Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
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
2  
@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

1 Answer 1

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.

share|improve this answer
    
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
1  
@Gerry: The answer was accepted, so it seems $\mathbb Z$ was indeed intended. –  joriki Nov 7 '12 at 15:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.