0
$\begingroup$

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?

  • 3
    But $24+25+26+27$ is not $400$.2012-11-07
  • 0
    It should be "*the* function", not "*a* function"; there's only one such function.2012-11-07
  • 0
    Why don't you set y and z equal to zero? I think that will give you your maximum.2012-11-07
  • 0
    @Ben x < y < z < w, so y and z cannot both be 0.2012-11-07
  • 2
    @Goku, what about 0 < 1 < 2 < 397? Or -2< -1 < 0 < 403? Experiment a bit, and then you might gain some insight?2012-11-07
  • 0
    In your edited question, $w>100$ and $x<100$ must be true.2012-11-07

1 Answers 1

3

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.

  • 0
    Yes. You don't suppose OP meant for the variables to be positive, do you?2012-11-07
  • 0
    @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 :-)2012-11-07
  • 1
    @Gerry: The answer was accepted, so it seems $\mathbb Z$ was indeed intended.2012-11-07