Marginal Analysis. Lecture 11.

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Christian Rodgers
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1 Marginal Analysis. Lecture 11. In many areas, particularly economics and business, the word marginal is used pretty much as a synonym for derivative. For example, suppose that the revenue function for some business is given by R(q) where q is the daily production. Then the derivative R (q) gives the rate at which the revenue changes with production, and in economics and business this is usually called the marginal revenue with respect to q. So, roughly speaking the marginal revenue at some value q tells approximately how much revenue will change with one unit increase in production. Let s try an example (lifted from Exercise 17 of Section 10.8): Example 1. The demand function for a certain tuna is given by p = 20, 000q 1.5 where q is the number of pounds of tuna that will sell per month at a price of p dollars per pound. Then the revenue function as a function of pounds produced in one month is R(q) = pq = 20, 000q 0.5. So the marginal revenue with respect to pounds of tuna produced is just the derivative R (q) = (0.5)(20000)q 1.5.

2 For example, the revenue realized from a monthly production of 400 pounds and the marginal revenue for that level of production are: R(400) = $1000 and R (400) = $1.25. So the actual revenue for 400 pounds per month is $1000, and the marginal revenue is $1.25 per pound. This means that at 400 pounds per month the revenue is $1000 and to produce 401 pounds would reduce the revenue by approximately $1.25. It s worth noting that at 401 pounds the actual revenue is R(401) = 20, 000(400) 0.5 = , so that extra pound of production cost $ So our estimate was pretty close. A second worthy observation is that the average revenue at a production of 400 pounds is, of course, the actual revenue divided by the total production, namely so at 400 pounds Average revenue at production q = R(q) q. Average revenue at 400 pounds = R(400) = = 2.50, which at least in this example is much more than the marginal revenue. 2

3 Example 2. [See Exercise 11 of Section 10.8.] A college paper sells for $0.90 per copy and its production costs per edition are: C(x) = x x 2 dollars per copy. (a) Find the revenue, profit, marginal revenue, marginal cost, marginal profit, and average cost functions R(x) = P (x) = R (x) = C (x) = P (x) = C(x) = (b) Evaluate each of these when the edition is 500 copies, and interpret what these mean for the paper. (c) At what production level would it make sense to stop producing the paper? Why? 3

4 Example 3. [See Exercise 23 of Section 10.8.] A network charges $1.6 million for a 30 second ad on a popular show, but offers a discount of $10, 000 x for x spots on the show. Assuming a company has $ 0.5 million in fixed costs, (a) Find the cost function, in terms of the number x of spots the company buys, the marginal cost function, and the average cost function: C(x) = C (x) = C(x) = (b) Find the marginal and average costs of the company purchases x = 3 spots. Discuss. 4

5 Example 4. [See Exercise 25 of Section 10.8.] For a firm the cost, in dollars, of reducing certain pollution emissions is given by C(q) = q 2 where q is the number of pounds reduced from some benchmark amount each day. In addition, a clean-air subsidy pays S(q) = 500q dollars for a reduction of q pounds per day. (a) What is the marginal cost of reducing emissions if the firm is currently reducing them by 10 pounds per day? (b) At what reduction level does the marginal cost surpass the marginal subsidy? What does that mean in practical terms? (c) What value of q gives the lowest net cost? How does that compare with your answer to part (a), and what can you read in your analysis? 5

6 Product and Quotient Rules. We warned earlier that we can not calculate the derivative of a product as the product of the derivatives. It is easy to see that this is so. Indeed the simple function f(x) = x 2 is the product of the two (equal) functions g(x) = x and h(x) = x. But the product of the derivatives of g(x) and h(x) is g (x)h (x) = 1 1 = 1, not even close to the derivative (= slope) of the parabola f(x) = x 2. So we need some other way to compute derivatives of products and quotients. Fortunately, this is fairly easy. The formal rules are just The Product Rule. d dx (fg) = ( d d f)g + f( dx dx g), or (fg) = f g + fg. The Quotient Rule. or ( ) d f dx g = ( d d f)g f( dx dx g), g 2 ( ) f = f g fg. g g 2 Justification for these two rules is not particularly difficult, but it is sufficiently time consuming that we shall omit it from class discussion. However, we do suggest that you check out the justification in the 6

7 text; understanding why the rule works is a major step in mastering the rule! Example 1. Consider the two functions f(x) = x 2 and g(x) = x 3. Then their product is the function f(x)g(x) = x 5, with derivative d dx (fg) = d dx x5 = 5x 4. Calculating this derivative using the Product Rule gives us d dx (fg) = (f (x))g(x) + f(x)(g (x)) = (2x)x 3 + (x 2 )(3x 2 ) = 5x 4, in complete agreement with the previous calculation! Example 2. Let s try a more serious example. Differentiate f(x) = (2x 3 x 2 1)(2 + 4x 5 ). Of course, one thing we could do is to multiply it out and then differentiate using our earlier rules. But this is the product of two functions, f(x) = g(x)h(x) where g(x) = 2x 3 x 2 1 and h(x) = 2 + 4x 5, so let s use the product rule: f (x) = = g (x)h(x) + g(x)h (x) = (2x 3 x 2 1) (2 + 4x 5 ) + (2x 3 x 2 1)(2 + 4x 5 ) = (6x 2 2x)(2 + 4x 5 ) + (2x 3 x 2 1)(20x 4 ) = 64x 7 28x 6 20x x 2 4x 7

8 Example 3. Find the slope of the line tangent to y = f(x) at x = 3 where f(x) = 5x + 1 x 2 1. This calls for the Quotient Rule. So with we have g(x) = 5x + 1 and h(x) = x 2 1 and f(x) = g(x) h(x), f (x) = h(x)g (x) h (x)g(x) (g(x)) 2 = (x2 1)(5) (2x)(5x + 1) (x 2 1) 2 = 5x2 + 2x + 5 (x 2 1) 2 Then f (3) = 7/8 and f(3) = 2, so the line tangent to y = f(x) at (3, 2) is y = 7 8 (x 3) + 2 = 7 8 x

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How To Understand The Theory Of Algebraic Functions Homework 4 3.4,. Show that x x cos x x holds for x 0. Solution: Since cos x, multiply all three parts by x > 0, we get: x x cos x x, and since x 0 x x 0 ( x ) = 0, then by Sandwich theorem, we get: x 0