If you could choose any diameter, what would be the dimensions for a 330 mil can that would use the least amount of aluminum?

Answer 1

Since this question is listed under the category of geometry, it can be assumed that calculus is not to be used. Consequently, when you get to the equation for surface area that Mark M lists in his answer to your question:

S = 2πr² + 660/r

You would need to use a graphing calculator or a graphing utility like desmos to graph this function and pick the radius from the graph that yields the minimum surface area. Don't forget to double the radius to get the diameter.

In answer to the question that James S introduces, since thickness is not mentioned in the problem, it can be assumed that thickness is held constant based on other criteria. In reality the thickness << radius so it would not effect the volume in any substantial way.

Answer 2

Volume = πr²h = 330. So, h = 330 / (πr²)

Minimize surface area = S = 2πr² + 2πrh = 2πr² + 2πr(330 / (πr²)) = 2πr² + 660/r

dS/dr = 4πr - 660 / r² = (4πr³ - 660) / r² = 0 when 4πr³ = 660. So, r³ = 165 / π.

Minimum surface area when diameter = 2r = 2(165/π)^1/3 ≈ 7.49 cm