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Unformatted text preview: 4. F ( x, y ) = 2 ye x-32 x-y 2 + 5 hence F x = 2 ye x-32 F y = 2 e x-2 y Setting these equal to zero, using F y = 0 we nd y = e x . Substituting into the rst equation, we nd 2 e 2 x = 32. Dividing by 2 and taking natural logarithms, we nd x = ln(4) and thus, y = 4. Computing second derivatives, 2 F x 2 = 2 ye x 2 F y 2 =-2 2 F xy = 2 e x Hence, the Hessian is D F =-4 ye x-4 e 2 x So that at the one potential relative extremum or saddle point, we have D F (ln(4) , 4) =-128 < 0. Hence, this point is a saddle point. 5. a. A,D,F; b. H; c. B; d. K; e. I...
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This note was uploaded on 05/12/2010 for the course CHEM 3A taught by Professor Fretchet during the Fall '08 term at University of California, Berkeley.
- Fall '08