# How do you find the dimensions (radius r and height h) of the cone of maximum volume which can be inscribed in a sphere of radius 2?

Feb 26, 2015

The answer is: $r = \frac{4}{3} \sqrt{2}$, $h = \frac{8}{3}$

We can imagine a vertical section of the figure, that would appear:

Let $r$ be the radius of the cone, $R$ be the radius of the sphere and $h$ be the height of the cone.

Let's put $D \hat{O} B = x$ with limitations $0 \le x \le \pi$.

In the right-angled triangle $D O B$:

$r = D B = R \sin x$, $O D = R \cos x$, than

$h = C D = O C + O D = R + R \cos x = R \left(1 + \cos x\right)$.

So the volume of the cone is:

$V = \frac{1}{3} \pi {r}^{2} h \Rightarrow V = \frac{1}{3} \pi {\left(R \sin x\right)}^{2} \cdot R \left(1 + \cos x\right) \Rightarrow$

$V = \frac{1}{3} \pi {R}^{3} {\sin}^{2} x \left(1 + \cos x\right)$

$V ' = \frac{1}{3} \pi {R}^{3} \cdot \left[2 \sin x \cdot \cos x \cdot \left(1 + \cos x\right) + {\sin}^{2} x \left(- \sin x\right)\right] =$

$= \frac{1}{3} \pi {R}^{3} \sin x \left[2 \cos x \left(1 + \cos x\right) - {\sin}^{2} x\right] =$

$= \frac{1}{3} \pi {R}^{3} \sin x \left(2 \cos x + 2 {\cos}^{2} x - {\sin}^{2} x\right) =$

$= \frac{1}{3} \pi {R}^{3} \sin x \left(2 \cos x + 2 {\cos}^{2} x - 1 + {\cos}^{2} x\right) =$

$= \frac{1}{3} \pi {R}^{3} \sin x \left(3 {\cos}^{2} x + 2 \cos x - 1\right)$

Now let's find the signum of the derivative, since $\sin x \ge 0$ for every $x$ in the limitations, than:

$V ' \ge 0 \Rightarrow 3 {\cos}^{2} x + 2 \cos x - 1 \ge 0 \Rightarrow$

$\frac{\Delta}{4} = {\left(\frac{b}{2}\right)}^{2} - a c = 1 + 3 = 4$

$\cos x = \frac{\left(- \frac{b}{2}\right) \pm \sqrt{\frac{\Delta}{4}}}{a} = \frac{- 1 \pm 2}{3}$,

So:

$\cos x \le - 1 \vee \cos x \ge \frac{1}{3}$

The first one has only the solution:

$x = \pi$,

the second:

$- \arccos \left(\frac{1}{3}\right) \le x \le \arccos \left(\frac{1}{3}\right)$, but for the limitations:

$0 \le x \le \arccos \left(\frac{1}{3}\right)$.

The function growths from zero to $\arccos \left(\frac{1}{3}\right)$, than it decreases.

So $x = \arccos \left(\frac{1}{3}\right)$ is the maximum requested.

Let's find now $r$ and $h$:

$r = R \sin x = R \sqrt{1 - {\cos}^{2} x} = 2 \sqrt{1 - {\left(\frac{1}{3}\right)}^{2}} = 2 \sqrt{1 - \frac{1}{9}} =$

$= 2 \sqrt{\frac{9 - 1}{9}} = 2 \sqrt{\frac{8}{9}} = 2 \cdot 2 \frac{\sqrt{2}}{3} = \frac{4}{3} \sqrt{2}$

$h = R \left(1 + \cos x\right) = 2 \left(1 + \frac{1}{3}\right) = 2 \left(\frac{3 + 1}{3}\right) = \frac{8}{3}$.