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what is the maxium area of a fixed perimeter closed graph?

can someone prove that the answer is a circle.

I can prove that the square have maxium area for a fixed perimeter rectangle.

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    If you restrict to sufficiently smooth curves, proving the circle is optimal is a standard problem in the calculus of variations.2012-12-25

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Consider a regular $n$-gon inscribed in a circle of radius $r$. Each side has length $2r \sin(\pi/n)$ giving the perimeter $P_n=2nr\sin \frac{\pi}{n}.$ And each of the $n$ triangles has area $r^2 \sin(\pi/n)\cos(\pi/n),$ giving the area $A_n=n r^2 \sin \frac{\pi}{n} \cos \frac{\pi}{n}.$ If we set $P=1$ for sake of comparisons, and solve for $r$ and plug into $A$ (and simplify) we have $A_n=\frac{1}{4n \tan(\pi/n)}.$ This increases as $n=3,4,5,...$ with the limit $1/(4\pi)=0.079577...,$ which is the area of a circle of perimeter 1. Note that for the square, $A_4=0.0625$, while for the hexagon, $A_6=0.0721687...$.

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    I thought so, I just wanted to point it out for the OP in case they thought otherwise. I very much like how intuitive this is, though! It is definitely worth remembering.2012-12-25