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 areas 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: $4x+2\pi r=a$
where $x$ is side length of the square and $r$ is radius of the circle
$A=x^2+\pi r^2=\frac{1}{16}(a-2\pi r{)}^2+\pi r^2$
$\frac{dA}{dr}=-\frac{1}{8}(a-2\pi r)\left(2\pi \right)+2\pi r$
$\frac{dA}{dr}=0,\text{give}r=\frac{a}{2(\pi +4)}$
for which $\frac{d^{2}A}{d{r}^2}=\frac{{\pi }^2}{2}+2\pi$ is positive and hence minimum.
$\Rightarrow 4x=a-2\pi r=a-\frac{a\pi }{\pi +4}=\frac{4a}{\pi +4}$
$\therefore x=\frac{a}{\pi +4}$
$\therefore A=x^2+r^2\pi =\frac{a^2}{4(\pi +4)}.$