Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 16

a.) Prove that of all the rectangles with a given area, the one with smallest perimeter is a square.
b.) Prove that of all the rectangles with a given, perimeter, the one with greatest area is a square.

a.) Let $A = xy$ and $P = 2x + 2y$ be the area and perimeter of the rectangle with dimensions $x$ and $y$.
If $A = xy$; $\displaystyle y = \frac{A}{x}$, then
$\displaystyle P = 2x + 2 \left( \frac{A}{x} \right) = 2x + \frac{2A}{x}$

If we take the derivative of the Perimeter function with respect to $x$... We get, $\displaystyle P'(x) =2 - \frac{2A}{x^2}$
when $P'(x) = 0$,
$\displaystyle 0 = 2 - \frac{2A}{x^2}$
$\displaystyle \frac{2A}{x^2} = 2$
$x^2 = A$

Then, the critical umber is $x = \sqrt{A}$
if $x = \sqrt{A}$, then
$\displaystyle y = \frac{A}{x} = \frac{A}{\sqrt{A}} = \sqrt{A}$
we got $x = y = \sqrt{A}$, hence, it is a square.

Let $P = 2x + 2y$ and $ A = xy$ be the perimeter and area of the square with dimensions $x$ and $y$.
$\displaystyle A' = 0 = \frac{P}{2} - 2x$
$\displaystyle 2x = \frac{P}{2}$
Then, the critical number is $\displaystyle x = \frac{P}{4}$

$
\begin{equation}
\begin{aligned}
\text{if } x &= \frac{P}{4}, \text{ then}\\
\\
y &= \frac{P - 2 \left( \frac{P}{4} \right)}{2} = \frac{P - \frac{P}{2}}{2} = \frac{P}{4}
\end{aligned}
\end{equation}
$

We got $\displaystyle x = y = \frac{P}{y}$, hence, it is a square.