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Unformatted text preview: f ( x,y ) = x 2 + 2 y 36 y . Solving f x = 2 x = 0 and f y = 6 y 26 y = 0, we have that the only critical points are (0,1) and (0,1). Since f xx = 2 ,f xy = f yx = 0 and f yy = 12 y , Hf ( x,y ) = f xx ( x,y ) f xy ( x,y ) f yx ( x,y ) f yy ( x,y ) = 2 0 12 y . Hence Hf (0 , 1) = 2 0 12 . Since 2 > 0 and det Hf (0 , 1) > 0, (0,1) is a local minimum. And Hf (0 ,1) = 212 . Since det Hf (0 ,1) < 0, (0,1) is a saddle point. 1...
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This note was uploaded on 09/02/2010 for the course MATH 241 taught by Professor Any during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 Any
 Math, Calculus

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