Question

Find the volume of the largest circular cone that can be inscribed in a sphere of radius \(r\).

Short answer

Thus, the volume of the largest right circular cone that can be inscribed in a sphere of radius \(r\) is \(\frac{{32}}{{81}}\pi {r^3}\).

Step-by-step solution

Draw the figure of right circular cone inscribed in a sphere

Let radius of the right circular cone \(R\) and O is the center of the sphere. \(x\) is the distance from the base of the cone to the center O. Let \(V\) be the volume of the sphere. Also, the height of a cone is given in the figure is \(AC = AO + OC = r + x\)

Then solve as follows:

By Pythagoras theorem,

\({R^2} = {r^2} - {x^2}\)

Volume of cone is obtained

Substitute \(h = r + x\) and \({R^2} = {r^2} - {x^2}\) in the volume of cone gives,

\(\begin{aligned}{c}V = \frac{1}{3}\pi {r^2}h\\ = \frac{1}{3}\pi \left( {{r^2} - {x^2}} \right)\left( {r + x} \right)\\ = \frac{1}{3}\pi \left( {{r^3} + {r^2}x - r{x^2} - {x^3}} \right)\end{aligned}\)

Now, differentiate the above volume as follows,

\(\begin{aligned}{c}\frac{{dV}}{{dx}} = \frac{d}{{dx}}\left( {\frac{1}{3}\pi \left( {{r^3} + {r^2}x - r{x^2} - {x^3}} \right)} \right)\\ = \frac{1}{3}\pi \left( {0 + {r^2} - 2rx - 3{x^2}} \right)\\ = \frac{1}{3}\pi \left( {{r^2} - 2rx - 3{x^2}} \right)\end{aligned}\)

Again, differentiate the volume gives,

\(\frac{{{d^2}V}}{{d{x^2}}} = \frac{1}{3}\pi \left( { - 2r - 6x} \right)\)

Obtain maximum and minimum value of the volume of cone

For a maximum and minimum value of volume,

\(\begin{aligned}{c}\frac{{dV}}{{dx}} = 0\\{r^2} - 2rx - 3{x^2} = 0\\\left( {r + x} \right)\left( {r - 3x} \right) = 0\\x = - r,\frac{r}{3}\end{aligned}\)

Take \(x = \frac{r}{3}\) then it gives,

\(\begin{aligned}{c}\frac{{{d^2}V}}{{d{x^2}}} = \frac{1}{3}\pi \left( { - 2r - 6\frac{r}{3}} \right)\\ = \frac{1}{3}\pi \left( { - 2r - 2r} \right)\\ = - \frac{{4\pi r}}{3} < 0\end{aligned}\)

Thus, the volume is maximum only when \(x = \frac{r}{3}\).

Maximum volume of cone in a sphere

The maximum volume of cone in a sphere is obtained as follows;

\(\begin{aligned}{c}V = \frac{\pi }{3}\left( {{r^2} - \frac{{{r^2}}}{9}} \right)\left( {r + \frac{r}{3}} \right)\\ = \frac{\pi }{3}\left( {\frac{{8{r^2}}}{9}} \right)\left( {\frac{{4r}}{3}} \right)\\ = \frac{{32}}{{81}}\pi {r^3}\\ = \left( {\frac{8}{{27}}} \right)\left( {\frac{4}{3}\pi {r^3}} \right)\\ = \frac{{32}}{{81}}\pi {r^3}\end{aligned}\)

Hence, the volume of the largest right circular cone that can be inscribed in a sphere of radius \(r\) is \(\frac{{32}}{{81}}\pi {r^3}\).