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.) Can someone help me get started on this? I have a picture drawn. What are the equations I need?

Answer 1

Is someone working on this?

Answer 2

yes, be patient.

Answer 3

Thank you. I wasn't sure. I appreciate the help; take your time.

Answer 4

If you imagine the line going from vertex of large cone to base as a linear line, then you can find the equation of that line
slope = rise/run = H/R
y = H - (H/R)x
the point (x,y) on that line could represent a point on the base of the inverted small cone
thus x = r and y = h
--> h = H -(H/R)r
Now plug that into equation of volume of a cone because we want to maximize the volume of the inverted cone

\[V = \frac{1}{3} \pi r^{2}h= \frac{1}{3}\pi r^{2} (H-\frac{H}{R}r) = \frac{1}{3}\pi Hr^{2}-\frac{H}{3R} \pi r^{3}\]
set derivative equal to 0 to optimize
\[\frac{dV}{dr} = \frac{2}{3}\pi Hr -\frac{H}{R}\pi r^{2} =0\]
solving for r
\[\rightarrow r = \frac{2}{3}R\]
plugging that back into the linear relationship
\[h = H-\frac{H}{R}(\frac{2}{3}R) = \frac{1}{3}H\]
Then plug these into Volume equation to get volume of inverted cone in terms of H,R
\[V = \frac{1}{3}\pi (\frac{1}{3}H)(\frac{2}{3}R)^{2} = \frac{4}{81}\pi HR^{2}\]
thus ratio of inverted cone to large cone is (4/81)/(1/3) = 4/27