Question

A window is in the form of a rectangle surmounted by a semi-circle opening. The perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

Answer 1

Let radius of semi-circle = r
$\therefore$ One side of rectangle = 2r. Let other side = x
$\therefore$ P = Perimeter = 10 (given)
$\Rightarrow 2x+2r+\frac{1}{2}\left(2\pi r\right)=10\Rightarrow 2x=10-r(\pi +2)$ ...(i)
Let A be area of the figure, then
A = Area of semi-circle + Area of rectangle $=\frac{1}{2}\pi r^2+2rx$
$\Rightarrow A=\frac{1}{2}\left(\pi r^2\right)+r[10-r(\pi +2\left)\right]$ [Using Eq. (i)]
$=\frac{1}{2}\left(\pi r^2\right)+10r-r^2\pi -2r^2=10r-\frac{\pi r^2}{2}-2r^1$
On differentiating twice w.r.t. r, we get
$\frac{dA}{dr}=10-\pi r-4r$ ...(ii)
and $\frac{d^{2}A}{d{r}^2}=-\pi -4$ ...(iii)
For maxima or minima, put $\frac{dA}{dr}=0\Rightarrow 10-\pi r-4r=0$
$\Rightarrow 10=(4+\pi )r\Rightarrow r=\frac{10}{4+\pi }$
On putting $r=\frac{10}{4+\pi }$ in Eq. (iii), we get $\frac{d^{2}A}{d{r}^2}=$ negative
Thus, A has local maximaum when $r=\frac{10}{4+\pi }$ ...(iv)
$\therefore$ Radius of semi-circle $=\frac{10}{4+\pi }$
and one side of rectangle $=2r=\frac{2\times 10}{4+\pi }=\frac{20}{4+\pi }$
and one side of rectangle i.e., x from Eq. (i) is given by
$x=\frac{1}{2}[10-r(\pi +2\left)\right]=\frac{1}{2}[10-(\frac{10}{\pi +4}\left)\right(\pi +2\left)\right]$ (from Eq.(iv))
$\therefore$ Radius of semi-circle $=\frac{10}{4+\pi }$
and one side of rectangle $=2r=\frac{2\times 10}{4+\pi }=\frac{20}{4+\pi }$
and other side of rectangle i.e., x from Eq. (i) is given by
$x=\frac{1}{2}[10-r(\pi +2\left)\right]=\frac{1}{2}[10-(\frac{10}{\pi +4}\left)\right(\pi +2\left)\right][fromEq.(iv\left)\right]=\frac{10\pi +40-10\pi -20}{2(\pi +4)}=\frac{20}{2(\pi +4)}=\frac{10}{\pi +4}$
Light is maximum when area is maximum
So, dimensions of the window are
length $=2r=\frac{20}{\pi +4}$,
breadth = $x=\frac{10}{\pi +4}$