A window is in the form of a rectangle surmounted by a semicircle. If…

The correct answer is Option (2) → Width = $\frac{20}{4+π}m$, Height of rectangle = $\frac{10}{4+π}m$

Let r metres be the radius of the semicircle and x metres be the side BC of the rectangle ABCD. Then perimeter of the window

$= 2r+ 2x + πr = 10$  (given)

$⇒x=\frac{1}{2}(10-2г-πг)$   ...(i)

Maximum light will be admitted through the window if the area of the window is maximum. Let A be the area of the window,

$A=2r.x+\frac{1}{2}πг^2$

$=2r.\frac{1}{2}(10-2г-πг)+\frac{1}{2}πг^2$  (using (i))

$=10r-2r^2-\frac{1}{2}πг^2$   ...(ii)

Differentiating (ii), w.r.t. r, we get

$\frac{dA}{dr}=10-4r-πг$ and $\frac{d^2A}{dr^2}=-4-π$

Now $\frac{dA}{dr}=0⇒10-4r-πг=0⇒r=\frac{10}{4+π}$.

Also when $r=\frac{10}{4+π},\frac{d^2A}{dr^2}=-(4+π)<0$

⇒ A is maximum when $r=\frac{10}{4+π}$,

then $x=\frac{1}{2}\left(10-(2+\pi).\frac{10}{4+π}\right)=\frac{10}{4+π}$

Therefore, the maximum light is admitted when the radius of semicircle = $\frac{10}{4+π}$ metres and the side BC of the rectangle = $\frac{10}{4+π}$ metres and side AB = $\frac{20}{4+π}$ metres.