Question

A window is the form rectangle surmounted by a semicircle, the the perimeter of the window is 30 cm, find the dimension of the window so that greatest possible light may be admitted.

Answer 1

Let $x$, $y$, be the length and breadth of the rectangle $ABCD$

Let $\frac{x}{2}$ be the radius of semi circle with center $o$.

And Let $P$ be the perimeter of figure.

$\therefore x+2v+\pi \frac{x}{2}=30$

$\begin{array}{c}\Rightarrow 2x+4v+\pi x=60\\ \Rightarrow 4v=60-\left(\pi +2\right)x\\ \Rightarrow v=\frac{60-\left(\pi +2\right)x}{4}........\left(1\right)\end{array}$

Now, Let $A$ be area of the figure

$\therefore A=xy+\frac{1}{2}\pi {\left(\frac{x}{2}\right)}^2=x\left[\frac{60-\left(\pi +2\right)x}{4}\right]+\frac{\pi x^2}{8}$ $\left[\because of\left(1\right)\right]$

$\begin{array}{c}\therefore A=\frac{1}{4}\left[60-\left(\pi +2\right)x^2\right]+\frac{\pi }{8}{x}^2\\ \therefore \frac{dA}{dx}=\frac{1}{4}\left[60-2\left(\pi +2\right)x\right]+\frac{\pi x}{4}\end{array}$

Now,

$\begin{array}{c}\frac{dA}{dx}=0\\ \Rightarrow \frac{1}{4}\left[60-2\left(\pi +2\right)x\right]+\frac{\pi x}{4}=0\\ \Rightarrow 60-2\left(\pi +2\right)+\pi x=0\\ \Rightarrow 60-2\pi x-4x+\pi x=0\end{array}$

$\begin{array}{c}\frac{d^{2}A}{d{x}^2}=\frac{1}{4}\left[0-2\left(\pi +2\right)\right]+\frac{\pi }{4}\\ =\frac{-2\left(\pi +2\right)}{4}+\frac{\pi +4}{4}\\ =-\frac{\pi +4}{4}\end{array}$

when, $x=\frac{60}{\pi +4},\frac{d^{2}A}{d{x}^2}=-\frac{\pi +4}{4}<0$

Therefore, $A$ is maximum when $x=\frac{60}{\pi +4}$

And,

$\begin{array}{c}v=\frac{60-\left(\pi +2\right).\frac{60}{\pi +4}}{4}\\ =\frac{60\pi +240-60\pi -120}{4\left(\pi +4\right)}\\ =\frac{30}{\pi +4}\end{array}$

Hence, area of figure is maximum and maximum light is admitted when length of rectangle $x=\frac{60}{\pi +4}$

Breadth of rectangle $v=\frac{30}{\pi +4}$

and radius of semi circle $\frac{x}{2}=\frac{15}{\pi +4}$