I have a positive definite, multidimensional quadratic form: $(x-x_o)^t M (x-x_o)$, where the "${}^t$" indicates transpose and $M$ is a positive definite matrix (in fact, it is a multidimensional guassian). So I know there is a single global maximum (at $(x_o,y_o)$). Problem is that I need the maximum subject to the constraint that all coordinates of the extremum must be non-negative. I know the desired maximum must have at least one zero coordinate (i.e. it is in the enclosing boundary). Does anybody know how to solve this?
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I am confused. If $M$ is positive definite, then the quadratic form is unbounded from above and has no maximum. It does has a global minimum at $x_o$. As for your question: Are you searching for the maximizer of $(x-x_o)^t M (x-x_o)$ subject to $x\ge 0$? – user251257 Jul 14 '15 at 23:18
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I'm sorry. You are right, It is the minimum I'm searching for, subject to x non-negative for each coordinate. – Luis A. Aguilar Jul 15 '15 at 00:41
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According to comments, the answer is about minimizing a strictly convex quadratic form:
If you just need a software package to solve it, most packages do that, like Matlab. Just look for quadratic programming.
If you need a method to compute a solution, there is for example the projected gradient or interior point method.
If you need a theoretical results, you should consider the Karush–Kuhn–Tucker conditions.
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