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Mathematics Question on Derivatives

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

A

πc6\frac {\pi ^c}{6}

B

πc4\frac {\pi ^c}{4}

C

4c{4}^c

D

2c{2}^c

Answer

2c{2}^c

Explanation

Solution

Let rr be the radius and θ\theta be the sectorial angle.
∴\therefore Perimeter of sector,
k=2r+0180∘πrk =2 r+\frac{0}{180^{\circ}} \pi r
⇒r=k2+θπ180∘\Rightarrow r=\frac{k}{2+\frac{\theta \pi}{180^{\circ}}}
∴\therefore Area of sector, A=θ360∘π r2A=\frac{\theta}{360^{\circ}} \pi \,r^{2}
=π360∘[θ×(k2+πθ180∘)2]=\frac{\pi}{360^{\circ}}\left[\theta \times\left(\frac{k}{2+\frac{\pi \theta}{180^{\circ}}}\right)^{2}\right]
⇒A=k2π360∘[θ×(2+πθ180∘)−2]\Rightarrow A =\frac{k^{2} \pi}{360^{\circ}}\left[\theta \times\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-2}\right]
On differentiating w.r.t. ' θ\theta, we get
dAdθ=k2π360∘[1×(2+πθ180∘)−2−2(2+πθ180∘)−3\frac{ dA}{d \theta}= \frac{k^{2} \pi}{360^{\circ}}\left[1 \times\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-2}-2\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-3}\right.
θ(π180∘)]\left.\theta\left(\frac{\pi}{180^{\circ}}\right)\right]
=k2π360∘(2+πθ180∘)−2[1−2πθ/180∘2+πθ/180∘]= \frac{k^{2} \pi}{360^{\circ}}\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-2}\left[1-\frac{2 \pi \theta / 180^{\circ}}{2+\pi \theta / 180^{\circ}}\right]
=k2π360∘(2+πθ180∘)−3[2−πθ180∘]= \frac{k^{2} \pi}{360^{\circ}}\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-3}\left[2-\frac{\pi \theta}{180^{\circ}}\right]
Put dAdθ=0 \frac{d A}{d \theta}=0
⇒2−πθ180∘\Rightarrow 2-\frac{\pi \theta}{180^{\circ}}
⇒θ=2×180∘π=2c\Rightarrow \theta=\frac{2 \times 180^{\circ}}{\pi}=2^{c}
At $\theta=2^{c}, \frac{d^{2} A}{d \theta^{2}}