Hello, galactus!

I *think* I've got a handle on this one . . .

We've all seen the classic 'sphere inside a cone', 'cylinder in a cone',

'cone in a sphere' problems, etc.

Here's one that is a little different if anyone would enjoy tackling it.

$\displaystyle \text{Find the ellipsoid of max volume which can be inscribed in a cone of radius }R$ $\displaystyle \text{ and height }H$

Let's orient the ellipsoid 'horizontally' . . . then we have: Code:

- *
: /|\
: / | \
: / | \
: / | \
H / ..*.. \
: /*::::|::::*\
: *-:-:-:+ - a -*
: / *::::b::::* \
- *- - - - * - - - -*
: R :

From similar right triangles, we have: .$\displaystyle \frac{a}{H-b} \:=\:\frac{R}{H}\quad\Rightarrow\quad b \:=\:\frac{H}{R}(R-a)\;\;{\color{blue}[1]}$

The volume of this ellipsoid of revolution is: .$\displaystyle V \;=\;\frac{4}{3}\pi a^2b\;\;{\color{blue}[2]}$

Substitute [1] into [2]: .$\displaystyle V \;=\;\frac{4\pi}{3} a^2\cdot\frac{H}{R}(R-a) \;=\;\frac{4\pi H}{3R}(Ra^2-a^3) $

Maximize: .$\displaystyle V' \;=\;\frac{4\pi H}{3R}(2Ra - 3a^2) \:=\:0\quad\Rightarrow\quad a(2R -3a) \:=\:0\quad\Rightarrow\quad\boxed{ a \:=\:\frac{2}{3}R}$

Substitute into [2]: .$\displaystyle b \:=\:\frac{H}{R}\left(R - \frac{2}{3}R\right)\quad\Rightarrow\quad\boxed{ b \:=\:\frac{1}{3}H}$