Applications Of Derivatives Question 329
Question: The ratio of height of cone of maximum volume inscribed in a sphere to its radius is
[Orissa JEE 2004]
Options:
A) $ \frac{3}{4} $
B) $ \frac{4}{3} $
C) $ \frac{1}{2} $
D) $ \frac{2}{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
Let diameter of sphere $ AE=2r $
Let radius of cone is x and height is y $ AD=y $ since $ BD^{2}=AD.DE $ or $ x^{2}=y(2r-y) $ …..(i) Volume of cone $ V=\frac{1}{3}\pi x^{2}y=\frac{1}{3}\pi y(2r-y)y=\frac{1}{3}\pi (2ry^{2}-y^{3}) $
therefore $ \frac{dV}{dy}=\frac{1}{3}\pi (4ry-3y^{2}) $
therefore $ \frac{dV}{dy}=0 $
therefore $ \frac{1}{3}\pi (4ry-3y^{2})=0 $
therefore $ y(4r-3y)=0 $
therefore $ y=\frac{4}{3}r,,0 $
Now $ \frac{d^{2}V}{dy^{2}}=\frac{1}{3}\pi (4r-6y) $ , put $ y=\frac{4}{3}r $
therefore $ \frac{d^{2}V}{dy^{2}}=\frac{1}{3}\pi ,( 4r-6\times \frac{4}{3}r ) $ = negative value
So, volume of cone is maximum at $ y=\frac{4}{3}r $
therefore $ \frac{Height}{Radius} $ = $ \frac{y}{r}=\frac{4}{3} $ .
BETA
