# Proof of extremising functionals

I'm having a really hard time knowing where to start with this, I'd appreciate if anyone could give me a hand!

Prove or disprove the following statement. The particular function that extremises a certain functional J among all the functions that render l to another functional K also gives an extremum to the functional K among all the functions that give J a precribed value.

Thanks

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As in: The unit square has minimal circumference among all rectangles of area $1$ and also maximal area among rectangles of circumference $4$. – Hagen von Eitzen Nov 22 '12 at 18:54

Assume $g$ has the property that $K(g)=1$ and that $K(f)=1$ implies $J(g)\ge J(f)$.
Let $c=J(g)$. If $f$ is any function with $J(f)=c$, let $y=K(f)$. Then $K(\frac1yf)=1$, hence $\frac cy=\frac1y J(f)=J(\frac1yf)\le J(g)=c$. We would like this to imply $y\ge 1$, however there are a few obstacles:
• What if $c<0$? No problem, we'd find $<\le 1$ instead - provided the other obstacles don't apply.
• What if $y<0$?
• In fact, what if $y=0$ and we mustn't divide by $y$?
• And what if $c=0$?