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galactus · Jun 16, 2009, 02:06 PM

Here is a fun optimization problem if anyone would like a whirl.

Find the ELLIPSOID of max volume that can be inscribed in a cone of radius R and height H

galactus · Jun 17, 2009, 07:33 AM

No takers, huh? It ain't that bad.

Use similar triangles as you would in any types of these things.

The volume of an ellipse of revolution is $V=\frac{4}{3}{\pi}a^{2}b$

Where, as you know, a and b are the major and minor axes.

Unknown008 · Jun 17, 2009, 07:51 AM

I'd love to take a try, you'll have to guide me though. Ok, what's the question asking really? A cone, in which an ellipse in 3D has to be scribed?

galactus · Jun 17, 2009, 10:01 AM

Yes, that's it. But don't over complicate it. These types of problems are all '3D'. Even if you

have, say, a sphere in a cone or a cone is a sphere or what not.

Using similar triangles, we get $\frac{a}{H-b}=\frac{R}{H}\Rightarrow b=\frac{H}{R}(R-a)$

Unknown008 · Jun 17, 2009, 11:10 AM

Oh gosh, now I've got more variables in the volume formula...

$V= \frac{4}{3}\pi a^2(\frac{H}{R}(R-a))$

Ok I found dV/da, solved to zero and got the optimum volume when

$a=\frac{2}{3}R$

galactus · Jun 17, 2009, 12:13 PM

Yep. That's it. Good deal. And $b=\frac{H}{3}$

Unknown008 · Jun 17, 2009, 09:00 PM

Ah, yup. Didn't know if b was required...

$b=\frac{H}{R}(R-a) = \frac{H}{R}(R-\frac{2}{3}R) = \frac{HR}{3R} = \frac{H}{3}$

Then, I guess max volume will be replacing the values of a and b?

$V=\frac{4}{3} \pi (\frac{2}{3}R)^2 (\frac{H}{3}) = \frac{16\pi R^2H}{81}$

galactus · Jun 18, 2009, 05:57 AM

See. I told ya' it wasn't that bad. Just a twist on the typical 'cone in sphere', 'sphere in a

Cylinder' type problems. Can you think of another? Maybe a tetrahedron in a sphere or a

Sphere in a tetrahedron?:).

Hey, how about this one:

What is the hyperboloid of max volume that can be inscribed inside a sphere of radius R? :confused:

Just made that up.

Unknown008 · Jun 18, 2009, 08:37 AM

You mean a one sheet hyperboloid, or the two sheet one? Anyway... can that be done? :p

Just made a search, the equation for the hyperboloid is

$x^2 - y^2 = k$

I guess the volume is

[Volume of cylinder radius r] - 2[Integral of curve from center to r] - 2[Curved volume of top or bottom surfaces]