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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 $x<y<z<w$. How can I find the maximum of $f$?

I think the key point is to use $x<y<z<w$. 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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But $24+25+26+27$ is not $400$. – Joe Johnson 126 Nov 7 '12 at 2:29
It should be "the function", not "a function"; there's only one such function. – joriki Nov 7 '12 at 2:42
Why don't you set y and z equal to zero? I think that will give you your maximum. – Ben Nov 7 '12 at 2:46
@Ben x < y < z < w, so y and z cannot both be 0. – amWhy Nov 7 '12 at 2:49
@Goku, what about 0 < 1 < 2 < 397? Or -2< -1 < 0 < 403? Experiment a bit, and then you might gain some insight? – amWhy Nov 7 '12 at 2:51
up vote 3 down vote accepted

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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Yes. You don't suppose OP meant for the variables to be positive, do you? – Gerry Myerson Nov 7 '12 at 6:34
@Gerry: They might, I don't know. It's certainly not unheard of that people pose problems that don't have a solution. Note that the OP didn't object to amWhy suggesting negative values. In any case, no harm done if the resulting experience is that badly posed questions lead to unhelpful answers :-) – joriki Nov 7 '12 at 12:23
@Gerry: The answer was accepted, so it seems $\mathbb Z$ was indeed intended. – joriki Nov 7 '12 at 15:13

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