Question

The perimeter of a sector is constant. If its area is to be maximum, the sectorial angle should be

• A. \(\dfrac{\pi}{6}\)
• B. \(\dfrac{\pi}{4}\)
• C. \(4^c\)
• D. \(2^c\)

Hint:

For maximum area of a sector with fixed perimeter, write radius in terms of the angle using the perimeter condition, then maximize the area function.

Answer 1

The Correct Option is D

Step 1: Let the radius and angle of sector be \(r\) and \(\theta\).
The perimeter of a sector is \[ P=2r+r\theta. \] Since the perimeter is constant, let \[ 2r+r\theta=P. \] So, \[ r(2+\theta)=P. \] Hence, \[ r=\frac{P}{2+\theta}. \]

Step 2: Write the area of sector.
The area of a sector is \[ A=\frac{1}{2}r^2\theta. \] Substitute \[ r=\frac{P}{2+\theta}. \] Then, \[ A=\frac{1}{2}\left(\frac{P}{2+\theta}\right)^2\theta. \] So, \[ A=\frac{P^2}{2}\cdot \frac{\theta}{(2+\theta)^2}. \] Since \(\frac{P^2}{2}\) is constant, we maximize \[ \frac{\theta}{(2+\theta)^2}. \]

Step 3: Differentiate with respect to \(\theta\).
Let \[ g(\theta)=\frac{\theta}{(2+\theta)^2}. \] Then, \[ g(\theta)=\theta(2+\theta)^{-2}. \] Differentiating, \[ g'(\theta)=(2+\theta)^{-2}-2\theta(2+\theta)^{-3}. \] Taking common factor, \[ g'(\theta)=\frac{(2+\theta)-2\theta}{(2+\theta)^3}. \] \[ g'(\theta)=\frac{2-\theta}{(2+\theta)^3}. \] For maximum area, \[ g'(\theta)=0. \] Therefore, \[ 2-\theta=0. \] Hence, \[ \theta=2. \]

Step 4: Final conclusion.
Therefore, the sectorial angle should be \[ \boxed{2^c} \]