Applications Of Derivatives Question 347
Question: The perimeter of a sector is p. The area of the sector is maximum when its radius is
[Karnataka CET 2002]
Options:
A) $ \sqrt{p} $
B) $ \frac{1}{\sqrt{p}} $
C) $ \frac{p}{2} $
D) $ \frac{p}{4} $
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Answer:
Correct Answer: D
Solution:
Perimeter of a sector = p. Let AOB be the sector with radius r. If angle of the sector be q radians, then area of sector $ (A)=\frac{1}{2}r^{2}\theta $ …………..(i)
Length of arc(s) = rq or $ \theta =\frac{s}{r} $ . Therefore perimeter of the sector $ =(p)=r+s+r=2r+s $ …………..(ii)
Substituting $ \theta =\frac{s}{r} $ in (i), A = $ ( \frac{1}{2}r^{2} ),( \frac{s}{r} )=\frac{1}{2}rs $
therefore $ s=\frac{2A}{r} $ .
Now substituting the value of s in (ii), we get $ p=2r+( \frac{2A}{r} ) $ or $ 2A=pr-2r^{2}. $
Differentiating with respect to $ r,, $ we get $ 2\frac{dA}{dr}=p-4r $ . We know that for the maximum value of area $ \frac{dA}{dr}=0 $ or $ p-4r=0 $ or $ r=\frac{p}{4} $ .
BETA
