Question

Show that the altitude of the right circular cone of maximum volume that can be described in a sphere of radius $r$ is $\frac{4r}{3}$. Also show that the maximum volume of the cone is $\frac{8}{27}$ of the volume of the sphere.

Answer 1

Let $R$ and $h$ be the radius and height of the cone.

$r$ be the radius of sphere.

To show $h=\frac{4r}{3}$

and Maximum volume of sphere = $\frac{8}{27}$ Volume of sphere

In $\Delta ABC$, AC = h - r

Therefore, using Pythagoras theorem,

$(h-r{)}^2+R^2=r^2$

$\Rightarrow R^2=r^2-(h-r{)}^2$

Volume of cone: V = $\frac{1}{3}\pi R^{2}h$

or $V=\frac{1}{3}\pi (r^2-(h-r{)}^2)h$

$V=\frac{1}{3}\pi [r^2-h^2-r^2+2hr]h$

$V=\frac{1}{3}\pi [2h^{2}r-h^3]$

For maxima or minima, $\frac{dV}{dh}=0$

Now, $\frac{dV}{dh}=\frac{1}{3}\pi [4hr-3h^2]$

Putting, $\frac{dV}{dh}=0$

we get

$4hr=3h^2$

$h=\frac{4r}{3}$

$\frac{d^{2}V}{d{h}^2}=\frac{1}{3}\pi [4r-6h]$

Putting $h=\frac{4r}{3}$

$\frac{d^{2}V}{d{h}^2}=\frac{1}{3}\pi [4r-\frac{6.4r}{3}]\Rightarrow -\frac{1}{3}\pi \left[4r\right]$

Which is less than zero, therefore $h=\frac{4r}{3}$ is a maxima and the volume of the cone at $h=\frac{4r}{3}$ will be maximum,

$V=\frac{1}{3}\pi R^{2}h$

$=\frac{1}{3}\pi [r^2-(h-r{)}^2]h$

$=\frac{1}{3}\pi [r^2-{(\frac{4r}{3}-r)}^2]\left[\frac{4r}{3}\right]$

$=\frac{1}{3}\pi \left[\frac{8r^2}{9}\right]\left[\frac{4r}{3}\right]$

$=\frac{8}{27}\left(\frac{4\pi r^3}{3}\right)$

$=\frac{8}{27}$xvolume of sphere