# Now if μ 1 μ 1 μ 1 0 so x 0 by complementary

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Now if μ 0 > 1, μ 1 = μ 0 - 1 > 0, so x = 0 by complementary slackness. It follows that py = 10 ( y = 10 /p ) by the budget constraint, and then 10 = py = 1 / 4 2 0 , so μ 0 = 1 / 40 p . This can only be a solution if 1 / 40 p < 1, which is equivalent to 1 > 40 p . But if μ 0 = 1, μ 1 = 1 - μ 0 = 0. The complementary slackness condition for x is satisfied, so x 0. Then y = 1 / 4 p 2 , and x = 10 - 1 / 4 p 0. This requires 40 p 1. Summing up, if 0 < p < 1 40 , ( x, y ) = (0 , 10) is the only maximum. If p = 1 40 , ( x, y ) = (0 , 10) is still the only maximum. Finally, if p > 1 40 , the maximum is (10 - 1 4 p , 1 4 p 2 ).

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MATHEMATICAL ECONOMICS FINAL, DECEMBER 11, 2000 Page 2 4. Find and classify all critical points of the function f ( x, y ) = ( x 2 + y 2 ) e - x 2 . Answer: The first-order conditions are [2 x - 2 x ( x 2 + y 2 )] e - x 2 and 2 ye - x 2 = 0. It follows that y = 0 and so 2 x - 2 x 3 = 0. The critical points are (1 , 0), (0 , 0), and ( - 1 , 0). The Hessian is: [(2 - 6 x 2 - 2 y 2 ) - 2 x (2 x - 2 x 3 - 2 xy 2 )] e - x 2 - 4 xye - x 2 - 4 xye - x 2 2 e - x 2 This yields: H (0 , 0) = 2 0 0 2 H (1 , 0) = H ( - 1 , 0) = - 4 e - 1 0 0 2 e - 1 . It follows that H (0 , 0) is positive definite because H 1 = 2 > 0 and H 2 = 4 > 0. Thus (0 , 0) is a local minimum (in fact, it is global). Moreover, H (1 , 0) = H ( - 1 , 0) is indefinite because H 2 = - 8 e - 1 < 0. Thus (1 , 0) and ( - 1 , 0) are saddlepoints. The function has no maximum. 5. Consider the matrix A = 3 2 1 3 1 0 3 3 1 a ) What is the rank of A ? Answer: The rank of A is 3. One way to see this is to compute the determinant. It is (3)(1)(1) + (3)(3)(1) + (3)(2)(0) - (3)(1)(1) - (3)(2)(1) - (3)(3)(0) = 3 + 9 - 3 - 6 = 3. Since it is non-zero, the matrix is invertible. As it has 3 rows, it has rank 3.
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