Cheapest can when the ends cost twice as much

A closed cylindrical can is to hold \(330\text{ cm}^3\). It has radius \(r\) cm and height \(h\) cm. The metal for the top and bottom costs \(2k\) cents per \(\text{cm}^2\) and the metal for the curved surface costs \(k\) cents per \(\text{cm}^2\), where \(k\) is a positive constant. Ignore the thickness of the metal and any overlap at the joins.

1. Show that the cost of the metal, \(C\) cents, is given by \(C=k\left(4\pi r^2+\dfrac{660}{r}\right)\).[2]
2. Find the value of \(r\) for which \(C\) is a minimum, and show that it gives a minimum.[4]
3. Show that, for the cheapest can, \(h=4r\). A manufacturer uses a can of radius \(3.2\) cm, with the height chosen so that it also holds \(330\text{ cm}^3\). Compare its cost with the minimum cost and suggest a reason for the manufacturer's choice.[3]