Maximizing and Minimizing

In the last post about the derivative we emphasized its utility with some examples about how it can be used to find some relationships between position, velocity, and acceleration or to maximize profit and minimize cost in numerous different practical situations. Well now I want to discuss that in more depth and provide an example. Last time we used the following definition for the derivative (or slope) using the limit:

With this we can find the exact slope at any point on a curve. Now consider the following graph. Each of the red lines represents the slope at that point with a tangent line

Consider what the slope of such a tangent line would be at a minimum or maximum on a curve. That’s right, the line would be flat and horizontal like this:

Therefore the slope at these points is zero. With this geometrical understanding of a minimum or maximum we can find the derivative and then determine where the derivative (or slope) is equal to zero.

Now let’s consider an example function that relates the production level of a manufacturer to the potential revenue per unit. Obviously we want to maximize revenue so the key will be to find the maximum. Here is the function:

Now you might ask, “Well how in the world did we know this tells us about this relationship?” The short answer is that after some experimenting by the company at various different production levels and measuring the resultant revenue, the corresponding data points were plotted and the following function was mathematically obtained as a best-fit approximation to describe those data points. (More on this in a future post.) So assuming this is the correct function that we are working with, let’s first find the derivative.

Now this derivative gives us the slope at any point along the curve of the original function above. Where does this new function equal zero? This is the same as finding the roots of the equation which is a typical problem in algebra. Here is how it’s done based on our example.

Now if we put these two production level values into the original function it will tell us the revenue produced. Here are the results.

Therefore production level 2 provides the maximum revenue potential. (The root zero is an interesting point called an inflection point. But that’s another topic.) To demonstrate that this process has in fact located the maximum, here is a table of revenue values at other production levels and a graph of the function.

Next time we will look more in depth at the derivative and some basic rules or short-cuts we can use so that we don’t have to go through the entire limit process every time we want to find the derivative.

About jonathanmcgehee

Forensic Engineer, Math Nerd, Husband of @rebekahmcgehee, Father of 4, Follower of Jesus

Posted on August 8, 2011, in Calculus, Mathematics and tagged , , , , , , . Bookmark the permalink. 5 Comments.

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