How do you apply similar triangles to this problem? I don't quite see a way, or then H and R should be involved... and I don't how that's useful.
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 base radius of the larger cone, and let h and r be the height and base radius of the smaller cone. Hint: Use similar triangles to get an equation relating h and r.)
I'm guessing I should start with this? I'm not sure how to express the function using only one variable, though.
$$ V = \pi r^ 2 \frac{h}{3} \\ r \in (0, R) \\ h \in (0, H) $$
Hello, maxpancho!
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 and be the height and base radius of the larger cone.
And let and be the height and base radius of the smaller cone.
Note thatCode:A - * - : /|\ : : / | \ H-h : / | \ : : / | r \ : H F *- - * - -* E - : / \ |G / \ : / \ h| / \ : / \ | / \ : / \|/ \ - B * - - - - * - - - - * C D R
Also that the height of is
We have: .
Solve for [1]
The volume of the smaller cone is: . [2]
Substitute [1] into [2]: .
Simplify: .
Can you finish now?