Geometry

A cone has diameter 12 and height 9. A cylinder is placed inside the cone so the base of the cylinder is concentric with the base of the cone and the upper base of the cylinder is contained in the surface of the cone. If the volume of the cone is nine times the volume of the cylinder, find the dimensions of the cylinder.

asked by Help Please
  1. Make a diagram showing the cross-section.
    let the radius of the cylinder be r and its height be h
    The small triangle to the right of the cylinder is similar to the triangle formed within the cone
    so h/(6-r) = 9/6 = 3/2
    2h = 18 - 3r, where r < 6
    h = (18-3r)/2

    volume of cone = (1/3)?(36)(9) = 108? cubic units
    volume of cylinder = ?(r^2)(h)
    = ?(r^2)(18-3r)/2

    108? = 9(?(r^2)(18-3r)/2)
    216? = 162?r^2 - 27?r^3
    divide by 27?
    8 = 6r^2 - r^3
    r^3 - 6r^2 + 8 = 0
    Here comes the hard part,
    This equation does not factor, I have no idea if you know how to solve a cubic. One method is Newton's Method.
    Another is to use a webbased method like Wolfram
    http://www.wolframalpha.com/input/?i=r%5E3+-+6r%5E2+%2B+8+%3D+0

    we get 2 positive solutions for
    r = 1.3054 , then h = 7.0419
    r = 5.7588 , then h = .3618

    both solutions are valid

    posted by Reiny

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