How to Minimize the surface area of a solid of revolution of a constant volume? I read this post here however, I want to know whether it would be possible to minimize the
surface area of a solid of revolution which is a non catenary.
Catenary curve for minimum surface of revolution
It is a 4 degree equation of
-1.810^-4x^4 + 6.52110^-3x^3-7.28810^-2x^2 + 2.5310^-1x + 2.775
it has a Surface area of 390 cm^2 and volume of 500cm^2
How do I minimize the surface area used while retaining the same volume?
Thanks in Advance!!
 A: I think you can consider a toy problem to see how this would work. Consider a right circular cylinder of radius $r$ and height $h$. We then have
$$
V=\pi r^2h\\
S=2\pi rh+2\pi r^2
$$
Thus, we can express
$$
S=\frac{2\pi V}{\pi r}+2\pi r^2\\
\frac{S}{V}=2\left(\frac{1}{r}+\frac{\pi r^2}{V} \right)
$$
We can minimize $S/V$ with respect to $r$,
$$
\frac{\partial (S/V)}{\partial r}=2\left(-\frac{1}{r^2}+\frac{2\pi r}{V} \right)=0\\
r=\sqrt[3]{\frac{V}{2\pi}}\\
h=\frac{V}{\pi r^2}
$$
As an example, let $V=2\pi$ so that by the above $r=1$ and $h=2$. Then $S=2\pi rh+2\pi r^2=6\pi$.
Suppose now that $r=1/2$. Then $h=8,\ V=2\pi$ and $S=17\pi/2$. For $r=2$ we get $h=1,\ V=2\pi$, and $S=12\pi$.
A: From the comments, it seems that we’re allowed to ignore the 4th degree equation that was given. Apparently, the OP just wants to find a shape with minimal surface area that has a volume of 500 cm^3.
The isoperimetric inequality states that a sphere has the smallest surface area for a given volume. That’s why soap bubbles are spherical. The radius of the desired sphere can be found by solving for $r$ in the equation $\tfrac43 \pi r^3 = 500$.
