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. $x = 2 r$
• 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'\left(a\right) = \frac{1}{16 \pi} \left[2a \pi+ 8a -16\right] $
$f'\left(a\right) = 0 $
$\Rightarrow 2a \pi+8a -16 = 0 $
$\Rightarrow 2 a \pi + 8 a = 16$
$ \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 $