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Why is a baked bean can this shape?. 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.
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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 surfacearea.
The volume of a cylinder is: лr2h The surface area of a cylinder is: 2лr2 + 2лrh
The volume of a tin is 330 cm3, so: лr2h = 330 So 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 = 0 And solve for r: r = 3√(660/4л)
Therefore the radius should be: r = 3.74cm And, substituting for height gives: h = 7.49cm Can you prove that, for any volume, minimum SA is given by h = 2r ?