# Question #af2a7

Oct 20, 2017

Option (D) is correct

#### Explanation:

Let us consider a sector of $X O Y$ of a circle of radius $r$ which has been formed by an arc of length $S$. This arc subtends angle $\theta$ radian at the center $O$ of the circle.

Now $\frac{S}{r} = \theta$

$\implies S = r \theta \ldots . \left[1\right]$

Let the perimeter of the sector $P = S + 2 r \ldots \ldots \left[2\right]$

Combining [1] and [2] we get

$P = r \theta + 2$

$\implies r = \frac{P}{\theta + 2} \ldots . . \left[3\right]$

Now area of the sector $A = \frac{\pi {r}^{2}}{2 \pi} \times \theta$

$\implies A = {r}^{2} / 2 \times \theta \ldots . . \left[4\right]$

Combining [3] and [4] we get

$A = \frac{1}{2} \times {P}^{2} / {\left(\theta + 2\right)}^{2} \times \theta$

Now

$A = {P}^{2} / 2 \times \frac{\theta}{{\theta}^{2} + 4 \theta + 4}$

$\implies A = {P}^{2} / 2 \times \frac{1}{\theta + 4 + \frac{4}{\theta}}$

$\implies A = {P}^{2} / 2 \times \frac{1}{{\left(\sqrt{\theta} - \frac{2}{\sqrt{\theta}}\right)}^{2} + 8} \ldots . . \left[5\right]$

By the given condition the perimeter $P$ of the sector is fixed or constant . So $A$ will be maximum when denominator of the RHS of [5] is minimum and this is possible only when

$\left(\sqrt{\theta} - \frac{2}{\sqrt{\theta}}\right) = 0$

$\implies \theta = 2$ radian

Hence option (D) is correct