum of AC. (b)(i) The marginal cost function MC is: MC = dt C dq = 9q 2 108q (ii) From (a)(ii), AC is minimum at q = 12 and is equal to: We also have: AC(12) = 3(12) = 500 MC(12) = 9(12) = 500 Thus, MC = AC when AC is at its minimum. (c) To nd the output at which marginal cost is at its minimum, I proceed as in (a)(ii) 1

2 by solving the FOC and checking the SOC. - FOC: - SOC (for maximization ot (*MC)): dmc -T: : 0 <+ Q:6 For any e) 0, fl#9<0<+for any Q ) 0, - 18 < 0 which is true for any positive q. Hence I conclude that MC is minimized at g:6, (d) Graphs of the marginal cost and average cost functions (on the same graph): Prc SC, Toa (e) Graph of the total cost function (o1 a difierent graph): fc TC 1 6coo looo a f- \tr 6 elo11-t\16rb20l (f) - If q changes from L to 1.1,, the approximate change in the value of 7C obtained q

3 by a linear approximation (or Taylor of order 1) is: T C(1.1) T C(1) + MC(1) 0.1 = [3(1) 3 54(1) (1) ] + [9(1) 2 108(1) + 500] 0.1 = If q changes from 1 to 1.1, the exact value of T C is obtained by evaluating T C at 1.1: T C(1.1) = 3(1.1) 3 54(1.1) (1.1) = The absolute (exact) error is therefore: e = = where the approximation overestimates the change. The percentage (exact) error is therefore: e 100 = Exercise 2: (a) (i) The maximization problem of the rm writes: max Π(K, L) = max [P K,L 0 K,L 0 Lα K α wl rk] The rst-order conditions for prot maximization are: { { Π = 0 P αl α K α 1 = r K Π = 0 P αl α 1 K α = w L (ii) Economic interpretation of the FOC: the rst term in each equation is the change in revenue for a change in the relevant input. This change in revenue contains two terms: the output price is multiplied by the marginal product of the input. This combination is often called the value of the marginal product or VMP, and the economic interpretation of the FOC condition for labor is: V MP L = w. In other words, marginal benet (the revenue generated from hiring an extra unit of labor) must equal marginal cost (the wage rate paid for an extra unit of labor). 3

4 (b) (i) The Hessian matrix writes of the prot function writes: ( ) Π KK Π KL H Π (K, L) = Π LK Π LL ( P α(α 1)L α K α 2 = P α 2 L α 1 K α 1 P α 2 L α 1 K α 1 P α(α 1)K α L α 2 ) (ii) The associated second-order conditions for prot maximization write: Π KK (K, L) < 0 P α(α 1)L α K α 2 α < 1 det(h Π )(K, L) > 0 P 2 α 2 (KL) 2α 2 () > 0 α < 1/2 (iii) Economics interpretation: - Π KK (K, L) is the price times the change in the marginal product of capital for a change in capital. Π KK (K, L) will be negative when α < 1, or in economic terms, when increases in capital cause the marginal product of capital, F K, to decline. A declining marginal product of capital is referred to as diminishing marginal returns to capital. - The second condition holds when α < 1/2 which is more restrictive than the rst condition, α < 1. The key to interpret this condition lies in examining returns to scale. Returns to scale are determined by the percentage change in output when all inputs are simultaneously increased by, say, x%. The three cases are increasing (output rises by more than x%), constant (output rises by exactly x%), or decreasing (output rises by less than x%). For our version of the Cobb-Douglas production function, we have: (xl) α (xk) α = x 2α Q So the second condition, 2α < 1, corresponds to decreasing returns to scale. (c) (i) We now solve the rst-order conditions found in (a)(i) in order to nd K and L the inputs that maximize the prot function of the rm: - From Π L = 0, I get K as a function of L, the exogenous variables and the parameters of the problem. K = ( w P α L1 α ) 1/α ( ) 4

5 - Now, to nd L, I substitute for K in Π K = 0. This yields: [ ( w ) ] 1/α α 1 P αl α P α L1 α = r - Rearranging and combining the exponents gives: L = (αp w α 1 r α ) 1/(1 2α) - Finally, I nd K by substituting L into the above equation ( ): K = (αp r α 1 w α ) 1/(1 2α) (ii) I now calculate the three partial derivatives of L with respect to the three exogenous variables, P, w and r: L 1 P = (αwα 1 r α ) 1/(1 2α) P [1/(1 2α)] 1 > 0 L α 1 w = (αp r α ) 1/(1 2α) w [(α 1)/(1 2α)] 1 < 0 L α = (αp w α 1 ) 1/(1 2α) r [ α/(1 2α)] 1] < 0 r The signs have been derived under the assumption that the SOC hold. Economic interpretation: L is increasing in the output price, but decreasing in both input prices. Note that since L and K are symmetric, similar comparative statics results can also be derive for K. (d) (i) The associated production function Q writes: Q = (L Q ) α = (α 2 P 2 w 1 r 1 ) α/(1 2α) (ii) The three partial derivatives of Q with respect to the three exogenous variables, P, w and r are: Q 2α = (α 2α w α r α ) 1/(1 2α) P 2α/(1 2α) 1 > 0 P Q α w = (α2α r α P 2α ) 1/(1 2α) w α/(1 2α) 1 < 0 Q α = (α 2α w α P 2α ) 1/(1 2α) r α/(1 2α) 1 < 0 r 5

6 The signs have been derived under the assumption that the SOC hold. Economic interpretation: increases in the product's price lead to increases in output, while increases in output prices lead to decreases in output. These results are unambiguous. Yet, we must keep in mind that they were derived using a specic example of a production function. 6