Verified Solution • 2025–26 Session moderate

QJEEM21Ma1a33: JEE Main 2021 Mathematics - Class 12 JEEM21Ma1a33 Solution: A wire of length 36 m is cut into two pieces, one of the pieces …

A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is k (meter), then (4π+1)k\left( {{4 \over \pi } + 1} \right)k is equal to _____________.

Step-by-Step Verified Solution

Let x + y = 36

x is perimeter of square and y is perimeter of circle side of square = x/4

radius of circle = y2π{y \over {2\pi }}

Sum Areas = (x4)2+π(y2π)2{\left( {{x \over 4}} \right)^2} + \pi {\left( {{y \over {2\pi }}} \right)^2}

=x216+(36−x)24π= {{{x^2}} \over {16}} + {{{{(36 - x)}^2}} \over {4\pi }}

For min Area :

x=144π+4x = {{144} \over {\pi + 4}}

⇒\Rightarrow Radius = y = 36 −- 144π+4{{144} \over {\pi + 4}}

⇒\Rightarrow k = 36ππ+4{{36\pi } \over {\pi + 4}}

(4π+1)k\left( {{4 \over \pi } + 1} \right)k = 36

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