Posted by **Mishaka** on Friday, December 16, 2011 at 7:43pm.

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 = 1/2pir^2h.)

- Calculus -
**Mishaka**, Friday, December 16, 2011 at 8:13pm
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)

- Calculus -
**JWAL**, Thursday, December 22, 2011 at 6:06pm
How did you get this? Please help I am stuck.

- Calculus -
**Steve**, Sunday, January 22, 2012 at 4:37pm
Draw the cones. If the large cone has height H and radius R, the small cone has height h and radius r, so that

r/R = 1 - h/H

The ratio of the two volumes is v/V = r^2h/R^2H = (r/R)^2 h/H

if we maximize that ratio as a function of u = h/H, we get

f(u) = (1-u)^2 * u

= (u - 2u^2 + u^3)

f'(u) = (1-4u+3u^2)

= (h-1)(3h-1)

Clearly, when u=1, f(u) = 0 a minimum

when u = 1/3, f(h) = 4/9*1/3 = 4/27

so, h/H = 1/3 for max volume ratio

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