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x x = 2 .
xn x2 + x2 + · · · + x2 = 1,
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
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
a22 · a2n ,
A = 21
an1 an2 · ann
a short calculation shows that at the point where f assumes its maximum,
= 2ai1 x1 + 2ai2 x2 + · · · + 2ain xn − 2λxi = 0,
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This document was uploaded on 01/12/2014.
- Winter '14