ANSWERS Exam 1 Practice Problems

Substitutinginwehave 2 0 1andthen

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Unformatted text preview: straint binds, so 0 can’t be true. Imply: 0, 1. In this case, the constraint would not bind, but the Note, we could also have a corner, such as objective function would then be 1, which is less than what we had before. If . , the function We are left then with a case where the constraint would bind and we have interior solutions, we would then have: 0 and 1 0 2 2 We would then have: So, 2 1 1 0 . Substituting in, we have: 2 0⇒ 1. And then 0. If the constraint binds, 0 1 0, 1. Here, 1 So, in this problem, the function achieves a maximum at 1 1 or 0 and 1 1 , 1 0. 0. , or , . is increasing in its two arguments provided each is above This makes sense. The function 1. Since they can’t both be above 1 (the constraint is that their sum can be no greater than 2) then it makes sense to either a) shut one of them down (set equal to 0) and set the other one as high as it can go, i.e., 2, or b) shut both of them down (set bother equal to 0). 4. Solve the following constrained optimization problems (be sure the appropriate second order conditions are satisfied). 3 , 4A) Minimize 2 5 subject to 0 10, 2 10 . ANSWER: This one is quite easy. is increasing in both of its arguments. Therefore, to minimize this , function, we would want to set the variables to their smallest values: 0 and 2. 2 , 4B) Maximize 3 4 , subject to 0 ANSWER: The Lagrangian for this problem is: 2 3 4 1 and 0 1 1 1. The first order conditions are: 2 6 0 and 2 6 0. 1 8 0 and 1 8 0. 0 and 1 0. 0 and 1 0. Suppose 0 and implies 6 2, or 0. But then, the condition 0 and 0. Then 0 implies 1 6 8 2 6 6 1, by the third condition, 0. But if this were true, the condition 0 and If both constraints do not bind ( 2 1, by the fourth condition, and 2 0 (the first condition) 0 implies . (Note, the 1, does not bind here. constraint, 0 Suppose 0. Then 0 and 1 8 , Fin...
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This document was uploaded on 02/07/2014.

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