Expert: Abe Mantell - 6/7/2009

Question
The figure shows two right curcular cones, one upside down inside the other.  The two bases are parallel and the vertex of the smaller cone lines at the center of the larger cones base.  What values of r and h will give the smaller cone the largest possible volume? (The diagram shows that the bigger cone has a height of 12 and a radius of 6).  I set up a proportion for the cones and got 2h=r.  And then I subbed that into the volume formula but the problem I am having is that when I solve for h it comes out as 0 which it cant be.  Help!

Answer
Hello,

Since the inner cone is upside down, we establish similar triangles with
the large triangle (side-view) and the triangle above the cone.  Using
y as the height and r as the radius, we get y/r=12/6 ==> y=2r, but the
height of the triangle we really want (from the inner cone) is 12-y.
So we get h=12-2r.  Thus, the volume is:
V=(1/3)pi*r^2*(12-2r)=(2/3)*pi*(6r^2-r^3)

Now differentiate: V'=(2/3)*pi*(12r-3r^2)=0 ==> r=0 or r=4

So, the inner cone of maximum volume has a radius of 4 and h of 4.

OK?

Abe