Question

A wire of length a is cut into two parts which are bent in the form of a square and a circle. The least value of the sum of the areas then formed is

• A. $\frac{a^2}{\pi +4}$
• B. $\frac{a^2}{2(\pi +4)}$
• C. $\frac{a^2}{3(\pi +4)}$
• D. $\frac{a^2}{4(\pi +4)}$

Answer 1

The correct option is D $\frac{a^2}{4(\pi +4)}$

Let L be the legth of side of square and r be the radius of circle.

Therefore, $4L+2\pi r=a$
Sum of area$=L^2+\pi r^2$
$Area=(\frac{a}{4}-\frac{\pi r}{2}{)}^2+\pi r^2$
$\frac{dArea}{dr}=2(\frac{a}{4}-\frac{\pi }{2}r)\left(\frac{-\pi }{2}\right)+2\pi r$
$0=-\pi (\frac{a}{4}-\frac{\pi }{2}r)+\pi r$
$\frac{a}{4}=\frac{\pi }{2}r=2r$

$\frac{a}{2(\pi +4)}=r$
$Area=\frac{a^2}{4(\pi +4)}$