I have to find out which shape of packaging for a fragile object has the most volume to fit the object and styrofoam packing. The sum of the height and the perimeter must be less than $100cm$.

There is a circular base (cylindrical package), a square base (rectangular package) and a rectangle base ($w = 2l$)(rectangular package).

The problem is I have no idea how to start. will the function represent the volume? the base? the height and perimeter? can I assume the y-axis is the volume?

Another question is: what is the perimeter of a cylinder? some one said (not on this site) that it was $2{\pi}r+2h$ but wouldn't it be $4{\pi}r+2h$ because there are two circumferences in a cylinder.

EDIT: this probably is a cubic function that is representing the volume

  • $\begingroup$ Do you mean surface area? $\endgroup$ Sep 19, 2014 at 20:20
  • $\begingroup$ no, the page specified that the perimeter plus the hight must be less than 100 cm. Surface area is not mentioned at all $\endgroup$ Sep 19, 2014 at 20:22
  • $\begingroup$ So the perimeter of the base? Or all of the faces? $\endgroup$ Sep 19, 2014 at 20:23
  • $\begingroup$ Please proofread your question. A base doesn't have a volume. You mean to say "which base should I choose so that the volume is maximum." $\endgroup$
    – David P
    Sep 19, 2014 at 20:26
  • $\begingroup$ I have no idea... $\endgroup$ Sep 19, 2014 at 20:26

1 Answer 1


There is no maximum volume, since the problem specifies that the sum must be less than $100$. So we assume that the regulations say $\le 100$.

To maximize volume, it is best to use the full $100$. For height $h$ and radius $r$, we have the constraint $h+2\pi r=100$. This is because the perimeter of the base is $2\pi r$.

The volume is $\pi r^2 h$. So we want to maximize $\pi r^2(100-2\pi r)$, where $0\le r\le \frac{100}{2\pi}$. To solve this problem, we can use the usual tools (derivative).

If we solve the problem of the previous paragraph, then whatever maximum volume $V$ we get, we can get arbitrarily close to $V$ if we take the less than seriously, but we cannot reach $V$.

The square base and rectangular base problems are handled in a similar way.

  • $\begingroup$ I think I've got it :) $\endgroup$ Sep 19, 2014 at 21:24

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