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Unformatted text preview: 3 36 2 + 1 2 = 4. Multiply both sides by 36 2 to get 3 + 36 = 4 36 2 , or = 39 12 . Thus, our values of ( x,y, ) are ( 2 39 , 12 39 , 39 12 ) and (2 39 ,12 39 , 39 12 ). Now, we can compute the Bordered Hessian: 6 x 2 y 6 x6 2 y2 Expanding over the rst row, we get6 x (6 x 2 )+2 y (2 y 6 ), which is equal to 36 x 2 + 24 y 2 = (36 x 2 + 24 y 2 ). The sign of the determinant is thus the same as the sign of . So, since for one of the points we found, was positive, that would be a local max. Since for the other, was negative, that would be a local min. (In fact, these are also global.) So, the local max occurs at ( 2 39 , 12 39 ). The local min occurs at (2 39 ,12 39 )....
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 Fall '11
 MichaelParzen

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