Extrema are the high and low points on a function’s graph.

It can be difficult to distinguish between the bumps and dips on a graph, the relative (or local) extrema, and the very highest and lowest points on a graph, the absolute extrema.

The Math-FAQ belows show how to find each type of extrema using derivatives.

As long as you have the profit function, you can find the maximum using the first derivative of the profit function.

Problem 1 The total profit P(x) (in thousands of dollars) from the sale of x units of a drug is

for x in [0, 10].

a. How many units should be sold to maximize profit?

b. What is the maximum profit?

If you are given the revenue and cost functions, you’ll need to combine those to get the profit function first. Then you can use the first derivative to find the maximum.

As the maximization and minimization problems become more complicated, using a table to organize examples of what is being modeled becomes more and more useful. Let’s look at another example of how this might work.

Problem A rectangular container with a square base, an open top, and a volume of 8788 ft^{3} is to be constructed of sheet metal. Find the dimensions of the tank that has the minimum surface area.

Batch size problems are easily solved if the total cost function is given to you. Since this is a minimization problem at its heart, taking the derivative to find the critical point and then applying the first of second derivative test does the trick. However, if we need to find the total cost function the problem is more involved. In Section 12.4 I pointed out a strategy in which we can apply our pattern recognition skills and a table to come up with the total cost function. Let’s look at an example:

Problem Every year, Danielle Santos sells 59,520 cases of her Delicious Cookie Mix. It costs her 2 dollars per year in electricity to store a case, plus she must pay annual warehouse fees of 4 dollars per case for the maximum number of cases she will store. If it costs her 747 dollars to set up a production run, plus 6 dollars per case to manufacture a single case, how many production runs should she have each year to minimize her total costs?

Solution Start the table by labeling the columns:

I have filled in the table with some examples of number of production runs and size of production run to establish the pattern in the last row. Now let’s fill in the first row. Examine each of the entries in the first row carefully.

You can click on the table to view a larger image.

This is mostly the same as in the text. However, the warehouse fees are new. Since the fees are 4 dollars times the maximum stores in the warehouse at any time, we get 59520 ∙ 4.

Now let’s fill in the second row:

I hope you see the pattern emerging. Let’s fill in the last row:

The total cost function in ? is the sum of all of the individual costs or

After simplifying this function, you can find its minimum by taking the derivative and setting equal it equal to zero. This will give you the critical number that you can then verify using the second derivative.

Points of Inflection are locations on a graph where the concavity changes. In the case of the graph above, we can see that the graph is concave down to the left of the inflection point and concave down to the right of the infection point. We can use the second derivative to find such points as in the MathFAQ below.

What is the significance of this point? On both sides of the inflection point, the graph is increasing. This means that as the number of connections increased, so did the revenue from those connections. However, on the left side of the inflection point, the increases in revenue due to increasing connections is getting smaller and smaller. On the right side of the point of inflection, increasing the connections results in larger and larger increases in revenue.