My Math Forum Extremum

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 December 16th, 2008, 08:39 AM #1 Newbie   Joined: Dec 2008 Posts: 26 Thanks: 0 Extremum I've got another interesting (I hope) problem and I hope you'll help me to solve it Well it goes like this: Determine all points, where function f has it's extremums, and define whether it is maximum or minimum, if: $f(x,y)=x^{2} - xy + y^{2} - 4ln|x| - 10ln|y|$
 December 16th, 2008, 10:19 AM #2 Senior Member   Joined: Dec 2008 Posts: 251 Thanks: 0 Re: Extremum There is a theorem for twice-differentiable functions f(x, y) that if $1. \,f_x\,=\,f_y\,=\,0$ at $(x,\,y)$ $2.\,f_{xx}f_{yy}\,-\,f_{xy}^2\,>\,0$ at $(x,\,y)$ $3.\,f_{xx}\,<\,0$ at $(x,\,y)$ then f(x, y) has a relative maximum at (x, y), and similarly a relative minimum if $f_{xx}\,>\,0$. (I believe it can be proved with Taylor's theorem.)
 December 16th, 2008, 10:25 AM #3 Guest   Joined: Posts: n/a Thanks: Re: Extremum Think about partial derivatives. $f_{x}=2x-\frac{4}{x}-y$...[1] $f_{y}=2y-\frac{10}{y}-x$...[2] Set each to 0. Solve [1] for y and sub into [2]. Then, solve for x. You can get y by subbing the x's back into the y you found. Check the critical points by using the second partials derivative test. It can be found in any calc book. $D=f_{xx}f(x,y)f_{yy}-f_{xy}^{2}(x,y)$
 December 16th, 2008, 11:13 AM #4 Newbie   Joined: Dec 2008 Posts: 26 Thanks: 0 Re: Extremum Thank you You are really helpful guys

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