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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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A hexagon beats a square, and generally an $n$-gon gets better as $n$ increases, approaching a circle. – coffeemath Dec 25 '12 at 6:23
@coffeemath can we prove this rigorously? – Charles Bao Dec 25 '12 at 6:32
That a hexagon beats a square is shown by a calculation. One has to find:[1] area for square of perimeter P, [2] area for regular hexagon of perimeter P. You may as well put P=1 to compare. – coffeemath Dec 25 '12 at 6:44
If you restrict to sufficiently smooth curves, proving the circle is optimal is a standard problem in the calculus of variations. – user7530 Dec 25 '12 at 8:00

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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This is pretty. But it only shows that a circle is better than any regular $n$-gon. There are still a lot of curves to be accounted for. – AndrewG Dec 25 '12 at 8:54
Andrew: I agree. I was responding to a comment by the OP in which he asked for a proof of my comment about hexagons beating squares, etc. The isoperimetric inequality is of course much deeper than comparing areas for polygons! – coffeemath Dec 25 '12 at 9:06
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. – AndrewG Dec 25 '12 at 9:08

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