More coni-mising

[Yoruichi-san]

Impressed with the alacrity with which you maximized your gold reward, High King Math offers you an additional reward: another (right circular) cone of gold, this time, one that must fit inside a sphere of radius R. What is your take-away this time?

[k-man]

Maximizing the volume of a cone inscribed in a sphere can be reduced to maximizing the area of a triangle inscribed in a circle. Since the largest area belongs to the equilateral triangle, the largest cone will have its height equal the diameter of the base. Therefore, the volume will be V=Pi * R

³ * sqrt(3)/4

[plasmid]

I'm not sure about that.

If two triangles have equal areas, I think the cones made by rotating those triangles about an axis do not necessarily have equal volume.

A small bit of area at a distance r from the axis of rotation will sweep through a volume of space proportional to r², so a patch of area that's far from the axis will sweep out a larger volume than an equal size area near the axis.

So maximizing the area of a triangle doesn't necessarily mean you've maximized the volume of the cone it will create.

[k-man]

plasmid, you're right. I didn't have the time to do the math and it seemed like this approach was worth a shot. Now, that I had some more time...

Representing as x the distance of the base of the cone from the center of the sphere we can express the square of the radius of the cone as r

²=R²-x² and the height of the cone is h=R+x. So the volume is expressed as a function V(x) = pi/3 * (R²-x²)(R+x). Taking a derivative and solving it, we find the extremum at x=R/3. It's a maximum because at V®=0 and V(0)>0. So, the largest cone has the height h=4R/3, radius r=R*sqrt(8)/3 and volume V=32*pi*R³/81