Question

A window is in the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is of constant length k then its maximum area is

• A. $\frac{k^2}{\pi +4}$ sq. unit
• B. $\frac{k}{\pi +4}$ sq. unit
• C. $\frac{k^2}{2(\pi +4)}$ sq. unit
• D. $\frac{k}{2(\pi +4)}$ sq. unit

Answer 1

The correct option is C $\frac{k^2}{2(\pi +4)}$ sq. unit

Perimeter of window when width is $x$ and $2\pi$ is length.

$\to 2x+2r+\frac{1}{2}2\pi r=K\Rightarrow 2x+r(2+\pi )=k$

A=Area of window $w=\frac{1}{2}\pi r^2+2rx$(area of rectangle + area of semicircle).

$A=\frac{1}{2}\pi r^2+r[k-r(2+\pi \left)\right]$

$A=r^2[-\frac{\pi }{2},-2]+kr$

For maximum area $\frac{dA}{dr}=0$ and $\frac{d^{2}A}{d{r}^2}<0$

$\frac{dA}{dr}=k-2r\left[\frac{\pi }{2}\right]=0\to [r=\frac{k}{\pi +4}]$

$A=\frac{\pi }{2}{\left[\frac{k}{\pi +4}\right]}^2+2\left[{\frac{k}{\pi +4}}^2\right]\to \frac{(\pi +4}{2}{\left[\frac{k}{\pi +4}\right]}^2$

$A=\frac{k^2}{2(\pi +4)}$