anonymous
  • anonymous
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?
Mathematics
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anonymous
  • anonymous
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?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Is someone working on this?
dumbcow
  • dumbcow
yes, be patient.
anonymous
  • anonymous
Thank you. I wasn't sure. I appreciate the help; take your time.

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dumbcow
  • dumbcow
|dw:1328479084743:dw| 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
anonymous
  • anonymous
Thank you for your time. I will work through this the best I can.
dumbcow
  • dumbcow
your welcome..hope it makes sense :)
anonymous
  • anonymous
One question about the whole slope thing...isn't the line going from the vertex to the base vertical? How can it have slope?
dumbcow
  • dumbcow
no think slant height, the line on outside of cone goes from vertex to point on edge of circular base
anonymous
  • anonymous
Oh, thank you. I understand now.

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