Calculus Related Rates/Maximizing Cones Problem

[SquirrelHunte]

Homework Statement

Given a right circular cone you put an upside down cone inside it so that its vertex is at the center of the base of the larger cone, and its base is parallel to the base of the larger cone. If you choose the upside-down cone to have the largest possible volume, what fraction of the volume of the larger cone does it occupy?

Let H and R be the height and radius of the larger cone, let h and r be the height and radius of the small cone.

Homework Equations

Volume of a cone= V=(1/3)pi r^2 h

The Attempt at a Solution

Obviously I want the end problem to look like this:

Maximum (largest) volume of the small/Volume of the larger cone

So first I wanted to use the right angles to get similar triangles relating h and r. So I got h/H=r/R.

thus h=Hr/R or (H/R)r

then I put that into the volume formula of the smaller cone: (1/3)pi r^2 ((H/R)r)..

you end up with (H/3R)(pi)(r^3)=V

First derivative: (3H/3R)(pi)(r^2)...then...(H/R)(pi)(r^2)=V'

To get my maximum I set my V' to zero so (H/R)(pi)(r^2)=0 but my problem is that the only way to get that equation to equal zero is if r=0, but it can't equal 0 so now I am stuck...

Help Please?

 

[dynamicsolo]

So first I wanted to use the right angles to get similar triangles relating h and r. So I got h/H=r/R.

thus h=Hr/R or (H/R)r

Be careful: I don't think I can agree that the vertical cross-sections of these cones are similar triangles. The vertex of the inner cone must touch the center of the base of the outer cone and could have any height from zero to the height of the outer cone. The opening angle of the inner cone is, then, not necessarily the same as that of the outer cone.

What you can safely say is that the triangle whose apex is the apex of the outer cone and whose base touches the base of the inner cone is similar to the vertical cross-section of the outer cone. So (H-h)/r = H/R . [It may turn out that the maximal inner cone is similar to the outer cone, but we don't know that yet...]

[EDIT: You'll get two critical points. If you do the Second Derivative Test or look at the graph of the volume function, you'll see that one is a local minimum because the volume function decreases gently to zero at that extremum...]