Optimization with calculus: a simple example

Mathematical optimization is concerned with finding the optimum value (either a maximum or a minimum) of a function given specific constraints which tell us the relationship between different variables. In this article I will explain how mathematical optimization is done using single-variable calculus. I will focus on a classical example that involves minimizing the amount of sheet metal required to enclose a specific volume of liquid.

Suppose you have a factory that produces soda cans. For simplicity, imagine these soda cans have infinitesimally small thickness and are perfectly cylindrical, with height h and diameter d. Then, suppose that I tell you that we need the volume to be exactly 0.25 m³, but that due to budget constraints we need to find a way to minimize the cost of producing the cans by minimizing the amount of sheet metal used, since more material = more production cost and therefore less profit for the company. Suppose that, based on the past 5 years history of selling these soda cans, that the cost of the sheet metal is $0.01/m².

Then, we consider this to be an optimization problem. First, we must figure out what exactly is to be minimized. Sure, material is one thing, but also cost. How do we quantify these two things at the same time? We consider the cost of a single can, which depends on the dimensions. We have $0.01/m² x surface area of the can, and since it is cylindrical, we can find the surface area as follows: the top and bottom contribute 2 circular areas, and given that the diameter…