# How do you find the dimensions of a rectangle whose area is 100 square meters and whose perimeter is a minimum?

Oct 8, 2014

Let $x$ and $y$ be the base and the height of the rectangle, respectively.

Since the area is 100 ${m}^{2}$,

$x y = 100 R i g h t a r r o w y = \frac{100}{x}$

The perimeter $P$ can be expressed as

$P = 2 \left(x + y\right) = 2 \left(x + \frac{100}{x}\right)$

So, we want to minimize $P \left(x\right)$ on $\left(0 , \infty\right)$.

By taking the derivative,

$P ' \left(x\right) = 2 \left(1 - \frac{100}{x} ^ 2\right) = 0 R i g h t a r r o w x = \pm 10$

$x = 10$ is the only critical value on $\left(0 , \infty\right)$

$y = \frac{100}{10} = 10$

By testing some sample values,

$P ' \left(1\right) < 0 R i g h t a r r o w P \left(x\right)$ is decreasing on $\left(0 , 10\right]$.

$P ' \left(11\right) > 0 R i g h t a r r o w P \left(x\right)$ is increasing on $\left[10 , \infty\right)$

Therefore, $P \left(10\right)$ is the minimum

I hope that this was helpful.

Hence, the dimensions are $10 \setminus \times 10$.