Question

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

Answer 1

The correct answer is $length=\frac{20}{π+4}m$ and $breadth=\frac{10}{π+4}m.$
Let $x$ and $y$ be the length and breadth of the rectangular window.
Radius of the semicircular opening$=\frac{x}{2}$

It is given that the perimeter of the window is $10m$
$∴x+2y+\frac{πx}{2}=10$
$⇒x(1+\frac{π}{2})+2y=10$
$⇒2y=10-x(1+\frac{π}{2})$
$⇒y=5-x(\frac{1}{2}+\frac{π}{4})$
∴Area of the window $(A)$ is given by,
$A=xy+\frac{π}{2}(\frac{x}{2})^2$
$=x[5-x(\frac{1}{2}+\frac{π}{4})+\frac{x^2}{8}]$
$=5x-x^2(\frac{1}{2}+\frac{π}{4})+\frac{x^2}{8}$
$∴\frac{dA}{dx}=5-2x(\frac{1}{2}+\frac{π}{4})+\frac{x}{4}$
$=5-x(1+\frac{π}{2})+\frac{π}{4x}$
$∴\frac{d^2A}{dx^2}=-(1+\frac{π}{2})+\frac{π}{4}=-1-\frac{π}{4}$
Now,$\frac{dA}{dx}=0$
$⇒5-x(1+\frac{π}{2})+\frac{π}{4}x=0$
$⇒5-x-\frac{π}{4}x=0$
$⇒x(1+\frac{π}{4})=5$
$⇒x=\frac{5}{(1+\frac{π}{4})}=\frac{20}{π+4}$
Thus,when $x=\frac{20}{π+4}\, then\, \frac{d^2A}{dx^2}<0$
Therefore, by second derivative test, the area is the maximum when length $x=\frac{20}{π+4}.$
Now,
$y=5-\frac{20}{π+4}(\frac{2+π}{4})$
$=5-\frac{5(2+π)}{π+4}=\frac{10}{π+4}m$
Hence, the required dimensions of the window to admit maximum light is given by $length=\frac{20}{π+4}m$ and $breadth=\frac{10}{π+4}m.$