# optimal size of conserve/tin cans

Given a volume V for a conserve can, using calculus we find that the dimensions that minimise the surface area (and hence the cost of material) is such that the height is equal to the diameter (hence it looks like a square when viewed from the side).

But when I go to my cupboard I find that many cans are not manufactured in this ratio. Why would manufacturers ignore the optimal solution? Or, am I missing something?

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Minimizing cost may not be a big consideration, since the cost is in any case low. Also, the cost of top and bottom bits, per square cm, is not the same as the cost of the side stuff, per square cm. Thicker, I think, also there is more wastage. – André Nicolas Nov 5 '12 at 18:37
I don't think this is a question that can be answered through mathematics. – Rahul Nov 5 '12 at 19:15
@Rahul I wasn't sure where I could ask, so I tried here. Thing is that maths gives us the result, the real-life doesn't seem to corroborate. – Geoff Nov 5 '12 at 19:17
This is an interesting question, but this is probably the wrong forum... Other considerations are standards, which are usually more limiting than isolated pure design considerations (metric vs. imperial, 120 vs. 220, left vs. right, etc.). Also, space for a nice label :-). – copper.hat Nov 5 '12 at 19:19
There is a cost $p$ per cm$^2$ of surface and a cost $q$ per cm of soldering along the rims. Taking the respective values of $p$ and $q$ into account will provide an optimum depending on $p$ and $q$. – Christian Blatter Nov 5 '12 at 19:38