Question

If the area of a rectangle is $64$ sq unit, find the minimum value possible for its perimeter.

Answer 1

Let the dimensions be $a$ and $b$
$\therefore$ $Area=ab$ $\Rightarrow$ $64=ab$
Perimeter $=2(a+b)$
$P=2(a+\frac{64}{a})$
$\Rightarrow$ $\frac{dP}{da}=2(1-\frac{64}{a^2})$

For maxima and minima, put $\frac{dP}{da}=0$
$\therefore$ $2(1-\frac{64}{a^2})=0$
$\Rightarrow$ $a^2=64$ $\Rightarrow$ $a=\pm 8$
Now

$\frac{d^{2}P}{d{a}^2}=2(+\frac{128}{a^3})$
At $a=8$
$\frac{d^{2}P}{d{a}^2}=2\left(\frac{128}{8^3}\right)=\frac{1}{2}>0$, minima
$\therefore$ Minimum value of $P\left(8\right)=2(8+\frac{64}{8})=32$