$A$ 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.

(N/A) Let $x$ and $y$ be the length and breadth of the rectangular part of the window. The radius of the semicircular opening is $r = \frac{x}{2}$.
The total perimeter of the window is given by $x + 2y + \pi r = 10$.
Substituting $r = \frac{x}{2}$,we get $x + 2y + \frac{\pi x}{2} = 10$.
$2y = 10 - x(1 + \frac{\pi}{2}) \Rightarrow y = 5 - x(\frac{1}{2} + \frac{\pi}{4})$.
The area $A$ of the window is the sum of the area of the rectangle and the semicircle:
$A = xy + \frac{1}{2} \pi r^2 = xy + \frac{1}{2} \pi (\frac{x}{2})^2 = xy + \frac{\pi x^2}{8}$.
Substituting $y$,we get $A = x(5 - x(\frac{1}{2} + \frac{\pi}{4})) + \frac{\pi x^2}{8} = 5x - \frac{x^2}{2} - \frac{\pi x^2}{4} + \frac{\pi x^2}{8} = 5x - x^2(\frac{1}{2} + \frac{\pi}{8})$.
To maximize area,find $\frac{dA}{dx} = 5 - 2x(\frac{1}{2} + \frac{\pi}{8}) = 5 - x(1 + \frac{\pi}{4}) = 0$.
$x(1 + \frac{\pi}{4}) = 5 \Rightarrow x = \frac{5}{\frac{4+\pi}{4}} = \frac{20}{4+\pi} \, m$.
Checking the second derivative: $\frac{d^2A}{dx^2} = -(1 + \frac{\pi}{4}) < 0$,so the area is maximum at $x = \frac{20}{4+\pi} \, m$.
Now,$y = 5 - \frac{20}{4+\pi}(\frac{2+\pi}{4}) = 5 - \frac{5(2+\pi)}{4+\pi} = \frac{20+5\pi - 10 - 5\pi}{4+\pi} = \frac{10}{4+\pi} \, m$.
Thus,the dimensions are length $x = \frac{20}{4+\pi} \, m$ and breadth $y = \frac{10}{4+\pi} \, m$.