JEE Main 2021 — Mathematics Calculus

Let the side of square be $x$x and radius of circle be $y$y.

$4x+2\pi y=36$4x+2πy=36 (given)

Area, $A={x}^{2}+\pi {y}^{2}$A=x2+πy2

$$\Rightarrow A={x}^{2}+\pi {(\frac{36-4x}{2\pi })}^{2}$$⇒A=x2+π(2π36−4x​)2

$$\Rightarrow \Rightarrow A={x}^{2}+\frac{1}{\pi }{(18-2x)}^{2}$$⇒⇒A=x2+π1​(18−2x)2

$$\Rightarrow \frac{dA}{dx}=2x+\frac{2}{\pi }(18-2x)(-2)$$⇒dxdA​=2x+π2​(18−2x)(−2)

$$\Rightarrow \frac{dA}{dx}=\frac{2\pi x+(36-4x)(-2)}{\pi }$$⇒dxdA​=π2πx+(36−4x)(−2)​

For critical points

$$\frac{dA}{dx}=0$$dxdA​=0

$$\Rightarrow 2\pi x-72+8x=0$$⇒2πx−72+8x=0

$$\Rightarrow x=\frac{36}{\pi +4}$$⇒x=π+436​

Now,

$$\frac{{d}^{2}A}{d{x}^{2}}=(2+\frac{8}{\pi })$$dx2d2A​=(2+π8​)

$$\Rightarrow \frac{{d}^{2}A}{d{x}^{2}}>0$$⇒dx2d2A​>0

Hence, area is minimized.

$$y=\frac{18}{\pi +4}$$y=π+418​

$K$K [circumference of circle)$=2\pi y\Rightarrow k=\frac{36\pi }{(\pi +4)}$=2πy⇒k=(π+4)36π​

$$\therefore (\frac{4+\pi }{\pi })\times k=36$$∴(π4+π​)×k=36