b) suppose thatr w 2, so that production cost in terms of K and L can be wrtten

Question

b) suppose thatr w 2, so that production cost in terms of K and L can be wrtten 2K+2L the isoquant slope MP /MP Is equal to -K/L, so that equating the isoquant slope to the -1 slope of the lsocost line yields K L. Substitute K- L in the production function Q (KL). Then use the resuitng equation to solve for L as a function of Q, using the exponent rules from above. This relationshlp gives the cost minim ng L as a function of Q. This function has the m l-b d where the multiplicative factor b- and the exponent d = use the abo e exponent les Since K-L the same function gves as a una on of Q c) Now substitute your solutions into the cost expression 2K+ 2L to get cost C as a function of Q. This function is given by CIa) ga", whereand h d) The average cost function AC Q s equal to cost d vided by output, or C Q Q Using your solution for Q t follows that AC Q = aQ , where a = and m = Graphing AC as a function of Q the result A, O B. ° C. a horizontal line an upward sloping curve a downward sloping curve Marginal cost MC Q S given by the der at e of C(Q), but it can also be der ved as the increase in cost when Q goes up by 1 unit using either approach, MC(Q "Q' here z = and r = The MC curve is O A. a downward sloping curve OB. an upward sloping curve O c. a horizontal ine higher than the AC curve D, a horizontal line that colncides with the horizontal AC curve

Explanation / Answer

part a. Substituting K=L in the production function: Q =( L2)1/2 ----> L = 1Q1. Thus, we have b = 1 and

part b. d = 1.

part c. We have K = L = Q. Cost Function C(Q) = 2K + 2L = 2Q+2Q = 4Q1. So, g = 4 and h =1

part d. AC(Q) = C(Q)/Q = 4Q/Q = 4Q0. We have, a = 4 and m = 0.

part e. When we graph the AC function as a function of Q (with Q on the horizontal axis and AC on the vertical axis) we will get a constant horizontal line = 4. So the answer is choice A.

part f. MC(Q) = dC(Q)/ dQ = 4Q0. Thus, z=4 and r=0.

part g: When we graph the MC function as a function of Q (with Q on horizontal axis and MC on vertical axis) we will get a constant horizantal line = 4. This is equal to the AC line. Thus, the answer choice is D (a horizontal line that coincides witht the horizontal AC curve).

part a.

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