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] $\displaystyle V_{cone}=\dfrac13 \pi r^2 \cdot h$ Volume of the cone
[2] $\displaystyle a=\pi r s~\implies~a = \pi r \sqrt{r^2+h^2}$ Area of the curved surface.
Solve [1] for r and plug in this term into [2]:
$\displaystyle a(h)= \dfrac{\sqrt{3V}}{\sqrt{h}} \cdot \sqrt{\dfrac{\pi h^3 + 3V}{h}}$
Now differentiate a wrt h (V is a constant!). You'll get:
$\displaystyle a'(h)=\dfrac{(\pi h^3-6V)\sqrt{3V}\cdot \sqrt{\dfrac{\pi h^3 + 3V}{h}}}{2h^{\frac32}(\pi h^3+3V)}$
Solve a'(h) = 0 for h. This is obviously only possible if
$\displaystyle \pi h^3-6V = 0$
Solve for h. Substitute V by the term of [1] and solve for h again. You'll get
$\displaystyle h^2 = 2 r^2~\implies~\boxed{\dfrac hr = \sqrt{2}}$
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.