Question

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 large cone, let h and r be the height and radius of the small cone. Use similar triangles to get an equation relating h and r. The formula for the volume of a cone is V =1/3 r^2h.)

Answer 1

Pretty sure I figured it out, 4/27. I found this by simplifying:

((1/3pi (h - 2/3 h))(4/9 r^2)) / (1/3 pi r^2 h)