Review question

# When is the volume of this cylinder inside a cone a maximum? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7785

## Solution

A truncated cone of height $h$ has circular ends of radii $2r$ and $r$. In this cone is inserted a circular cylinder having its axis along the axis of the cone. One end of the cylinder lies in that face of the cone which is of radius $2r$ and the circumference of the other end lies in the curved surface of the cone (see diagram). Given that the radius of the base of the cylinder is $s$, show that the volume of the cylinder is

$\frac{\pi h s^2(2r - s)}{r}.$

The volume of a cylinder is given by (area of the base)$\times$(height).

Since the radius of the cylinder is $s$, the area of its base is $\pi s^2$.

We now need to find the height of the cylinder which we shall call $a$. For this, consider the following cross section of the diagram above.

From the two similar triangles at the left hand side we can see that \begin{align*} \frac{h}{r}&=\frac{a}{2r-s} \\ \iff\quad a&=\frac{h}{r}(2r-s) \end{align*}

Thus the volume of the cylinder is $\frac{\pi s^2 \times h(2r - s)}{r}$ as required.

If $s$ is allowed to vary, find, in terms of $h$ and $r$, the maximum volume of the cylinder.

We have \begin{align*} V &= \frac{\pi s^2 h (2r - s)}{r} \\ &= (2 \pi h)s^2 - \frac{\pi h}{r}s^3 \\ \frac{dV}{ds} &= (2 \pi h) 2s - \frac{\pi h}{r}3s^2\\ &= \pi h s \left(4-\frac{3s}{r}\right). \end{align*}

Putting $\dfrac{dV}{ds} = 0$, we have $s = 0 \quad \text{ or } \quad 4-\frac{3s}{r} = 0.$

The second case is clearly the one we require, giving $s = \dfrac{4}{3}r$. This yields $V_{max} = \frac{32\pi r^2 h}{27}.$