Prove that the square has the smallest perimeter of all rectangles of a given area

Prove that for a fixed area, the rectangle with the minimal perimeter is a square.

Proof. Let $x$ and $y$ denote the sides of the rectangle. If the area is fixed, then $xy$ is a constant, say $xy = A$ . The perimeter of the rectangle is then a function $P(x,y) = 2x + 2y$ . But, since

$xy = A \quad \implies y = \frac{A}{x},$

we can write $P$ as a function of $x$ alone,

$P(x) = 2x + 2 \frac{A}{x}.$

So, to find the value of $x$ at which $P(x)$ is minimal we take the derivative,

$P(x) &= 2x + 2\frac{A}{x} \quad \implies \quad P'(x) = 2 - \frac{2A}{x^2}.$

This has a zero at $x = \sqrt{A}$ and we have

$\begin{align*} P'(x) &< 0 & \text{when} && x &< \sqrt{A} \\ P'(x) &> 0 & \text{when} && x &> \sqrt{A}. \end{align*}$

Therefore, $P(x)$ is decreasing for $x < \sqrt{A}$ and increasing for $x > \sqrt{A}$ . Hence, $P(x)$ has its minimum at $x = \sqrt{A}$ . Since $x = \sqrt{A}$ implies $y = \sqrt{A}$ as well, we have that the perimeter is minimal when $x = y$ , i.e., when the rectangle is a square $. \qquad \blacksquare$