why is a baked bean can this shape?
Post on 04-Jan-2016
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The cans must be cylindrical for best packaging, so we need to know what the best height and width would be for a fixed volume.
In order to use the smallest amount of packaging, we need to
minimise the surface
area.
The volume of a cylinder is:
лr2hThe surface area of a cylinder is:
2лr2 + 2лrh
The volume of a tin is 330 cm3, so:
лr2h = 330So the height can be written in terms of the radius:
h = 330/(лr2)
Now we can write the surface area (SA) in terms of the radius only:
SA = 2лr2 + 660лr/(лr2)Which can be simplified to:
SA = 2лr2 + 660/r
In order to determine when the surface area is at a minimum, we need to differentiate…
dSA/dr = 4лr – 660/r2
And put the result equal to 0 to find any stationary points:
4лr – 660/r2 = 0
To solve this, we need to multiply through by r2…
4лr3 – 660 = 0And solve for r:
r = 3√(660/4л)
Therefore the radius should be:
r = 3.74cmAnd, substituting for height gives:
h = 7.49cm
Can you prove that, for any volume,
minimum SA is given by h = 2r ?
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