Step 1:

Perimeter of the window when the width of window is $x$ and $2r$ is the length.

$\Rightarrow 2x+2r+\large\frac{1}{2}$$\times 2\pi r$$=10$[Given]

$2x+2r+\pi r=10$

$2x+r(2+\pi)=10$------(1)

For admitting the maximum light through the opening the area of the window must be maximum.

$A=$Sum of areas of rectangle and semi-circle.

Step 2:

Area of circle=$\pi r^2$

Area of rectangle=$ l\times b=2\times r\times x$

$A=2rx+\large\frac{1}{2}\pi r^2$

$\;\;=r[10-(\pi+2)r]+\large\frac{1}{2}\pi r^2$

$\;\;=10r-(\large\frac{1}{2}$$\pi+2)r^2$

For maximum area $\large\frac{dA}{dr}$$=0$ and $\large\frac{d^2A}{dr^2}$ is -ve.

$\Rightarrow 10-(\pi+4)r=0$

$(\pi+4)r=10$

$r=\large\frac{10}{\pi+4}$

Step 3:

$\large\frac{d^2A}{dr^2}$$=-(\pi+4)$[Differentiating with respect to r]

(i.e)$\large\frac{d^2A}{dr^2}$ is -ve for $r=\large\frac{10}{\pi+4}$

$\Rightarrow A$ is maximum.

From (1) we have

$\Rightarrow\:10=(\pi+2)r+2x$

Put the value of $r$ in (1)

$10=(\pi+2)\times \big(\large\frac{10}{\pi+4}\big)$$+2x$

$10=\large\frac{10(\pi+2)}{\pi+4}$$+2x$

$\;\;\;\;=\large\frac{10(\pi+2)+2x(\pi+4)}{\pi+4}$

$10(\pi+4)=10(\pi+2)+2x(\pi+4)$

$10(\pi+4)-10(\pi+2)=2x(\pi+4)$

$10\pi+40-10\pi-20=2x(\pi+4)$

$20=2x(\pi+4)$

$10=x(\pi+4)$

$x=\large\frac{10}{\pi+4}$

Step 4:

Length of rectangle$=2r=2\big(\large\frac{10}{\pi+4}\big)$

$\qquad\qquad\qquad\qquad\quad=\large\frac{20}{\pi+4}$

breadth=$\large\frac{10}{\pi+4}$