Applications Of Derivatives Question 218
Question: The radius of the cylinder of maximum volume, which can be inscribed in a sphere of radius R is
[AMU 1999]
Options:
A) $ \frac{2}{3}R $
B) $ \sqrt{\frac{2}{3}}R $
C) $ \frac{3}{4}R $
D) $ \sqrt{\frac{3}{4}}R $
Show Answer
Answer:
Correct Answer: B
Solution:
If r be the radius and h the height, then from the figure,
$ r^{2}+{{( \frac{h}{2} )}^{2}}=R^{2} $
therefore $ h^{2}=4(R^{2}-r^{2}) $
Now, $ V=\pi r^{2}h $ = $ 2\pi r^{2}\sqrt{R^{2}-r^{2}} $
\ $ \frac{dV}{dr}=4\pi r\sqrt{R^{2}-r^{2}}+2\pi r^{2}.\frac{1}{2}\frac{(-2r)}{\sqrt{R^{2}-r^{2}}} $
For max. or min., $ \frac{dV}{dr}=0 $
therefore $ 4\pi r\sqrt{R^{2}-r^{2}}=\frac{2\pi r^{3}}{\sqrt{R^{2}-r^{2}}} $
therefore $ 2(R^{2}-r^{2})=r^{2} $
therefore $ 2R^{2}=3r^{2} $
therefore $ r=\sqrt{\frac{2}{3}}R $
therefore $ \frac{d^{2}V}{dr^{2}}=-ve $ .
Hence V is max., when $ r=\sqrt{\frac{2}{3}}R $ .
BETA
