right circular cone

**anilpadavil** · October 21st 2009, 09:02 AM

Hello,
please help me,
Show 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?

**earboth** · October 21st 2009, 11:44 AM

Originally Posted by anilpadavil

Hello,
please help me,
Show 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.

**anilpadavil** · October 21st 2009, 05:27 PM

anyone knows a simpler way to solve the question

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