This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Page 1 Assignment 7 100 points total plus 10 bonus points Section 15.3 Problem 6 : Our objective function is ? ? , ? = ?? and our equation of constraint is ? ? , ? = 4 ? 2 + ? 2 = 8 . Their gradients are ? ? , ? = ? + ? ? ? , ? = 8 ? + 2 ? So the equation ? = ? becomes ? + ? = 8 ? + 2 ? . This gives 8 ? = ? and 2 ? = ? Multiplying, we get 8 ? 2 = 2 ? 2 If = 0 , then ? = ? = 0 , which does not satisfy the constraint equation. So and we get 2 ? 2 = 8 ? 2 ? = 2 ? To find ? , we substitute for ? in our equation of constraint. 4 ? 2 + ? 2 = 8 4 ? 2 + 4 ? 2 = 8 ? = 1 So our critical points are 1,2 , 1, 2 , ( 1,2) and 1, 2 . Since the constraint is closed and bounded, maximum and minimum values of ? subject to the constraint exist. Evaluating ? ? , ? at the critical points, we have ? 1,2 = ? 1, 2 = 2 ? 1, 2 = ? 1,2 = 2 Thus, the maximum value of ? on ? ? , ? = 8 is 2, and the minimum value is 2 . Problem 10 : The gradients of the objective and the constraint functions are ? = 2 + + 4 ? Page 2 ? = 2 ? + ? + 2 ? So, we have ? = ? ? = 1, ? = 2 Going back to the constraint function, we can solve for = 11 . This gives us one critical point 1,11,2 . ? 1,11,2 = 21 This is the maximum value of ?? , , ? on ?? , , ? = 16 . To see this, we note that = 16 ?...
View
Full
Document
This note was uploaded on 03/21/2010 for the course AMS 261 taught by Professor Fortmann during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 FORTMANN

Click to edit the document details