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Unformatted text preview: Questions and Answers from Econ 210A Final: Fall 2008 I have gone to some trouble to explain the answers to all of these questions, because I think that there is much to be learned by working through them. Please let me know if you find mistakes or inadequate explanations. Ted Bergstrom Question 1) Firm A has production function f ( x 1 ,x 2 ) = x 1 / 2 1 + x 1 / 2 2 2 k where k > 0. a) For what values of k is f a quasiconcave function? For what values of k is f a concave function? Explain your answers. Firm B has production function g ( x 1 ,x 2 ) = ( x 2 1 + x 2 2 ) k/ 2 where k > 0. b) For what values of k is g a quasiconcave function? For what values of k is g a concave function? Explain your answers. Answers to Question 1 Answer to 1a: The easiest way to check for quasiconcavity of f is to remem ber that a function is quasiconcave if and only if every monotonic increasing transformation of that function is quasiconcave. In particular, the function f ( x 1 ,x 2 ) = x 1 / 2 1 + x 1 / 2 2 2 k is a monotonic increasing transformation of the function g ( x 1 ,x 2 ) = x 1 / 2 1 + x 1 / 2 2 , But it is easy to verify that g is quasiconcave. You could do this by checking the bordered Hessian conditions, which are now pretty simple because there are no offdiagonal terms in the regular Hessian part of the bordered Hessian. You could make the problem even easier by checking that g is a concave function and then using the fact that a concave function must be quasiconcave. Or you could alternatively just note that the isoquants for g have diminishing marginal rate of substitution. When is the function f concave? We could check this by writing out the Hessian matrix and making sure that it is negative semidefinite. But there is an easier way. We see that the function f is homogeneous of degree k . We know that it is quasiconcave, and our textbook tells us that if a function is quasi concave and homogeneous of degree 1 it is concave. Suppose that k < 1. Define 1 the function h ( x ) = f ( x ) 1 /k . We know that h ( x ) is homogeneous of degree one and quasiconcave, so it is concave. But now f ( x ) = h ( x ) k where k < 1. This means that f is a concave function of a concave function and hence must be concave. What if f is homogeneous of degree k > 1? Then it will not be a concave function. There are increasing returns to scale. It is easy to check that f is not a concave function. One could prove this by showing a single example of two points that violate the condition for concavity For example let x = (0 , 0) and x = (1 , 0). Then f ( x ) = 0, f ( x ) = 1. Now for any t (0 , 1), tx + (1 t ) x = ( t, 0). We have (1 t ) f ( x ) + tf ( x ) = t . But f ((1 t ) x + tx ) = f ( t, 0) = t k ....
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