Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 6

First the dimension of a rectangle with area $1000 m^2$ whose perimeter is as small as possible.
Let $A$ and $P$ be the area and the perimeter of the rectangle respectively.
So, $A = xY$; $xy = 1000$
We have, $\displaystyle x = \frac{1000}{y} \quad \Longleftarrow \text{(Equation 1)}$

Also, $P = 2x + 2y$
Substitute Equation 1

$\begin{equation} \begin{aligned} P &= 2 \left( \frac{1000}{y} \right) +2y\\ \\ P &= \frac{2000}{y} + 2y \end{aligned} \end{equation}$

when $P' = 0$,

$\begin{equation} \begin{aligned} P' &= \frac{-2000}{y^2} +2\\ \\ 0 &= \frac{-2000}{y^2} +2\\ \\ y^2 &= \frac{2000}{2}\\ \\ y &= \sqrt{1000}\\ \\ y &= 10\sqrt{10}m \end{aligned} \end{equation}$

So,

$\begin{equation} \begin{aligned} x &= \frac{1000}{y} = \frac{1000}{10\sqrt{10}}\\ \\ x &= 10 \sqrt{10}m \end{aligned} \end{equation}$

Therefore, the rectangle has length $x = 10 \sqrt{10}m$ and width $y = 10 \sqrt{10}m$