Design a Tin Can (harder pre-cal but easy calc)

Design a tin can so that the total area including a top and bottom is minimized for a required volume. The can has radius r and height h.

A={{2\,V}\over{r}}+2\,\pi\,r^2\mbox{; let } r=au

The reason for changing variables was to choose a nice a which will make u unitless and make the problem generic since the numerical value of V has nothing to do with the essence of the problem. Also, we can get rid of the π so that the function to minimize is easy to study. The value of a landen picked is not as arbitrary as it looks at first and there was some experimentation with the substitution using Maxima ( to find a good one.

a={{2^{{{2}\over{3}}}\,V^{{{1}\over{3}}}}\over{2\,\pi^{{{1}\over{3  }}}}}
A={{2^{{{1}\over{3}}}\,\pi^{{{1}\over{3}}}\,\left(u^3+2\right)\,V^{  {{2}\over{3}}}}\over{u}}

Now all we need to do is minimize:


A little playing around with a calculator suggest that the minimum value is f(u) = 3 and that this happens when u = 1. So we rewrite f(u) to emphasize 3 and 1 and we hit the jackpot.


In this form it is obvious that the minimum value of f(u) = 3 and this happens when u = 1.

r={{2^{{{2}\over{3}}}\,V^{{{1}\over{3}}}}\over{2\,\pi^{{{1}\over{3  }}}}}
h={{2^{{{2}\over{3}}}\,V^{{{1}\over{3}}}}\over{\pi^{{{1}\over{3}}}  }}=2r