# Proving that a function has an absolute maximum/minimum

I am stuck with the following problem. I can find the absolute minimum or maximum of a function with 2 variables or more, but I can't prove that absolute values exist.

Assume we have the following function:

$$f(x,y)=\frac{x^2-y^2}{\pi^{x^2-y^2}}$$

I need to show that the absolute maximum and minimum values exist

My idea: For a function to have absolute min/max value, the domain set should be compact: that is, closed and bounded and the function be continuous. But I am really stuck here. I dont know how to show that the domain set is compact. Could anyone please help me with this matter?

• This is a one variable problem. Oct 27 '14 at 0:11
• So how should I approach to the problem then? Could you please elaborate a little? Oct 27 '14 at 0:12

We are maximizing/minimizing $\frac{t}{\pi^t}$. The derivative is $(1-t\ln \pi)/\pi^t$. So the function is increasing for a while then decreasing. It follows that the max exists.
There is no minimum, since when $t$ is large negative, then the top is large negative and the bottom is close to $0$, so our function is very large negative.
a note: the domain need not be compact for a function to have an absolute minimum or maximum. Take the trivial function $g(x)=0$. This function has an absolute minimum equal to its absolute maximum, which is 0.
As Andre Nicolas mentions, you can make this into a one variable problem. Let $r=x^2+y^2$. Then let $\tilde{f}(r)=\frac{r}{\pi^r}$. Try taking the derivative and see where this leads you.
• I think $r$ can be negative if $x<y$ Oct 27 '14 at 0:22