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Unformatted text preview: 3 36 λ 2 + 1 λ 2 = 4. Multiply both sides by 36 λ 2 to get 3 + 36 = 4 · 36 λ 2 , or λ = ± √ 39 12 . Thus, our values of ( x,y,λ ) 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 λ 2 y2 λ ± ± ± ± ± ± Expanding over the ﬁrst row, we get6 x (6 x ·2 λ )+2 y (2 y ·6 λ ), which is equal to 36 x 2 λ + 24 y 2 λ = λ (36 x 2 + 24 y 2 ). The sign of the determinant is thus the same as the sign of λ . So, since for one of the points we found, λ was positive, that would be a local max. Since for the other, λ 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
 Optimization, Harvard University, Mathematical optimization, local max, Rachel Epstein

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