Basics

• Optimization is a way to find the maxima or minima of a function -- the points on an interval where the function is at its biggest or smallest. Optimization may be unconstrained, in which case there are no constraints on the function you are studying, or it may be constrained, in which case your function is subject to a constraint. This constraint is usually a condition that you can describe with another mathematical function.
  
  

Calculus

• If you want to find a maximum or minimum, you could find the output value for the function at every point along the x-axis. This approach, however, would be extremely time-consuming. Calculus makes it simpler with a mathematical tool called the derivative. When a function is at a maximum or minimum, it will have no slope -- in other words, its graph will be flat at that point. In calculus, the derivative of a function will give you the slope of the function at any given point.
  
  

Derivatives

• You can find where maxima and minima should be by taking the derivative of a function and setting it equal to zero. But how do you figure out which points are maxima and which points are minima? You can solve this conundrum by taking the derivative of the derivative, which gives you the second derivative of the function. At a maximum, the second derivative will be negative, while at a minimum, it will be positive.
  
  

Example

• The best way to understand optimization is by looking at an example. Let's say you want to enclose a rectangular area of land with a chain-link fence, and you want to enclose as much land as you possibly can. The only limitation is that you have only 1,000 feet of fencing material. You can turn this problem into a pair of mathematical equations. The first would be the area enclosed by the fence: A = xy. The second would be your maximum amount of fencing material: 1,000 = 2x + 2y. You can start by solving the second equation for y, which gives you: 500 - x = y. Next, you substitute this solution for y in the area equation, and you get: A = (500 - x) x. Multiply through the values and you get: A = 500x - x^2. The first derivative will be A' = 500 - 2x, where the ' symbol indicates "derivative of A," and the second derivative will be A" = -2. If you set A' = 0 = 500 - 2x, you'll find that A' = 0 when x is 250. The second derivative is negative at that point (as everywhere else), so it's a maximum. If you plug x = 250 back into the equation for the maximum amount of fencing material, you get 500 - 250 = y, so y = 250. That means you'll encircle the maximum possible amount of land when you build a square enclosure.