Question

A window is in the form of a rectangle surmounted by a semi-circular opening. The total perimeter of the window is $10m$.
Statement I One of the dimension of the window to admit maximum light through the whole opening is $\frac{20}{\pi +4}m$
Statement II One of the dimension of the window to admit maximum light through the whole opening is $\frac{10}{\pi +4}m$.

• A. Only statement I is true
• B. Only statement II is true
• C. Both the statements are true
• D. None of the above

Answer 1

Answer: (C)

Let radius of semi-circle $=r$
$\therefore$ One side of rectangle $=2r$. Let other side $=x$
$\because P=$ Perimeter $=10$ (given)
$\Rightarrow 2x+2r+\frac{1}{2}(2\pi r)=10$
$\Rightarrow 2x=10-r(\pi +2)......$(i)
Let $A$ be area of the window, then
$A=$ Area of semi-circle $+$ Area of rectangle $=\frac{1}{2}\pi r^2+2rx$
$\Rightarrow A=\frac{1}{2}\left(\pi r^2\right)+r[10-r(\pi +2)]$ [using Eq. (i)]
$=\frac{1}{2}\left(\pi r^2\right)+10r-r^2\pi -2r^2$
$=10r-\frac{\pi r^2}{2}-2r^2$

On differentiating twice w.r.t. $r$, we get
$\frac{dA}{dr}=10-\pi r-4r.....$(ii)
and $\frac{d^{2}A}{d{r}^2}=-\pi -\mathrm{4.....}$(iii)
For maxima or minima, put $\frac{dA}{dr}=0$
$\Rightarrow 10-\pi r-4r=0$
$\Rightarrow 10=(4+\pi )r$
$\Rightarrow r=\frac{10}{4+\pi }$
On putting $r=\frac{10}{4+\pi }$ in Eq. (iii), we get $\frac{d^{2}A}{d{r}^2}<0$
Thus, $A$ has local maximum when
$r=\frac{10}{4+\pi }......$(iv)
$\therefore$ Radius of semi-circle $=\frac{10}{4+\pi }$
and one side of rectangle $=2r=\frac{2\times 10}{4+\pi }=\frac{20}{4+\pi }$
and other side of rectangle i.e., $x$ from Eq. (i) is given by
$x=\frac{1}{2}[10-r(\pi +2)]$
$=\frac{1}{2}\left[10-\left(\frac{10}{\pi +4}\right)(\pi +2)\right]$[from Eq. (iv)]
$=\frac{10\pi +40-10\pi -20}{2(\pi +4)}$
$=\frac{20}{2(\pi +4)}$
$=\frac{10}{\pi +4}$
Light is maximum when area is maximum.
So, dimensions of the window are length $=2r=\frac{20}{\pi +4}$, breadth $=x=\frac{10}{\pi +4}$.
So, both the statements are true.