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Let $f$ be the function such that $f(x,y,z,w)=x+w, \quad x,y,z,w\in{\Bbb Z}$ where $ x+y+z+w=400, $ and $x. How can I find the maximum of $f$?

I think the key point is to use $x. I guess $98<99<101<102$ should be a choice. But I have no idea about how to give a proof.


[EDITED:] According to answers, $\max f=+\infty$. What's the minimum of $f$? I think there should be a bound. Playing around the examples, I think $\min f$ should be given by $(98,99,101,102)$. Any examples "better" than this?

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    In your edited question, w>100 and x<100 must be true.2012-11-07

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Since the sum of all four variables is constant, maximizing $x+w$ is equivalent to minimizing $y+z$. Since you can make $x,y,z$ as negative as you like and then use $w=400-(x+y+z)$, $f$ is unbounded and has no maximum.

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    @Gerry: The answer was accepted, so it seems $\mathbb Z$ was indeed intended.2012-11-07