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

Answers (1)

Let ABCD be a window of rectangular form surmounted by a semicircular with diameter AB
given, perimeter of the window (i) = 10m.
Let the length and breadth of the rectangle be 2x and 2y respectively.

Since p = 10m
ie $2x+4y+\pi x= 10$
$\Rightarrow 4y= 10-2x-\pi x$
$\Rightarrow 4xy= 10x-2x^{2}-\pi x^{2}---(i)$
Now, area of the window
$A= \left ( 2x \right )\left ( 2y \right )+\frac{1}{2} \pi x^{2}$
$\Rightarrow A=4xy+\frac{1}{2}\pi x^{2}$
$\Rightarrow A=10x-2x^{2}-\pi x^{2}+\frac{1}{2}\pi x^{2} \left [ using(i) \right ]$
$\Rightarrow A=10x-2x^{2}-\frac{1}{2}\pi x^{2}$
On differentiating A w.r.t x, we get
$\frac{dA}{dx}= 10-4x-\pi x$
The critical numbers of x are given by $\frac{dA}{dx}= 0$
$\Rightarrow 10-4x-\pi x= 0$
$\Rightarrow -x\left ( 4+\pi \right )= -10$
$\Rightarrow x= \frac{10}{4+\pi }$
Now, $\frac{dA}{dx}= 10-4x-\pi x$
Differentiating w.r.t x we get
$\frac{d^{2}A}{dx^{2}}= -4-\pi = -\left ( 4+\pi \right )< 0$
Thus A is maximum when $x= \frac{10}{4+\pi }m$
Now, length of the window, $2x= \frac{20}{4+\pi }m$ and width of the window
$2y= \frac{10-2x-\pi x}{2} \left [ using (i) \right ]$
$= 5-\frac{10}{4+\pi }-\frac{\pi }{2}\; \frac{10}{4+\pi }$
$= \frac{10\left ( 4+\pi \right )-20-10\pi }{2\left ( 4+\pi \right )}= \frac{40+10\pi -20-10\pi}{2\left ( 4+\pi \right )}$
$= \frac{10}{4+\pi }m$
Also radius of the semicircle is $= \left ( \frac{10}{4+\pi } \right )m$
Hence, the dimensions of the rectangular point of the window are $= \frac{20}{4+\pi }m\; and\; \frac{10}{4+\pi }m$ .