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 ?