ANSWERS Exam 1 Practice Problems

Convexityanecessaryandsufficientconditionafunction

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Unformatted text preview: he other derivatives, however, we see: 2 ; 2; 2. Notice, , , , 22 0, so this approach is insufficient to establish strictly convexity of the 2nd principal minor, in this case, is 22 the function. Convexity: (A necessary and sufficient condition). A function is concave if and only if its Hessian 0 and , positive semi‐definite over the entire domain. Here, we need is 0. This function satisfies both criteria. 10 , 2B) Show the function is strictly concave. ANSWER: Continuing with what we did in 2A, we need to show that the principal minors of the Hessian matrix are negative definite. This means they will alternate in sign. , 2 ; principal minor, 2 ; , 2 , , 2; and , 0. Hence: The first , 2 0 0 and the second principal minor, 0 2 4 0, so this function satisfies the conditions needed for strict concavity. , 2 and b) , 3. Consider the following function: subject to the constraints a) 1 1 . Suppose we wanted to maximize 0. 3A) Using the techniques we studied in class, do the constraints bind? Explain. 3B) Find the optimum for this problem. 1 ANSWER: Form the Lagrangian: We have, as first‐order conditions: a) 0 and 1 0. b) 0 and 1 0. 1 2 . , c) 0 and 2 0 . Consider first, corner solutions. a) Suppose 0. Then, since we need 1 0, the assumption that 0 implies 0 1 0⇒ 1 . We restrict 0. If 0 (the inequality constraint, 0, we have 2. With 2, we have from the condition binds) then from c) and the assumption 1 0 that 1, which is consistent with the constraint binding. 2 , 0. Since the objective function and constraint are symmetric in the variables, the same would apply if we set In each case, the objective function obtains a value of 2. 0 (the constraint does not bind). In this case, the conditions Now suppose a) 0 and 1 0. b) 0 and 1 0. 1, 0. Suppose both 1. If this is the case, the con...
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This document was uploaded on 02/07/2014.

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