Statistics Solutions

Solution: The values of h/r that we found in Problems 1 and 2 are a little closer to the ones that actually occur on supermarket shelves, but they still

Question: The values of h/r that we found in Problems 1 and 2 are a little closer to the ones that actually occur on supermarket shelves, but they still don't account for everything. If we look more closely at some real cans, we see that the lid and the base are formed from discs with radius larger than r that are bent over the ends of the can. If we allow for this we would increase h/r. More significantly, in addition to the cost of the metal we need to incorporate the manufacturing of the can into the cost. Let's assume that most of the expense is incurred in joining the sides to the rims of the cans. If we cut the discs from hexagons as in Problem 2, then the total cost is proportional to

\[4\sqrt{3}{{r}^{2}}+2\pi rh+k\left( 4\pi r+h \right)\]

where k is the reciprocal of the length that can be joined for the cost of one unit area of metal. Show that this expression is minimized when

\[\frac{\sqrt[3]{V}}{k}=\sqrt{\frac{\pi h}{r}}\left( \frac{2\pi -h/r}{\pi h/r-4\sqrt{3}} \right)\]

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