A window is in the shape of a rectangle surmounted by a semicircle. If the perimeter of the window be $20ft$. find the maximum area.

Let ‘$x$’ be the radius of semi circle then $2x$ is length of rectangle. Let ‘$y$’ be the breadth of rectangle

Given perimeter $=20$

$2x+y+\pi x+y=20$

$2x+2y+\pi x=20$

$2y=20-\pi x-2x\to \left(1\right)$

$\mathrm{Area} \mathrm{of} \mathrm{window} = \mathrm{area} \mathrm{of} \mathrm{rectangle} + \mathrm{area} \mathrm{of} \mathrm{semicircle}$

$A=2xy+\frac{1}{2}\pi x^2$

$=x(20-\pi x-2x)+\frac{1}{2}\pi x^2[\because \text{ from }(1\left)\right]$

$=20x-\pi x^2-2x^2+\frac{1}{2}\pi x^2=20x-\frac{\pi }{2}{x}^2-2x^2$

Let $f\left(x\right)=20x-\frac{\pi }{2}{x}^2-2x^2$

$f^{\mathrm{\prime }}\left(x\right)=20-\frac{\pi }{2}\left(2x\right)-4x=20-\pi x-4x$

$f^{\prime \prime }\left(x\right)=-\pi -4<0$

Cleary it has maximum area

Consider $f^{\mathrm{\prime }}\left(x\right)=0$

$20-\pi x-4x=0\Rightarrow 20=(\pi +4)x\Rightarrow x=\frac{20}{\pi +4}$

Maximum area $\left(A\right)=20x-\frac{\pi }{2}{x}^2-2x^2$

$=20\left(\frac{20}{\pi +4}\right)-\frac{\pi }{2}{\left(\frac{20}{\pi +4}\right)}^2-2{\left(\frac{20}{\pi +4}\right)}^2$

$=\frac{400}{\pi +4}-\frac{\pi }{2}\frac{400}{(\pi +4{)}^2}-2\frac{400}{(\pi +4{)}^2}$

$=\frac{400(\pi +4)-200\pi -800}{(\pi +4{)}^2}$

$=\frac{400\pi +1600-200\pi -800}{(\pi +4{)}^2}$

$=\frac{200\pi +800}{(\pi +4{)}^2}=\frac{200(\pi +4)}{(\pi +4{)}^2}=\frac{200}{(\pi +4)}\text{ sq.feet }$