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Calculus

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. The formula for the volume of a cone is V = 1/2pir^2h.)

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Mishaka

  1. Pretty sure I figured it out, 4/27. I found this by simplifying:
    ((1/3pi (h - 2/3 h))(4/9 r^2)) / (1/3 pi r^2 h)

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    Mishaka

  2. How did you get this? Please help I am stuck.

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    JWAL

  4. Draw the cones. If the large cone has height H and radius R, the small cone has height h and radius r, so that

    r/R = 1 - h/H

    The ratio of the two volumes is v/V = r^2h/R^2H = (r/R)^2 h/H

    if we maximize that ratio as a function of u = h/H, we get

    f(u) = (1-u)^2 * u
    = (u - 2u^2 + u^3)
    f'(u) = (1-4u+3u^2)
    = (h-1)(3h-1)

    Clearly, when u=1, f(u) = 0 a minimum
    when u = 1/3, f(h) = 4/9*1/3 = 4/27

    so, h/H = 1/3 for max volume ratio

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    Steve

  5. Okay, letting x be height and b radius of smaller,
    then x/b=h/r, and
    volumeof smaller is 1/3pi(h-x)b^2, or
    1/3pi(h-x)(x^2r^2/h^2=
    1/3pi((x^2r^2/h)-(x^3r^2/h^2).
    Now differinate with respect to x,
    and set to 0, so
    ((2xr^2/h)-(3x^2r^2/h^2)=0,
    so you get x=2/3h.
    So, fraction is ((1/3pi(h-2/3h)(4/9r^2)/(1/3pir^2h)=
    4/27

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    :)

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