Me-mo-re-try

For one year of middle school, my math teacher—let's call her W—was also my P.E. teacher. I remember little from the P.E. class except that W loved volleyball and forced us to play often. When I once protested—outside, in the winter cold—that hitting the ball hurt my forearms, W told me that I just needed to "toughen up" and push past the bruising. I wasn't sure what offended me more—the suggestion that I wasn't tough, or the idea that I should injure my creamy, delicate forearms.

I still remember some geometry from W, like

Proposition: among all rectangles with fixed perimeter \(p\) in the plane, the one with maximum area is the square with side length \(\tfrac{p}{4}\).

To prove this, W used a little algebra. Let \(x\) and \(y\) be the side lengths of a rectangle with perimeter \(p=2x+2y\). The area of the rectangle is \[A=xy=x(\tfrac{p}{2}-x)=-x^ 2+\tfrac{p}{2}x\] This is a quadratic in \(x\) which is known to attain its maximum value when (and only when) \(x=\tfrac{p}{4}\), when also \(y=\tfrac{p}{4}\).

I was recently reminded that the proposition is also a consequence of the arithmetic-geometric mean inequality, which W did not cover. It says that the geometric mean of any numbers is always less than or equal to the arithmetic mean of the numbers.

Theorem (AM-GM inequality): For \(a_1,\ldots,a_n\ge 0\), \[\sqrt[n]{a_1\cdots a_n}\le\frac{a_1+\cdots+a_n}{n}\] and equality holds if and only if \(a_1=\cdots=a_n\).

For \(n=2\), the inequality is equivalent to \[xy\le\left(\frac{2x+2y}{4}\right)^ 2\] Applied to W's geometry problem, this tells us that the area \(A=xy\) of any rectangle with side lengths \(x\) and \(y\) and perimeter \(p=2x+2y\) is at most the area of the square with side length \(\tfrac{p}{4}\); moreover, only the square achieves the maximum area, because only it has \(x=y\). That's exactly what we needed. Neat!

We can also ask the dual question.

Question: among all rectangles with fixed area \(A\) in the plane, which one minimizes perimeter?

Again the AM-GM inequality provides the answer. It says that \[4\sqrt{xy}\le 2x+2y\] So the perimeter \(p=2x+2y\) of any rectangle with side lengths \(x\) and \(y\) and area \(A=xy\) is at least the perimeter of the square with side length \(\sqrt{A}\); moreover, only the square achieves the minimum perimeter.

I don't recall whether W ever covered this dual result, which doesn't fit quite as easily into a middle school lesson about quadratics. But both results are nice illustrations of the AM-GM inequality, and the general principle that symmetry is often associated with optimality.