Mathematics - Application of Derivative Question with Solution | TestHub

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

$4x+2\pi y=36$ (given)

Area, $A=x^2+\pi y^2$

$\Rightarrow A=x^2+\pi {\left(\frac{36-4x}{2\pi }\right)}^2$

$\Rightarrow \Rightarrow A=x^2+\frac{1}{\pi }{\left(18-2x\right)}^2$

$\Rightarrow \frac{dA}{dx}=2x+\frac{2}{\pi }\left(18-2x\right)\left(-2\right)$

$\Rightarrow \frac{dA}{dx}=\frac{2\pi x+(36-4x)(-2)}{\pi }$

For critical points

$\frac{dA}{dx}=0$

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

$\Rightarrow x=\frac{36}{\pi +4}$

Now,

$\frac{d^{2}A}{d{x}^2}=\left(2+\frac{8}{\pi }\right)$

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

Hence, area is minimized.

$y=\frac{18}{\pi +4}$
$K$ [circumference of circle)$=2\pi y\Rightarrow k=\frac{36\pi }{(\pi +4)}$
$\therefore \left(\frac{4+\pi }{\pi }\right)\times k=36$