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Math Help - Maximize cone inside a cone

  1. #1
    Member maxpancho's Avatar
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    Maximize cone inside a cone

    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.

    $$ V = \pi r^ 2 \frac{h}{3} \\ r \in (0, R) \\ h \in (0, H) $$
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  2. #2
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    Re: Maximize cone inside a cone

    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.
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  3. #3
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    Re: Maximize cone inside a cone

    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 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.

    Code:
                      A
        -             *         -
        :            /|\        :
        :           / | \      H-h
        :          /  |  \      :
        :         /   | r \     :
        H      F *- - * - -* E  -
        :       / \   |G  / \
        :      /   \ h|  /   \
        :     /     \ | /     \
        :    /       \|/       \
        - B * - - - - * - - - - * C
                      D    R
    Note that \Delta ABC \sim \Delta AFE.
    Also that the height of \Delta AFE is H-h.

    We have: . \frac{H-h}{r} \:=\:\frac{H}{R}

    Solve for h\!:\;\;h \:=\:\frac{H}{R}(R-r) [1]

    The volume of the smaller cone is: . V \:=\:\frac{\pi}{3}r^2h [2]

    Substitute [1] into [2]: . V \:=\:\frac{\pi}{3}r^2\left(\frac{H}{R}(R-r)\right)

    Simplify: . V \;=\;\frac{\pi H}{3R}(Rr^2-r^3)

    Can you finish now?
    Thanks from maxpancho
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  4. #4
    Member maxpancho's Avatar
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    Re: Maximize cone inside a cone

    Hum, I actually tried this myself, too, but not sure what to do about these H and R?
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  5. #5
    Member maxpancho's Avatar
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    Re: Maximize cone inside a cone

    Or do we treat them like constants? I haven't thought about that.
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