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Question

Mathematics Question on Application of derivatives

A circular sector of perimeter 6060 metre with maximum area is to be constructed. The radius of the circular arc in metre must be

A

10

B

15

C

5

D

20

Answer

15

Explanation

Solution

Perimeter of sector = 2r+rθ2r + r \theta
60=2r+rθ\Rightarrow \, 60 = 2r + r\theta (given)

  θ=602rr\Rightarrow \; \theta = \frac{60 - 2r}{r}
Area of sector, A=πr2θ360A = \frac{\pi r^2 \theta}{360^{\circ}}
=πr2(602r)r360= \frac{\pi r^{2} \left(60 -2r\right)}{r 360}
=πr180(30r)= \frac{\pi r}{180} \left(30 -r\right)
dAdr=π180(302r)\Rightarrow \frac{dA}{dr} = \frac{\pi}{180} \left(30 -2r\right)
For maximum area, dAdr=0\frac{dA}{dr} = 0
302r=0\Rightarrow 30-2r=0
r=15\Rightarrow r = 15
$\therefore \frac{d^{2}A}{dr^{2}} = \frac{\pi}{180} \left(0-2\right) = \frac{-\pi}{90}