Question

A wire of length $a$ is cut into two parts which are bent, respectively, in the form of a square and a circle. The least value of the sum of the area so formed is

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

Answer 1

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

Given $4x+2\pi r=a$

$A=x^2+\pi r^2=\frac{1}{16}{(a-2\pi r)}^2+\pi r^2$

$\frac{dA}{dr}=0\Rightarrow r=\frac{a}{2(\pi +4)}$

$\frac{d^{2}A}{d{r}^2}$ is +ve and hence minimum.

Now, $4x=a-2\pi r=a-\frac{a\pi }{\pi +4}=\frac{4a}{\pi +4}\Rightarrow x=\frac{a}{\pi +4}$

$\Rightarrow A=x^2+r^2\pi =\frac{a^2}{4(\pi +4)}$