# 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.

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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