Question

A wire of length $2$ units is cut into two parts which are bent respectively to form a square of side $=x$ units and a circle of radius $=r$ units. If the sum of the areas of the square and the circle so formed is minimum, then :

• A. $2x=(\pi +4)r$
• B. $(4-\pi )x=\pi r$
• C. $x=2r$
• D. $2x=r$

Answer 1

Answer: (C)

Let length of two parts be ‘$a$’ and ‘$2-a$’
As per condition given, we write
$a=4x$ and $2-a=2\pi$r
$\therefore x=\frac{a}{4}$ and $r=\frac{2-a}{2\pi }$
$\therefore$ A (square) = ${\left(\frac{a}{4}\right)}^2=\frac{a^2}{16}$ and
A (circle) $=\pi {\left[\frac{\left(2-a\right)}{2\pi }\right]}^2=\frac{\pi \left(4+a^2-4a\right)}{4{\pi }^2}$
$=\left(\frac{a^2-4a+4}{4\pi }\right)$
$f\left(a\right)=\frac{a^2}{16}+\frac{a^2-4a+4}{4\pi }$
$\therefore f\left(a\right)=\frac{a^2\pi +4a^2-16a+16}{16\pi }$
$\therefore f^{\prime }\left(a\right)=\frac{1}{16\pi }\left[2a\pi +8a-16\right]$
$f^{\prime }\left(a\right)=0$
$\Rightarrow 2a\pi +8a-16=0$
$\Rightarrow 2a\pi +8a=16$
$\therefore 2a\left(\pi +4\right)=16$
$\Rightarrow a=\frac{8}{\pi +4}$
$x=\frac{a}{4}=\frac{2}{\pi +4}$ and $r=\frac{2-a}{2\pi }=\frac{2-\frac{8}{\pi +4}}{2\pi }$
$=\frac{2\pi +8-8}{2\pi \left(\pi +4\right)}=\frac{1}{\pi +4}$
$\therefore x=\frac{2}{\pi +4}$ and $r=\frac{1}{\pi +4}$
$\Rightarrow x=2r$