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Math Help - Cone Optimization

  1. #1
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    Cone Optimization

    Heres a calc problem.

    I have to find the maximum volume of the upside down cone inside the larger cone.

    Thanks
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  2. #2
    MHF Contributor
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    Hello

    Here is a cut on a plane



    Volume of "small" cone is v = 1/3 pi rē h
    It is possible to express r and h function of theta, R and H
    First
    h\;=\;\frac{r}{tan \theta}

    Then
    \frac{r}{tan \alpha} + \frac{r}{tan \theta}\;=\;H

    After simplifications
    r\;=\;\frac{RHtan \theta}{R+Htan \theta}

    Thus
    v\;=\;\frac{1}{3}\pi R^3 H^3 \frac{(tan \theta)^2}{(R+Htan \theta)^3}

    After differentiating I find an extremum for
    tan \theta\;=\;\frac{H+\sqrt{H^2+6RH}}{3H}

    With R=6' and H=12'

    tan \theta\;=\;1

    \theta\;=\;45°

    Corresponding volume is
    v\;=\;\frac{1}{3}\pi\frac{R^3H^3}{(R+H)^3}

    which is about 15% of the volume of the "big" cone
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  3. #3
    Eater of Worlds
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    This is a little different than the usual cylinder in a cone and what not.

    But, like most others we can use similar triangles. Here it is in general. You can add the constants H=12 and R=6

    Let H=height of big cone, R=radius of big cone, r=radius of small cone, h=height of small cone.

    \frac{H-h}{r}=\frac{H}{R}

    r=\frac{(H-h)R}{H}

    Plug this into the cone volume formula:

    V=\frac{\pi}{3}\left(\frac{(H-h)R}{H}\right)^{2}\cdot h

    Now differentiate w.r,t h, set to 0 and solve for h. H=12, R=6 are constants

    You should get h=\frac{H}{3} and go from there.
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  4. #4
    MHF Contributor
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    Much more simple than mine !
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