I posted ave_rate_verizon_wksht to help you work on calculating average rates of change. Let’s look at what students did in class on the problem in this worksheet.

Here are several examples where the average rate of change is calculated from some type of economics formula.

Problem 1 The demand for a particular product is given by

where p is the unit price in dollars.

a. Find the average rate of change of demand with respect to price between a price of 5 dollars and 7 dollars.

b. Find the instantaneous rate of change of demand at a price of 5 dollars.

(Sorry for the shaky cam…too much caffeine!)

The average rate tells us that for each increase in price of 1 dollars between 5 dollars and 7 dollars, the demand for the product drops by 26 items. The instantaneous rate of change tells us that at a price of 5 dollars, the demand is dropping by 22 items per dollar.

Problem 2 The demand for a particular product is given by

where p is the unit price in dollars.

a. Find the average rate of change of demand with respect to price between a price of 5 dollars and 7 dollars.

b. Find the instantaneous rate of change of demand at a price of 5 dollars.

Problem 3 The profit (in thousands of dollars) for selling x hundred units of compressors is

a. Find the average rate of change of profit with respect to compressors from x = 10 to x = 11.

b. Find the exact profit from the 1001^{st} compressor.

c. Find the instantaneous rate of change of profit with respect to compressors at x = 10.

Problem 4 The profit (in thousands of dollars) for selling x hundred units of graphics displays is

a. Find the average rate of change of profit with respect to displays from x = 10 to x = 11.

b. Find the exact profit from the 1001^{st} display.

c. Find the instantaneous rate of change of profit with respect to displays at x = 10.

In these last two problems, pay careful attention to the units on the rates. They are all in thousands of dollars per hundred units. This simplifies to tens of dollars per unit since one thousand divided by one hundred is ten.

Also note that to find the profit from the 1001st item, we need to find the profit at a production level of 1001 and subtract the profit at a production level of 1000. This quantity is called the marginal profit at a production level of 1000. As noted in the text, it is approximately equal to the instantaneous rate of change at a production level of 1000.

A few of you had trouble with the problems referring to slopes of tangent lines. Remember, the slope of a tangent line to a function is given by the derivative. So by setting the derivative equal to the desired slope value, we can solve for the x value.

Problem 1 For the function .

Find where the tangent line is horizontal.

Find where the tangent line’s slope is equal to -1.

Problem 2 For the function .

Find where the tangent line is horizontal.

Find where the tangent line’s slope is equal to 1.

Problem 3 For the function .

Find where the tangent line is horizontal.

In this last example, you need to factor the quadratic to find the two places where the tangent line is horizontal (has a slope of zero). In some instances, you might need to use the quadratic formula to solve the equation.

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.

In Monday’s class, student found many antiderivatives using the Substitution Method. The basic process is illustrated below for the antiderivative

.

Let’s look at the basic steps.

Choose the expression for u. This is generally the inside part of a composition in the integrand. Use the derivative to find an other expression for du.

Match the integrand with u and du. All variables in the original integrand must change to u.

Change the integrand so that it is written in terms of u.

Work out the antiderivative in terms of u.

Put in the expression for u so that the antiderivative is written in terms of the original variable.

Now let’s look at the examples carried out in class.