Question: A sphere of radius $r$ is inscribed in a circular, right cone.

What is the minimum radius and height of the circular cone? (Thus, volume)

very crude diagram of the problem

Because the answer would specifically be proportional to the radius given $r$,

I have set the base radius of the cone $ar$, and would like to solve for $a$.

Because triangle AB'O, and AC'O are congruent, I have labelled the angles $\theta$

Then, $\tan(\theta)= \frac{1}{a}$

The double angle formula is:

$\tan(2\theta)= \frac{2\tan(\theta)}{1-\tan^{2}(\theta)}$

Plugging in the values, it results in:

$\tan(2\theta)= \frac{2a}{a^2-1}$

For simplicity, let's set the radius 1, such that the base has a radius length $a$, then the height becomes

$a\times\frac{2a}{a^2-1} = \frac{2a^2}{a^2-1}$

This makes perfect sense, because if we extend the base radius to infinity, then the resulting cone's top vertex would end up on the top of the sphere.

$\lim\limits_{a \to \infty}\frac{2a^2}{a^2-1} = 2$

(We have set the sphere's radius to 1)

Thus, we have the variables to actually solve for the minimum value.

The volume of the right circular cone is given by:

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

We have set $r = a$ and $h = \frac{2a^2}{a^2-1}$


$V = \frac{1}{3}\pi a^2 \frac{2a^2}{a^2-1} = \frac{2}{3}\pi \frac{a^4}{a^2-1} $

Neglecting the coefficients, deriving V with respect to a and solving for 0,

We get

$a = \sqrt{2}$ and $h = 4$


$a = \sqrt{2}r$ and $h = 4r$

Apparently though, I'm wrong. The answer is:

$a = 2r$ and $h = 4r$

But I don't understand what's wrong with my argument. Any ideas?

  • $\begingroup$ I don't think you have made any mistakes, the book must have a typo, if a=2r and h=4r the sides of the cone would not be tangent to the sphere. $\endgroup$
    – WW1
    Apr 21, 2015 at 0:34
  • $\begingroup$ It is weird how the question was asked the cone's radius to height ratio would have a minimum value of zero ( when a=r and $h\to \infty$ ) . I don't see how minimizing that ratio is the same as minimizing the volume of the cone. $\endgroup$
    – WW1
    Apr 21, 2015 at 0:39
  • $\begingroup$ @WW1 oh, that actually does not refer to the ratio, I just meant radius AND height. Sorry, that is definitely my mistake. $\endgroup$
    – VladeKR
    Apr 21, 2015 at 0:43
  • $\begingroup$ "radius and height" is a very peculiar thing to be asked to minimize - does the original question include " thus volume" ? $\endgroup$
    – WW1
    Apr 21, 2015 at 0:56
  • $\begingroup$ @WW1, the original question asks for the radius, and height required for the minimum volume; for a minimum volume there has to be a base radius, "a", and because of geometry, the height is also dependent on the chosen "a." $\endgroup$
    – VladeKR
    Apr 21, 2015 at 1:54

1 Answer 1


As user WW1 says, the book must have a typo - your work is correct. Here's another way we could have calculated it giving the same answer:

The radius of the incircle of a triangle is $\frac{\triangle}{s}$ where $\triangle$ is the area of the triangle and $s$ is the semi-perimeter. Let's call $R$ the radius of the cone and $h$ the height of the cone. If $r$ is the radius of the sphere this gives us

$$\begin{align}&r=\frac{Rh}{\sqrt{h^2+R^2}+R} \\\implies &r\sqrt{h^2+R^2}+rR=Rh \\\implies &r\sqrt{h^2+R^2}=R(h-r) \\\implies &r^2\left(h^2+R^2\right)=R^2(h^2-2hr+r^2) \\\implies &r^2h^2=R^2\left(h^2-2hr\right) \\\implies &\frac{r^2h^2}{h^2-2hr}=R^2 \\\implies &\frac{r^2h}{h-2r}=R^2 \end{align}$$

Now the volume of the cone is $$V=\dfrac{\pi R^2h}{3}=\frac{\pi r^2h^2}{3(h-2r)}$$

Hence $$\frac{dV}{dh}=\frac{\pi\left(r^2h^2-2r^2h(h-2r)\right)}{3(h-2r)^2}=\frac{\pi\left(4r^3h-r^2h^2\right)}{3(h-2r)^2}$$

Setting the derivative to $0$, we get $$4r^3h-r^2h^2=0\hspace{5mm}\implies\hspace{5mm}4r-h=0\hspace{5mm}\implies\hspace{5mm}h=4r$$

Plugging this into our equation for $R^2$ we get $$R^2=\frac{r^2(4r)}{4r-2r}=\frac{4r^3}{2r}=2r^2$$ Hence $R=\sqrt{2}r$ and $h=4r$ minimizes the volume as you found.


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