When does this cone have a maximum volume?

When does this cone have a maximum volume?

Question

A hollow container is to be made out of a fixed total area $\pi a^2$ of sheet metal and its shape is to be that of a right circular cone completed by the circular base on which it stands. Find the radius of the base when the container encloses maximum volume.

Suggestion

Drawing a diagram will probably help.

Can we write down or work out a formula for the surface area of a cone?

Can we derive formulae for the height and volume of the cone from this?

Solution

The surface area $A$ of a cone with radius $r$ and height $h$ is given by $A=\pi r^2 + \pi s r$, where $s=\sqrt{h^2 + r^2}$ is the sloped side of the cone.

The $\pi r^2$ part comes from the base of the cone. The other part when laid flat is a sector with radius $s$ and arclength $2\pi r$.

If the angle in the sector is $\theta$, then $s\theta = 2\pi r$. The area of the sector is $\dfrac{1}{2}s^2 \theta$, which on eliminating $\theta$ is $\pi s r$, as above.

We know that this area is $\pi a^2$, hence

$$\begin{align*} \pi a^2 &= \pi r^2 + \pi s r\\ a^2 -r^2 &= \sqrt{h^2r^2 + r^4}\\ \sqrt{\frac{a^4}{r^2} - 2a^2} &= h. \end{align*}$$

Now the volume of a cone is

$V=\dfrac{\pi}{3} h r^2$

, which gives

$$\begin{align*} V=\frac{\pi}{3} \sqrt{a^4r^2 - 2a^2r^4}. \end{align*}$$

To find when the volume is a maximum, we differentiate and set the first derivative equal to zero. We’ll use the substitution

$u=a^4r^2 - 2a^2r^4$

and the

Chain Rule

.

$$\begin{align*} V &= \frac{\pi}{3}u^{\frac{1}{2}} \\ \frac{dV}{dr} = \frac{dV}{du}\times\frac{du}{dr} &= \frac{\pi}{6}u^{-\frac{1}{2}}\left(2a^4r-8a^2r^3\right) \\ &= \frac{\pi}{3}\frac{a^4r-4a^2r^3}{\sqrt{a^4r^2-2a^2r^4}} \end{align*}$$

This is zero when

$$\begin{align*} a^4r-4a^2r^3 &= 0\\ a^2r\left(a^2-4r^2\right) &= 0\\ r=0 \text{ or } \frac{a}{2}. \end{align*}$$

Note that we can ignore the negative solution, $r=-\frac{a}{2}$, because of the physical context of the question.

Finding the second derivative of the volume (to show we have a maximum) would be difficult.

Instead we could argue that $V=0$ for $r=0$ and $r=\dfrac{a}{\sqrt{2}}$, and is positive for all values of $r$ in between. The turning point at $r=\dfrac{a}{2}$ must therefore be a maximum.