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Unformatted text preview: 1). Now, the second derivatives work out to be f xx =2 xye( x 2 + y 2 ) / 2xy (1x 2 ) e( x 2 + y 2 ) / 2 , and f yy =2 xye( x 2 + y 2 ) / 2xy (1y 2 ) e( x 2 + y 2 ) / 2 , and f xy = (1x 2 )(1y 2 ) e( x 2 + y 2 ) / 2 . Thus, we can evaluate the second derivatives at the critical points to get. • D (0 , 0) = 0 21 2 =1 < 0 implies a saddle point at (0 , 0). • D (1 , 1) = D (1 ,1) = (2 e1 ) 22 > 0, and f xx (1 , 1) = f xx (1 ,1) = 2 e1 > 0 implies a local minimum at (1 , 1) and at (1 ,1). • D (1 ,1) = D (1 , 1) = (2 e1 ) 22 > 0, and f xx (1 ,1) = f xx (1 , 1) =2 e1 < 0 implies a local maximum at (1 ,1) and at (1 , 1)....
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This note was uploaded on 01/12/2010 for the course MATH 241 taught by Professor Teleman,c during the Winter '08 term at University of California, Berkeley.
 Winter '08
 Teleman,C
 Derivative

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