VIDEO: The optimal can

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The optimal can - an extreme value problem

Manufacturers want to use as little material as possible for cans and beer cans should be handy. So how do they have to Dimensions a cylindrical can with a capacity of 0.5 l should be selected so that as little material as possible is needed? And do the manufacturers adhere to these optimal dimensions at all? This task sounds nonsensical at first, because a glance at the can shelf shows that the manufacturers On the whole, make the cans uniform, i.e. the same height and diameter Select. But is this perhaps only due to the standard filling machines? Or because the cans are easy to handle in the chosen shape?

  1. These questions can be checked in mathematics. In short, the task is: what diameter (or radius) and what height do you need for the can cylinder choose so that the can holds a volume of 0.5 l and the surface (that is the material consumption) as small as possible will.
  2. This is an extreme value problem with a main condition (the surface should be minimal) with a secondary condition (the volume is 0.5 = 500 cm³).
  3. When dealing with such problems, you must first set up both the main and secondary conditions as equations. In this case, the radius r of the cylinder circle and the height h of the cylinder are the two unknowns (which you want to calculate).
  4. You can look up the formulas for the volume V and the surface F of a cylinder in the formulary. Notice that the surface of a cylinder consists of the two circles and a rectangle (the cylinder jacket).
    5. Calculate cylinder height

    You know some sizes of a cylinder such as diameter or ...

  5. The following applies: V = ¶ r² * h = 500 cm³ as a secondary condition and F = 2 ¶ r² + 2 ¶ r * h as the main condition that should be minimal.
  6. The main condition initially contains the two unknowns r and h. From the secondary condition you can now separate one of the two unknowns (h is useful because it is easier to calculate) and insert it into the main condition. The procedure is similar to substituting two equations with two unknowns. Only here you have it Functions to do.
  7. You get h = 500 / ¶ r² (the cm³ are left out for the further calculation; the result is then calculated in the unit "cm") and put this in the surface F.
  8. F (r) = 2 ¶ r² + 2 ¶ r * (500 / ¶ r²) = 2 ¶ r² + 1000 / r, that is, the surface of your can now only depends on the radius.
  9. According to the task, the surface should be minimal, so you are looking for an extreme value of this function.
  10. To do this, derive F (r) according to the variable r and set the derivative to zero.
  11. You calculate F '(r) = 4 ¶ r - 1000 / r² (you can look up the derivative of 1 / r in the formulary if you don't know).
  12. The following applies to an extremum: 4 ¶ r - 1000 / r² = 0.
  13. From this you calculate r³ = 250 / ¶ and r = 4.3 cm (third root on TR). Your minimal box has a diameter of almost 9 cm.
  14. You can now calculate the height h of the can from the secondary condition (cf. Point 8.) to h = 8.6 cm. The diameter and height therefore match.

Mathematics and reality - critically question the result

But does a beer can really look like this, about as high as it is wide? Everyday life contradicts the result of the mathematics Clearly, the cans are relatively higher, so narrower and of course more manageable. It remains uncertain whether the customer's wishes are in the foreground here. And something else should be taken into account: Beer cans are not filled to the top, i.e. larger than 500 ml. In addition, the ideal cylinder shape is of course given.

  • However, something was not taken into account when it came to material consumption: there is waste! It is created when the circles are cut. It is not known whether it will be melted down again or disposed of. In any case, it is a loss to the company. Perhaps you will recalculate the extreme value task of the optimal can taking this waste into account.
  • Then you don't need two circles for the surface, but two squares in addition to the rectangular cylinder surface. The result for this case is r = 4 cm and h = 10 cm, so the can becomes narrower and taller. That’s astonishing!
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