Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is  4r/3. Also find the maximum volume  of cone.

 

 

 

 
 
 
 
 

Answers (1)

   Let R be the radius of cone.
    Let OA= OB= r\left ( radius of sphere \right )                                                              
\frac{AC= r+r}{Let \, v} -\left ( i \right )

                   
height of the cone be the volume of cone
To prove:AC= \frac{4r}{3}
Now, \frac{r= \frac{\pi }{3}\left ( BC \right )^{2}\left ( AC \right )}{r= \frac{\pi }{3}\left ( BC \right )^{2}\left ( r+x \right )}-\left ( ii \right ) from fig
Now in \Delta OBC
\left ( OB \right )^{2}= \left ( OC \right )^{2}+BC^{2}
r^{2}= x^{2}+R^{2}
r^{2}-x^{2}= \left ( BC\right )^{2} -\left ( iii \right )
\therefore v= \frac{\pi }{3}\left ( r^{2}-x^{2} \right )\left ( r+x \right )
= \frac{\pi }{3}\left ( r+x \right )^{2}\left ( r-x \right )

            i        ii       
Diifferantitaing both side w.r.t x we get
\frac{dv}{dx}= \frac{\pi }{3}\left [ \left ( r+x \right )^{2} \times \left ( -1 \right )+\left ( r-x \right ).2\left ( r+x \right )\right ]
= \frac{\pi }{3}\left [ r+x \right ]\left [ -r-x+2r-2x \right ]
= \frac{\pi }{3}\left ( r+x \right )\left ( r-3x \right )
for maximum and minimum volume
\frac{dv}{dx}= 0
\Rightarrow \frac{\pi }{3}\left ( r+x \right )\left ( r-3x \right )= 0
\Rightarrow x=- r \ or \ \ x= \frac{r}{3}
x= -r is not possible
Now again differentiating w.r.t x,we get
\frac{d^{2}v}{dx^{2}}= \frac{\pi }{3}\left [ \left ( r+x \right )\left ( -3 \right ) +\left ( r-3x \right )1\right ]
\Rightarrow \frac{d^{2}v}{dx^{2}}= \frac{\pi }{3}\left [ -3r-3x+r-3x \right ]
\Rightarrow \frac{d^{2}v}{dx^{2}}= \frac{\pi }{3}\left ( -2r-6x \right )
\Rightarrow \frac{d^{2}v}{dx^{2}}_{ at x= \frac{r}{3}}= \frac{\pi }{3}\left ( -2r-\frac{6r}{3} \right )
\Rightarrow \frac{-4\pi r}{3}< 0
so,volume of cone is maximum when x= \frac{r}{3}
Now,put x= \frac{r}{3}  is  x= \frac{r}{3}\left ( i \right )
AC= r+x
= r+\frac{r}{3}
(attitude) AC= \frac{4 r}{3}
\therefore the \: attitude \: of \: cone \: is \: \frac{4 r}{3} \: when volume is maximum
from equation (iii)
r^{2}-x^{2}= \left ( BC \right )^{2}
\left ( BC \right )^{2}= r^{2}\left ( \frac{r}{3} \right )^{2}= r^{2}-\frac{r^{r}}{9}
\left ( BC \right )^{2}= \frac{8r^{2}}{9}
maximum volume of cone = \frac{\pi }{3}\left ( \frac{8r^{2}}{9} \right )\left ( \frac{4 r}{3} \right )  using (ii)
\frac{4\pi r^{3}}{3}\times \frac{8}{27}= \frac{32\pi r^{3}}{81} that can be volume= \frac{8}{27}\times volume of sphere 




 

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