Question

A cone with height \(h\) is inscribed in a larger cone with height\(H\)so that its vertex is at the center of the base of the larger cone. Show that the inner cone has maximum volume when \(h = \frac{1}{3}H\).

Short answer

It is proved that inner cone has maximum volume when \(h = \frac{1}{3}H\).

Step-by-step solution

Solution

Draw the diagram:

Let\(R\)be the radius of larger cone and\(H\)be the height of the larger cone.

And\(r\)be the radius of the smaller cone and\(h\)be the height of the smaller cone.

By the property of similar triangles

\(V'\left( r \right) < 0\)

\(\begin{aligned}{l}\frac{H}{R} &= \frac{{H - h}}{r}\\ \Rightarrow H - h &= \frac{{rH}}{R}\\ \Rightarrow h &= H - \frac{{rH}}{R} &= \frac{H}{R}\left( {R - r} \right)\end{aligned}\)

The volume of the smaller cone is

\(\begin{aligned}{l}V &= \frac{1}{3}\pi {r^2}\frac{H}{R}\left( {R - r} \right)\\V\left( r \right) &= \frac{{\pi {r^2}H}}{3} - \frac{{\pi {r^3}H}}{{3R}}\\V'\left( r \right) &= \frac{{2\pi rH}}{3} - \frac{{\pi {r^2}H}}{R}\\V'\left( r \right) &= \frac{{\pi rH}}{{3R}}\left( {2R - 3r} \right)\end{aligned}\)`

Find when the volume of smaller cone will be maximum

Now\(V'\left( r \right) = 0\)when\(r = 0\)or\(2R - 3r = 0\)

Or when \(r = 0\)or \(r = \frac{{2R}}{3}\)but we cannot take\(r = 0\),

Now since\(V'\left( r \right) > 0\)for\(r < \frac{{2R}}{3}\)and\(V'\left( r \right) < 0\)for \(r > \frac{{2R}}{3}\)\(r = \frac{{2R}}{3}\)

Therefore, volume of the smaller cone will be maximum when\(r = \frac{{2R}}{3}\).

Then the height of the smaller cone

Or,

It is proved that inner cone has maximum volume when \(h = \frac{1}{3}H\).