# Is there a “simple” proof of the isoperimetric theorem for squares?

"Simple" means that it doesn't use any integral or multivariable calculus concepts.

A friend of mine who's taking a differential calculus course came up with the problem

Prove that among all the quadrilaterals with a given perimeter, the one with the biggest area is the square.

I solved the problem with Lagrange multipliers, using $2h + 2b$ as the function and $hb = A$ as the constraint. But I'm the one who took the multivariable calculus course, not him.

So I'd like to know if there's a way of proving this theorem using differential calculus concepts, or even geometry and trigonometry.

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Isn't the problem you solved different from the one you pose? You solved a maximization of the perimeter given a fixed area. –  Jose27 Aug 13 '12 at 1:46
Well, $A$ is an arbitrary constant and $h$ and $b$ are variables. I got $h = b$ as the only solution for this maximization, shouldn't that be enough for proving it? –  user1002327 Aug 13 '12 at 1:51
See this. The isoperimetric theorem is discussed a bit later, after the inequality shown on that page. –  Timmy Turner Aug 13 '12 at 1:52
@user1002327: Yes, but you're solving a different problem: You're maximizing perimeter with a given area as constrain. On the other hand you're asking for maximization of the area given a perimeter. –  Jose27 Aug 13 '12 at 1:57
Oh, you're right. –  user1002327 Aug 13 '12 at 1:59

In the case of rectangles, here's a solution of the dual problem: find the rectangle of smallest perimeter for a given area. The shapes should be the same.

Just complete the square (if you can pardon the expression). The width is $w$; the height is $A/w$ (where $A$ is the area). So the semi-circumference is $$w + \frac A w = \left(w - 2\sqrt{A}+ \frac A w\right) + 2\sqrt{A} = \left(\sqrt{w} - \sqrt{\frac{A}{w}}\right)^2 + 2\sqrt{A}.$$ This is as small as possible when the expression that gets squared is $0$. So that $=0$ when $w=\text{what?}$

Later edit: Now let's try it more directly. The perimeter is $4\ell$. You have a rectangle with two opposite sides of length $k$ and and two of length $2\ell-k$. The area is \begin{align} A & = k(2\ell-k) = -k^2 + 2k\ell = -\Big(k^2 - 2k\ell\Big) = -\Big(k^2 -2k\ell + \ell^2\Big) +\ell^2 \\[8pt] & = -\Big(k-\ell\Big)^2 + \ell^2. \end{align} This is as big as possible when $k=\ell$, so you have a square.

We still have the case of non-rectangles to deal with.

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The second proof is very simple, looks like we can use that. And I don't think they're going to deal with the other quadrilaterals in a differential calculus course for engineers, so I'm marking your answer as the correct one. Thanks. –  user1002327 Aug 13 '12 at 2:03

This is essentially the arithmetic-geometric mean (AM-GM) inequality for lists of 2 numbers. http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Geometric_interpretation

In higher dimensions, the inequality says that the hypercube has the most volume among all boxes where the sum of length + width + depth + ... is fixed.

The wiki page above has a bunch of proofs, many of which don't use calculus.

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Since you mentioned being interested in a geometric proof, here's one that should be easy to understand (albeit with one sneaky detail swept under the rug). Start with a $w\times h$ rectangle, with $w\gt h$; we'll prove that there's another rectangle of the same perimiter but greater area. Since the area of the rectangle is twice the area of the triangle of base $w$ and height $h$, we can just consider the triangle's area. But now consider adding some small amount $x$ to the height $h$ and subtracting the same amount from the width $w$. This doesn't change the perimeter, since we're just redistributing a small segment, but we can see what it does to the area:

The area of the original triangle with base $w$ and height $h$ is the sum of the pink and green triangles, while the area of the new triangle with base $w-x$ and height $h+x$ is the sum of the pink and blue triangles. But as long as $w-x\gt h+x$, the blue triangle will have a greater area than the green one: they both have the same base ($x$) and the blue one has a greater height. This implies that as long as $w\gt h$, we can increase the area of the triangle (and thus the rectangle) by trading off some amount of width for the same amount of height, and that in turn implies that the maximum area must be achieved by the square.

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I really like this one, I'm very tempted to choose yours as the correct answer. –  user1002327 Aug 13 '12 at 22:27
Thank you! As I said, it does have one subtle catch - I don't have a quick, easy geometric proof as to why the height of the blue triangle is larger other than the intuition that the intersection point lies on the 'right' side of the $45^\circ$ bisector through the right angle. An algebraic proof is easy, but loses a lot of the charm of the core proof; still, if you present this one, most students probably won't catch you on it! –  Steven Stadnicki Aug 14 '12 at 4:15