A window in the form of a rectangle is surmounted by a semi-circular opening.

Solution:

Let the dimensions of the rectangular part be $x$ and $y$.

Radius of semi-circle $=\frac{x}{2}$

Total perimeter $=10$

$\Rightarrow(x+2 y)+\pi\left(\frac{x}{2}\right)=10$

$\Rightarrow 2 y=\left[10-x-\pi\left(\frac{x}{2}\right)\right]$

$\Rightarrow y=\frac{1}{2}\left[10-x\left(1+\frac{\pi}{2}\right)\right]$             ......(1)

Now,

Area, $A=\frac{\pi}{2}\left(\frac{x}{2}\right)^{2}+x y$

$\Rightarrow A=\frac{\pi x^{2}}{8}+\frac{x}{2}\left[10-x\left(1+\frac{\pi}{2}\right)\right]$        $[$ From eq. $(1)]$

$\Rightarrow A=\frac{\pi x^{2}}{8}+\frac{10 x}{2}-\frac{x^{2}}{2}\left(1+\frac{\pi}{2}\right)$

$\Rightarrow \frac{d A}{d x}=\frac{\pi x}{4}+\frac{10}{2}-\frac{2 x}{2}\left(1+\frac{\pi}{2}\right)$

For maximum or minimum values of $A$, we must have

$\frac{d A}{d x}=0$

$\Rightarrow \frac{\pi x}{4}+\frac{10}{2}-\frac{2 x}{2}\left(1+\frac{\pi}{2}\right)=0$

$\Rightarrow x\left[\frac{\pi}{4}-1-\frac{\pi}{2}\right]=-5$

$\Rightarrow x=\frac{-5}{\left(\frac{4-x}{4}\right)}$

$\Rightarrow x=\frac{20}{(\pi+4)}$

Substituting the value of $x$ in eq. $(1)$, we get

$y=\frac{1}{2}\left[10-\left(\frac{20}{\pi+4}\right)\left(1+\frac{\pi}{2}\right)\right]$

$\Rightarrow y=5-\frac{10(\pi+2)}{2(\pi+4)}$

$\Rightarrow y=\frac{5 \pi+20-5 \pi-10}{(\pi+4)}$

$\Rightarrow y=\frac{10}{(\pi+4)}$

$\frac{d^{2} A}{d x^{2}}=\frac{\pi}{4}-\frac{\pi}{2}-1$

$\Rightarrow \frac{d^{2} A}{d x^{2}}=\frac{\pi-2 \pi-4}{4}$

$\Rightarrow \frac{d^{2} A}{d x^{2}}=\frac{-\pi-4}{4}<0$

Thus, the area is maximum when $x=\frac{20}{\pi+4}$ and $y=\frac{10}{\pi+4}$.

So, the required dimensions are given below :

Length $=\frac{20}{\pi+4} \mathrm{~m}$

Breadth $=\frac{10}{\pi+4} \mathrm{~m}$