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Show that a right-circular cylinder of greatest volume that can be inscribed in a right-circular cone that has a volume that is 4/9 the volume of the cone.

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  1. Let the cone radius = R
    cone height = H
    Let cylinder height = h
    then cylinder radius, r = R(1-h/H)

    Volume of cylinder, V
    =πr²h
    =πR²(1-h/H)²h

    For maximum volume,
    dV/dh=0
    π(1-h/H)^2R^2-(2πh(1-h/H)R^2)/H=0
    which simplifies to:
    (H^2-4hH+3h^2)=0
    which when solved for h gives
    h=H/3 or h=H
    h=H will give a zero volume (min.) so reject.
    for h=H/3,
    Volume of cylinder
    =πR²(1-(H/3)/H)²(H/3)
    =πR²(2/3)²(H/3)
    =(2/3)²(πR²H/3)
    =4/9 volume of the cone.

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