Minimize surface of cone with given volume without using derivatives

Question

Given volume of cone equal to V, i need to minimize side surface area of given cone without using derivatives. Exactly i'm trying to use AM-GM inequality.

First of all i tried to do it like it was done here: Maximise right circular cone volume with fixed surface area using inequalites , but got stucked - case in link is inverse:given surface area and we need to minimize volume. In addition in url OP works with whole surface of a cone(i need only side surface). Here are my calculations:

Given $V; V_{cone} = \frac{1}{3}\pi R^2 h; A = \pi R l, A\rightarrow min. $ Since $l = \sqrt{h^2 + R^2} \;and\; h = \frac{3V}{\pi R^2}:$ $$l = \sqrt{ \frac{3V}{\pi R^2} + R^2} \Rightarrow $$ $$\Rightarrow A = \pi R \sqrt{ \frac{3V}{\pi R^2} +R^2} = \pi \sqrt{ \frac{3V}{\pi} +R^4}$$ And then i got stucked. Any ideas how to solve it?

UPD: I've also expressed $R$ on $h$ value. Then: $R = \sqrt{\frac{3V}{\pi h} }, \; l = \sqrt{\frac{3V}{\pi h} + h^2 } \Rightarrow $ $$\Rightarrow A = \pi \sqrt{\frac{3V}{\pi h} } \sqrt{\frac{3V}{\pi h} + h^2 }.$$ So, $$A = \pi \sqrt{\frac{9V^2}{\pi^2 h^2} + \frac{3Vh}{\pi} } = \pi \sqrt{\frac{3V}{\pi}\Big(\frac{3V}{\pi h^2} +h \Big) }\rightarrow \min. $$ Hence $ \min is \; achieving \; \iff $ $h = -\frac{3V}{\pi h^2} \Rightarrow h^3 = -\frac{3V}{\pi}.$

BUT A is equal to zero in that case. Will it be the right solution?

Answer 1

Ok i got this by myself already.

What we need is to express volume and side area on $R$ and $h$ values.

Then consider $A^2 = \pi^2 R^2 l^2.$

Because $l^2 = h^2 + R^2 :$ $$ A^2 = \pi^2 R^2 (h^2 +R^2) = \pi^2 R^2h^2 +\pi^2 R^4.$$

Express terms of equality on $V \Rightarrow$ $$\Rightarrow A^2 = \frac{9V^2}{h^2} + 3V\pi h.$$ Use AM-GM inequality: $$A^2 = 2 \frac{\frac{9V^2}{h^2} + 3V\pi h}{2} > 2\sqrt{27V^3 \pi \frac{1}{h} }.$$ This inequality will turn to equality $\iff \frac{9V^2}{h^2} = 3V\pi h \iff$ $$\iff h^3 = \frac{3V}{\pi} \Rightarrow h = \sqrt[3]{\frac{3V}{\pi}}.$$ Now substitute it to area's formula and get answer: $$A_{min} = \pi R \sqrt{R^2+ \sqrt[3]{\Big(\frac{3V}{\pi}} \Big)^2 } = \sqrt{\pi^2 R^4 + \pi^{\frac{4}{3} } (3V)^{\frac{2}{3} }}.$$

Answer 2

At first like inlink before we have derivative LM method:

$$ V= \dfrac{\pi R^2 h}{3};\, A= \pi R\sqrt {R^2+h^2};$$ We can remove constants by LM method $$ V_1= R^2 h\,;\, A_1= R\sqrt {R^2+h^2};$$

$$ \dfrac{2Rh}{R^2}=\dfrac{\sqrt {R^2+h^2}+R^2/\sqrt {R^2+h^2}}{R\cdot \dfrac{2h}{2\sqrt {R^2+h^2}}}$$ which simplifies to $$ h= \sqrt{2} R$$ and semi-vertical angle of cone $$ \tan \alpha = \dfrac{R}{h}=\dfrac{1}{\sqrt{2}}$$

In terms of $R,$ the area and volume are

$$ A= \pi \sqrt 3 R^2;\,V= \dfrac{\pi \sqrt 2 R^3}{3} ;$$

are related by a non-dimensional number

$$ \dfrac{A^3}{R^2}=\dfrac{27 \pi \sqrt 3}{2}$$

Next the AM/GM method becomes relevant here.. to see this please give the most elementary situation that can be pulled into this problem as well.