Sol11 - SOLUTION TO ASSIGNMENT 11 - MATH 251, WINTER 2007...

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SOLUTION TO ASSIGNMENT 11 - MATH 251, WINTER 2007 1. It is known that a differentiable function f : R 2 R has a maximum at a point P if ∂f/∂x = ∂f/∂y = 0 at P and the 2 × 2 symmetric matrix - ˆ 2 f ∂x 2 2 f ∂x∂y 2 f ∂x∂y 2 f ∂y 2 ! is positive definite; minimum at a point P if ∂f/∂x = ∂f/∂y = 0 at P and the 2 × 2 symmetric matrix ˆ 2 f ∂x 2 2 f ∂x∂y 2 f ∂x∂y 2 f ∂y 2 ! is positive definite; saddle point at P if the 2 × 2 symmetric matrix ˆ 2 f ∂x 2 2 f ∂x∂y 2 f ∂x∂y 2 f ∂y 2 ! has one negative eigenvalue and one positive eigenvalue. If P is either a maximum, minimum or saddle point, we call it a simple critical point. Determine the nature of the simple critical point of the following functions at the origin (0 , 0) f ( x,y ) = 2 x 2 + 6 xy + y 2 , f ( x,y ) = x sin( x ) - cos( y ) - xy. (You may view the graphs and rotate them in Maple using plot3d(2*xˆ 2+6*x*y+yˆ 2, x= -4. .4, y = -4. .4); plot3d(x*sin(x) - cos(y) -x*y, x= -4. .4, y = -4. .4, numpoints=3000); ). Remark. This criterion can be generalized to functions f : R n R . If all the first partials vanish at a point P and the matrix of mixed derivatives is positive definite (resp. negative definite) at P , then the function has a minimum (resp. maximum) at P . Solution. The function f ( x,y ) = 2 x 2 + 6 xy + y 2 . We have ∂f ∂x = 4 x + 6 y, ∂f ∂y = 6 x + 2 y. Both partials vanish at zero. The Hessian matrix is ˆ 2 f ∂x 2 2 f ∂x∂y 2 f ∂x∂y 2 f ∂y 2 ! = ± 4 6 6 2 . The eigenvalues are 3 ± 37. One is negative and the other positive. Thus, zero is saddle point. 1
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2 SOLUTION TO ASSIGNMENT 11 - MATH 251, WINTER 2007 The function f ( x,y ) = x sin( x ) - cos( y ) - xy . We have ∂f ∂x = x cos( x ) + sin( x ) - y, ∂f ∂y = sin( y ) - x. Both partials vanish at zero. The Hessian matrix is ˆ 2 f ∂x 2 2 f ∂x∂y 2 f ∂x∂y 2 f ∂y 2 ! = ± 2cos(
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Sol11 - SOLUTION TO ASSIGNMENT 11 - MATH 251, WINTER 2007...

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