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.

Let $x$ and $y$ be respectively the length and breadth of the rectangle.

Radius of the semi-circle is given by $= \cfrac{x}{2}$

Circumference of the circular part $= \cfrac{{\pi x}}{2}$

Perimeter of the window $= AB + BC + AD + DC$
$\Rightarrow x + 2y + \cfrac{{\pi x}}{2} = 10$ (Given)

$\Rightarrow 2x + 4y + \pi x = 20 \Rightarrow y = \cfrac{{20 - (2 + \pi )x}}{4}$ …(i)

Area of the window

$=$area of rectangle$+$area of semi-circle

$A = xy + \cfrac{1}{2}\pi {\left( {\cfrac{x}{2}} \right)^2} = x\left( {\cfrac{{20 - (2 + \pi )x}}{4}} \right) + \cfrac{{\pi {x^2}}}{8}$ (from

(i)) $\Rightarrow A = {\rm{ }}\cfrac{{20x - (2 + \pi ){x^2}}}{4} + \cfrac{{\pi {x^2}}}{8}$
$\Rightarrow \cfrac{{dA}}{{dx}} = \cfrac{{20 - (2 + \pi )2x}}{4} + \cfrac{{2\pi x}}{8}$

To find the maximum or minimum

$A,\cfrac{{dA}}{{dx}} = 0$

$\Rightarrow \cfrac{{20 - (2 + \pi )2x}}{4} + \cfrac{{2\pi x}}{8} = 0$

$\Rightarrow 20 - (2 + \pi )2x + \pi x = 0 \Rightarrow 20 + x(\pi - 4 - 2\pi ) = 0$

$\Rightarrow 20 - x(4 + \pi ) = 0 \Rightarrow x = \cfrac{{20}}{{4 + \pi }}$

$\cfrac{{{d^2}A}}{{d{x^2}}} = \cfrac{{ - (2 + \pi )2}}{4} + \cfrac{{2\pi }}{8} = \cfrac{{ - 4 - 2\pi + \pi }}{4} = \cfrac{{ - 4 - \pi }}{4} \Rightarrow \cfrac{{{d^2}A}}{{d{x^2}}} < 0\,\forall \,x$

Hence, maximum light enters the window when
$x =$length $= \cfrac{{20}}{{4 + \pi }}m$

Therefore, breadth $y = \cfrac{{20 - (2 + \pi )\cfrac{{20}}{{4 + \pi }}}}{4} = \cfrac{{80 + 20\pi - 40 - 20\pi }}{{4(4 + \pi )}}$

$= \cfrac{{40}}{{4(4 + \pi )}} = \cfrac{{10}}{{4 + \pi }}m$