Bordered Hessian problem

Bordered Hessian problem - 3 36 2 1 2 = 4 Multiply both...

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Harvard University, Math 20 Fall 2011, Instructor: Rachel Epstein 1 Bordered Hessian Find the local maxima and minima of f ( x,y ) = x + 2 y subject to the constraint 3 x 2 + y 2 = 4. The solution is on the next page. (Note: your answers will be a bit ugly. Also note that the hard part of a question like this is finding the possible points using the Lagrange multiplier method. Once you know what the Bordered Hessian is, it is easy to apply it to figure out what is a local max or min. If you want to do more practice problems, you could take any of the constrained optimization problems for 2 variables that we have seen, or make one up, and write out the Bordered Hessian without bothering to find the possible points.)
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Harvard University, Math 20 Fall 2011, Instructor: Rachel Epstein 2 L ( x,y ) = x + 2 y - λ (3 x 2 + y 2 - 4). Set all derivatives equal to 0 to get: 1 - 6 = 0 2 - 2 = 0 3 x 2 + y 2 - 4 = 0. Now, We can solve for x and y to get x = 1 6 λ and y = 1 λ . Plugging these into the third equation (which is the constraint) gives us:
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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 x-6 λ 2 y-2 λ ± ± ± ± ± ± Expanding over the first row, we get-6 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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Bordered Hessian problem - 3 36 2 1 2 = 4 Multiply both...

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