JEE Main · 2016 · 03 Apr 2016 · Offline · Application of Derivatives

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:

Answer: (B)

Solution

The square's perimeter

4 x

and the circle's circumference

2 π r

use up the 2 units of wire:

4 x + 2 π r = 2

, so

π r = 1 - 2 x

. The total area is

A = x^{2} + π r^{2} = x^{2} + \frac{( 1 - 2 x ) ^{2}}{π}

. Differentiating,

A^{'} (x) = 2 x - \frac{4 ( 1 - 2 x )}{π}

; setting it to zero gives

π x = 2 (1 - 2 x)

, so

x (π + 4) = 2

and

x = \frac{2}{π + 4}

. Then

r = \frac{1 - 2 x}{π} = \frac{1}{π + 4}

, and comparing the two,

x = 2 r

at the minimum.

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option