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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!!

RobPratt
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Happy Rabbit44
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    Since you’re giving us the equation of the profile curve, the shape is fixed, and so are the area and volume — there’s nothing to vary, so minimizing is impossible. – bubba Mar 12 '22 at 08:53
  • So you are saying that I won't be able to find the shape that gives the minimum surface area while retaining the same volume? – Happy Rabbit44 Mar 12 '22 at 09:43
  • What about dividing it into frustums, then minimising that? – Happy Rabbit44 Mar 12 '22 at 09:44
  • Again, if the equation is fixed (your 4th degree equation), then the shape is fixed, not variable, so the volume and surface area are fixed. If you want to minimize, there has to be some paramter that we can tweak in order to make the surface area smaller. What is the parameter that we’re allowed to tweak?? – bubba Mar 12 '22 at 11:33
  • could you elaborate a bit more on that? examples or something? Initially, the shape is fixed however, I don't mind the shape. I only want to mathematically find any shape with a volume of 500cm^3 that takes the smallest surface area as possible. – Happy Rabbit44 Mar 12 '22 at 14:01
  • If we’re allowed to ignore the 4th degree equation you gave, then the shape with minimal surface area is a sphere. – bubba Mar 13 '22 at 03:22
  • And what would be the best way to derive that? – Happy Rabbit44 Mar 14 '22 at 11:32
  • The isoperimetric inequality states that a sphere has the smallest surface area for a given volume https://en.wikipedia.org/wiki/Isoperimetric_inequality#In_%7F'%22%60UNIQ--postMath-00000019-QINU%60%22'%7F – bubba Mar 15 '22 at 04:54
  • That’s why soap bubbles are spherical. – bubba Mar 15 '22 at 04:57

2 Answers2

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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 $S=12\pi$.

Cye Waldman
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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 $\tfrac43 \pi r^3 = 500$.

bubba
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