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Provide a convincing argument that a firm using with g(K,L) =min{K/4, L/5} would select K*= 128...

Provide a convincing argument that a firm using with g(K,L) =min{K/4, L/5} would select K*= 128 in the Long Run given P=$96, w=$16 and r=$4??

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Answer #1

The production function is given as Q = min (K/4 , L/5) , which is a Leontiff production function having L-shaped kinked isoquant. The cost of production is given as c=WL + TK or c = 16 L + 4 K .

The optimal combination of inputs would be where the corner of the isoquant lies, which is where K/4 = L/5 or K= 0.81 . Putting this in the production function as Q = min ((0.8 L)/4 , L/5) or Q = min (L/5 , L/5) or Q = L/5 or L^* = 5Q , and since K= 0.81 , we have K^* = 0.8 L^* or K^* = 0.8 (5Q) or K^* = 4 Q . These are the required conditional input demand.

In the long run, the firm must have zero economic profit. For K*=128, since K^* = 4 Q we have 128 = 4 Q or Q = 32 . Also, since L^* = 5Q , we have L^* = 5*32 or L^* = 160 . For the given price, the total revenue at this quantity would be TR = PQ = 96*32 = 3072 dollars. The total cost would be as TC = 16 L^* + 4 K^* = 2560 + 512 = 3072 dollars. The profit here would be \pi = TR - TC or \pi = 3072 - 3072 = 0 . Hence, the firm would employ K*=128 in the long run for the given price and input prices.

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OPTIONAL : Note however that the question asks for a convincing arguement which is provided above, but this would be true for any Q>0. The reason being that the cost function is 96Q, and marginal cost of $96 is equal to price of $96 for all Q>0. For example, take Q=10, we would have K*=40, L*=50, and hence \pi = TR - TC = PQ - (wL^* + rK^*) = 960 - (16*50 + 4*40) = 960 - 960 = 0 .

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