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Math Help - right circular cone

  1. #1
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    Question right circular cone

    Hello,
    please help me,
    S
    how that the right circular cone of least curved surface and given volume has an altitude equal to 2^(1/2) time the radius of the base
    ?
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  2. #2
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    Quote Originally Posted by anilpadavil View Post
    Hello,
    please help me,
    S
    how that the right circular cone of least curved surface and given volume has an altitude equal to 2^(1/2) time the radius of the base
    ?
    [1] V_{cone}=\dfrac13 \pi r^2 \cdot h Volume of the cone

    [2] a=\pi r s~\implies~a = \pi r \sqrt{r^2+h^2} Area of the curved surface.

    Solve [1] for r and plug in this term into [2]:

    a(h)= \dfrac{\sqrt{3V}}{\sqrt{h}} \cdot \sqrt{\dfrac{\pi h^3 + 3V}{h}}

    Now differentiate a wrt h (V is a constant!). You'll get:

    a'(h)=\dfrac{(\pi h^3-6V)\sqrt{3V}\cdot \sqrt{\dfrac{\pi h^3 + 3V}{h}}}{2h^{\frac32}(\pi h^3+3V)}

    Solve a'(h) = 0 for h. This is obviously only possible if

    \pi h^3-6V = 0

    Solve for h. Substitute V by the term of [1] and solve for h again. You'll get

    h^2 = 2 r^2~\implies~\boxed{\dfrac hr = \sqrt{2}}

    Remark: Especially the differentiation of a is really horrible. So I hope for you that someone knows a simpler, shorter way to solve the question.
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  3. #3
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    anyone knows a simpler way to solve the question
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