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Math Help - Volume

  1. #1
    Member SengNee's Avatar
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    Volume

    A right circular cone is inscribed in a sphere of radius a with the vertex and the circumference of the base of the cone touching the surface of the sphere.
    Show that the volume of the cone is given by I=\frac{1}{3}\pi (2ah^2-h^3) where h is the height of the cone.
    Hence, find the height of the cone when its volume is maximum.
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by SengNee View Post
    A right circular cone is inscribed in a sphere of radius a with the vertex and the circumference of the base of the cone touching the surface of the sphere.
    Show that the volume of the cone is given by I=\frac{1}{3}\pi (2ah^2-h^3) where h is the height of the cone.
    Hence, find the height of the cone when its volume is maximum.
    In a right triangle the product of the two hypotenuse segments equals the square of the height. Euclid's theorem (That's the name how this theorem is called in Germany - maybe you know this theorem under a different name)

    In comparison with the two sketches you see:

    q = h ..... of the cone
    p = 2a - h ..... of the cone

    The volume of a cone is:

    V=\frac13 \pi r^2 \cdot h

    Therefore you get:

    V=\frac13 \pi (h(2a-h)) \cdot h~\implies~V=\frac13 \pi\left(2ah^2 - h^3\right)

    To find the maximum volume calculate

    V'(h) = \frac13 \pi \left(2ah - 3h^2\right)

    V'(h) = 0~\implies~h = 0~\vee~\boxed{h = \frac23 a}
    Attached Thumbnails Attached Thumbnails Volume-kegel_inkugel.gif  
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