A wire of length 36m is cut into two pieces, one of the pieces is...

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

$4x + 2\piy = 36$ (given)

Area, $A = {x}^{2} + \pi{y}^{2}$

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

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

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

$⇒ \dfrac{dA}{dx} = \dfrac{2\pix + ( 36 - 4x ) ( - 2 )}{\pi}$

For critical points

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

$⇒ 2\pix - 72 + 8x = 0$

$⇒ x = \dfrac{36}{\pi + 4}$

Now,

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

$⇒ \dfrac{{d}^{2}A}{d{x}^{2}} > 0$

Hence, area is minimized.

$y = \dfrac{18}{\pi + 4}$
$K$ [circumference of circle)$= 2\piy ⇒ k = \dfrac{36\pi}{( \pi + 4 )}$
$∴ \left(\dfrac{4 + \pi}{\pi}\right) \times k = 36$