Question

Find the dimensions of a rectangle with area 2500 m2 whose perimeter is as small as possible. List the dimensions in non-decreasing order.

Answer 1

$\bf \textit{area of a rectangle}=A=lw \\\\\\ \textit{perimeter of a rectangle}=P=2w+2l\\\\ -----------------------------\\\\ A=2500\implies 2500=lw\implies \cfrac{2500}{w}=\boxed{l} \\\\\\ P=2w+2\boxed{l}\implies P=2w+2\left( \cfrac{2500}{w} \right)\implies P=2w+\cfrac{5000}{w} \\\\\\ P=2w+5000w^{-1}\\\\ -----------------------------\\\\ now \\\\\\ \cfrac{dP}{dw}=2+5000(-1w^{-2})\implies \cfrac{dP}{dw}=\cfrac{2w^2-5000}{w^2}$


$\bf \textit{critical points will be at } \begin{cases} 0=w^2\\\\ 0=\cfrac{2w^2-5000}{w^2} \end{cases}$

now, there are 3 critical points, two of them do not apply, since the perimeter cannot be either, but one will apply

then run a first-derivative test on that one, to the left-region of it and the right-region of it, to see if it's a minimum

if it's a minimum, then, that's the lowest P you can have for that Area