Continuing in this way we nally obtain n mutually

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Unformatted text preview: 1 x x = 2 . · xn x2 + x2 + · · · + x2 = 1, 1 2 n or xT x = 1, where We let g (x) = g (x1 , x2 , . . . , xn ) = xT x − 1, so that the equation of the sphere is given by the constraint equation g (x) = 0. Our approach consists of finding of finding the point on S n−1 at which the function f (x) = f (x1 , x2 , . . . , xn ) = xT Ax assumes its maximum values. To find this maximum using Lagrange multipliers, we look for “critical points” for the function H (x, λ) = H (x1 , x2 , . . . , xn , λ) = f (x) − λg (x). These are points at which ∇f (x1 , x2 , . . . , xn ) = λ∇g (x1 , x2 , . . . , xn ), and g (x1 , x2 , . . . , xn ) = 0. In other words, these are the points on the sphere at which the gradient of f is a multiple of the gradient of g , or the points on the sphere at which the gradient of f is perpendicular to the sphere. If we set a11 a12 · a1n a a22 · a2n , A = 21 · · · · an1 an2 · ann a short calculation shows that at the point where f assumes its maximum, ∂H = 2ai1 x1 + 2ai2 x2 + · · · + 2ain xn − 2λxi = 0, ∂xi o...
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This document was uploaded on 01/12/2014.

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