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# T9 - z,x 2 y-z-2 x-y 3 z 2-1 Therefore critical points...

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ERG2018 Tutorial 9th November 10, 2008 Optimization of Unconstrained Multivariate Functions Critical Points and First Order Condition: f ( x ) = 0 . Classiﬁcation of Critical Points and Second Order Conditions: A critical point is LocalMaximum, if eigenvalues of the Hessian matrix 2 f ( x ) are nonpositive; LocalMinimum, if eigenvalues of the Hessian matrix 2 f ( x ) are nonnegative; SaddlePoint, if eigenvalues of the Hessian matrix 2 f ( x ) have both strictly positive and negative elements. To ﬁnd the eigenvalue of a matrix A , solve the equation system det ( λI - A ) = 0 . Example: Determine the critical points of f ( x,y,z ) = x 2 + xy + y 2 - 2 xz - yz + z 3 - z and classify them as local maxima, local minima, or saddle. 1. First, the gradient of f ( x,y,z ) is f ( x,y,z ) = (2 x + y - 2
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Unformatted text preview: z,x + 2 y-z,-2 x-y + 3 z 2-1) . Therefore, critical points ( x,y,z ) can be identiﬁed with equation:    2 x + y-2 z = 0 x + 2 y-z = 0-2 x-y + 3 z 2-1 = 0 Solving this, we get two group of solutions: (1 , , 1) and (-1 / 3 , ,-1 / 3) , which are the critical points of function f ( x,y,z ) . 2. The Hessian Matrix of f ( x,y,z ) is Hf ( x,y,z ) =   2 1-2 1 2-1-2-1 6 z   . For the ﬁrst critical point, the eigenvalues are 2 , 4-√ 10 , 4 + √ 10 , which are all positive, therefore it is a local minimum point. For the second critical point, the eigenvalues are-2 . 89 , 1 . 09 , 3 . 80 , therefore it is a saddle point. 1...
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