01test3 - Mathematical Economics Final 1 Consider the...

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Mathematical Economics Final, December 13, 2001 1. Consider the function f ( x, y ) = x 3 + xy + y 2 - x/ 2 + y . Find and classify (maximum, minimum, saddlepoint) all critical points of f . Answer: The derivative is df = (3 x 2 + y - 1 / 2 , x +2 y +1). Setting this to zero, we find 3 x 2 + y - 1 / 2 = 0 and x + 2 y = - 1. The second equation can be rewritten y = - (1 + x ) / 2. Substituting in the first equation yields 3 x 2 - x/ 2 - 1 = 0. This has solutions x = 2 / 3 and x = - 1 / 2. The critical points are then (2 / 3 , - 5 / 6) and ( - 1 / 2 , - 1 / 4). The Hessian is: H = d 2 f = 6 x 1 1 2 . Thus H 1 = 6 x and H 2 = 12 x - 1. When x = 2 / 3, H 1 = 4 > 0 and H 2 = 7 > 0. The Hessian is positive definite, and so (2 / 3 , - 5 / 6) is a local minimum. (There is no global maximum or minimum since f ( x, 0) = x 3 - x/ 2 is unbounded above and below.) When x = - 1 / 2, H 1 = - 3 < 0 and H 2 = - 2 < 0. The Hessian is indefinite, and so ( - 1 / 2 , - 1 / 4) is a saddlepoint. 2. A consumer has utility function u ( x, y ) = ln x + ln y . The consumer consumes non-negative quantities of both goods, subject to two budget constraints: 3 x + 2 y 6 and 2 x + 3 y 6. Find ( x * , y * ) that maximizes utility subject to the above four constraints. Be sure to check the constraint qualification and second-order conditions. Answer: This sort of problem can arise when one of the goods is rationed via rationed coupons, and there is a market for ration coupons where the relative price for coupons is different than for goods. We first consider constraint qualification. There are four constraints: 3 x + 2 y - 6 0, 2 x + 3 y - 6 0, - x 0, and - y 0. The matrix of derivatives of the constraints is: 3 2 2 3 - 1 0 0 - 1 .
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