Question

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

Answer 1

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

Since the area is $100m^2$.

$xy=100\Rightarrow y=\frac{100}{x}$

The perimeter $P$ can be expressed as

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

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

By taking the derivative,

$P^{\prime }\left(x\right)=2(1-\frac{100}{x^2})=0\Rightarrow x=\pm 10$

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

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

By testing some sample values,

$P^{\prime }\left(1\right)<0\Rightarrow P\left(x\right)$ is decreasing on $(0,10]$.

$P^{\prime }\left(11\right)>0\Rightarrow P\left(x\right)$ is increasing on $[10,\infty )$

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

I hope that this was helpful.

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