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I want to calculate the minimum surface area of a (closed) box for a given volume. So let’s say I have a given volume V (e.g. V=10m^3). And I need a box where all the surface area is as minimal as possible. This would be a great starting point if I knew how to calculate that.

To make things more complicated: Let’s say I have a given volume V but also I have a limit for the height of the box. So the height would be a certain measurement h (or higher), length and width would still be variable.

If you could answer my first question that would already be great. I think for the second problem I could also just do the same calculation/minimization and just work with a given area A (A = V/h).

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Say that the Surface area is given by

$$A=2(ab+bc+ca).$$

Then, from the property that the Geometric Mean is always less that or equal to the Arithmetic Mean ($AM-GM$), we get

$$\frac{ab+bc+ca}{3}\geq\sqrt[3]{(abc)^2}.$$

Multiplying by $6$ gives

$$2(ab+bc+ca)\geq 6\sqrt[3]{(abc)^2},$$

where

$$abc=10m^3.$$

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  • $\begingroup$ Maybe I misunderstood your answer but I meant having a given volume and trying to find the minimum surface area. If I am not mistaken you assumed it the other way around. $\endgroup$
    – chocolanto
    May 7 '18 at 13:47
  • $\begingroup$ No the Minimum Surface area is given by $$6\sqrt[3]{100}$$ as i have written it above! $\endgroup$ May 7 '18 at 13:49
  • $\begingroup$ Aah I understand. Thank you $\endgroup$
    – chocolanto
    May 7 '18 at 13:54
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HINT

Indicating with $x,y,z$ the sides of the box we have

  • $S=2(xy+yz+zx)$ surface to minimize

with the constraint

  • $V=xyz=10$
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