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Math Help - right circular cone

  1. #1
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    Red face right circular cone

    Hello,
    please help me ,
    Show that semi - vertical angle of right circular cone of given surface area and maximum volume is sin-1(1/3)?
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  2. #2
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    Hello anilpadavil
    Quote Originally Posted by anilpadavil View Post
    Hello,
    please help me ,
    Show that semi - vertical angle of right circular cone of given surface area and maximum volume is sin-1(1/3)?
    If the semi-vertical angle = \theta, the radius = r, the surface area = S and the volume = V:

    The vertical height, h = \frac{r}{\tan\theta} and the slant-height, l = \frac{r}{\sin\theta}.

    \Rightarrow S = \pi r^2 + \pi r l
    = \pi r^2 + \frac{\pi r^2}{\sin \theta}
    \Rightarrow \sin\theta = \frac{\pi r^2}{S-\pi r^2}

    \Rightarrow \tan\theta = \frac{\pi r^2}{\sqrt{S^2 - 2\pi S r^2}}

    So V = \tfrac13\pi r^2 h
    = \frac{\pi r^3}{3\tan\theta}

    = \frac{\pi r^3\sqrt{S^2-2\pi S r^2}}{3\pi r^2}

    =\tfrac13(S^2r^2-2\pi Sr^4)^{\frac12}
    \Rightarrow 3\frac{dV}{dr}=\tfrac12(S^2r^2-2\pi Sr^4)^{-\frac12}.(2S^2r-8\pi Sr^3)
    = 0 when 2S^2r-8\pi Sr^3=0
    \Rightarrow r = 0 or S - 4\pi r^2 = 0

    r = 0 gives a minimum volume; and the volume is a maximum when \pi r^2 = \tfrac14S

    For this value of r, \sin\theta = \frac{\pi r^2}{S-\pi r^2}= \frac{\tfrac14S}{S-\tfrac14S}=\frac13

    Grandad
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  3. #3
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    thank you very much
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