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 \(breadth=\frac{10}{π+4}m.\)