# Thread: twist on an old optimization problem

1. ## twist on an old optimization problem

We've all seen the classic 'sphere inside a cone', cylinder in a cone', cone in a sphere' problems and etc.

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

"Find the ellipsoid of max volume which can be inscribed in a cone of radius R and height H".

Any willing participants?. The old, cliche, max min and related rates problem have become rather worn out for most of us, I would assume. Though, they are still challenging for calc students. I just thought I would add a twist. Something a little different.

2. 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.

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

Let's orient the ellipsoid 'horizontally' . . . then we have:
Code:
    -           *
:          /|\
:         / | \
:        /  |  \
:       /   |   \
H      /  ..*..  \
:     /*::::|::::*\
:    *-:-:-:+ - a -*
:   /  *::::b::::*  \
-  *- - - - * - - - -*
:    R   :

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

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

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

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

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

3. Cool, Soroban.