Question

A rectangle has area $50c{m}^2$ Find the dimensions when it perimeter is the least.

Answer 1

Area of a rectangle $=50{cm}^2$

let the length be $x$

and breadth be $y$

$\therefore$ $xy=50⟶\left(1\right)$

Now, perimeter $=2x+2y⟶\left(2\right)$

From $\left(1\right)$,

$x=50/y⟶\left(3\right)$

Substituting $\left(3\right)$ in $\left(2\right)$ we get

$P=2\left(\frac{50}{y}\right)+2y$

$=\frac{100}{y}+2y$

$\therefore$ $\frac{dP}{dy}=\frac{-100}{y^2}+2=0$

$\Rightarrow 2=\frac{100}{y^2}$

$\Rightarrow y=\sqrt{50}=5\sqrt{2}cm$

$\therefore$ $x=5\sqrt{2}cm$

$\therefore$ For the perimeter to be least for a given area the rectangle would have to be a square.