Question

A window is in the form of a 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

Let the length and breadth of the rectangular part of the window be $2l$ and $2b$ respectively. Thus the radius of the semicircular arc will be $l$. The total perimeter of the window will be $1$ length of rectangle, $2$ breadths of rectangle and the length of the semicircular arc.

$\Rightarrow l+2b+\pi l=10$

$\Rightarrow (1+\pi )l+2b=10$

$\Rightarrow b=\frac{1}{2}[10-(1+\pi \left)l\right]$

To admit maximum light through the window implies that the area of the window is maximum.

Area of the window will be $A$ $=\left(2l\right)\left(2b\right)+\frac{\pi l^2}{2}=4lb+\frac{\pi l^2}{2}=2l[10-(1+\pi \left)l\right]+\frac{1}{2}\pi l^2=20l-2l^2-2\pi l^2+\frac{1}{2}\pi l^2=20l-2l^2-\frac{3}{2}\pi l^2$

For area to be maximum, $\frac{dA}{dl}=0$

$\Rightarrow 20-4l-3\pi l=0$

$\Rightarrow l=\frac{20}{4+3\pi }$

$b=\frac{1}{2}[10-(1+\pi \left)l\right]=\frac{1}{2}[10-(1+\pi \left)\frac{20}{4+3\pi }\right]=\frac{1}{2}\left[\frac{40+30\pi -20-20\pi }{4+3\pi }\right]=\frac{10+5\pi }{4+3\pi }$

Hence, the breadth of the rectangular window is $2b=\frac{20+10\pi }{4+3\pi }$, the length of the window is $2l=\frac{40}{4+3\pi }$ and the radius of the semicircular arc is $l=\frac{20}{4+3\pi }$