Question

Show that of all the rectangles with a given area the one with smallest perimeter is a square.

Answer 1

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

$xy=A⟹y=\frac{A}{x},$

we can write P as a function of x alone,

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

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

$P\left(x\right)=2x+2\frac{A}{x}⟹P^{\prime }\left(x\right)=2-\frac{2A}{x^2}$.

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

$\begin{array}{cccccc}{P}^{\prime }\left(x\right)& <0& \text{when}& & x& <\sqrt{A}\\ P^{\prime }\left(x\right)& >0& \text{when}& & x& >\sqrt{A}.\end{array}$

Therefore, P(x) is decreasing for$x<\sqrt{A}$ and increasing for $x>\sqrt{A}.$ Hence, $P\left(x\right)$ 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.