4 Techniques of Differentiation with Applications Copyright © Cengage Learning. All rights reserved.

Copyright © Cengage Learning. All rights reserved. 4.2 A First Application: Marginal Analysis Copyright © Cengage Learning. All rights reserved.

A First Application: Marginal Analysis We have considered linear cost functions of the form C(x) = mx + b, where C is the total cost, x is the number of items, and m and b are constants. The slope m is the marginal cost. It measures the cost of one more item. Notice that the derivative of C(x) = mx + b is C(x) = m. In other words, for a linear cost function, the marginal cost is the derivative of the cost function.

A First Application: Marginal Analysis Marginal Cost A cost function C specifies the total cost as a function of the number of items x, so that C(x) is the total cost of x items. The marginal cost function is the derivative C of the cost function C. Thus, C(x) measures the rate of change of cost with respect to x. Units The units of marginal cost are units of cost (dollars, say) per item.

A First Application: Marginal Analysis Interpretation We interpret C(x) as the approximate cost of one more item. Quick Example If C(x) = 400x + 1,000 dollars, then the marginal cost function is C(x) = $400 per item (a constant).

Example 1 – Modeling Cost Suppose that the cost in dollars to manufacture portable music players is given by C(x) = 150,000 + 20x – 0.0001x2 where x is the number of music players manufactured. Find the marginal cost function C and use it to estimate the cost of manufacturing the 50,001st music player.

Example 1 – Solution Since C(x) = 150,000 + 20x – 0.0001x2 the marginal cost function is C(x) = 20 – 0.0002x. The units of C(x) are units of C (dollars) per unit of x (music players). Thus, C(x) is measured in dollars per music player. The cost of the 50,001st music player is the amount by which the total cost would rise if we increased production from 50,000 music players to 50,001.

Example 1 – Solution cont’d Thus, we need to know the rate at which the total cost rises as we increase production. This rate of change is measured by the derivative, or marginal cost, which we just computed. At x = 50,000, we get C(50,000) = 20 – 0.0002(50,000) = $10 per music player. In other words, we estimate that the 50,001st music player will cost approximately $10.

A First Application: Marginal Analysis Marginal Revenue and Profit A revenue or profit function specifies the total revenue R or profit P as a function of the number of items x. The derivatives, R and P, of these functions are called the marginal revenue and marginal profit functions. They measure the rate of change of revenue and profit with respect to the number of items.

A First Application: Marginal Analysis Units The units of marginal revenue and profit are the same as those of marginal cost: dollars (or euros, pesos, etc.) per item. Interpretation We interpret R(x) and P (x) as the approximate revenue and profit from the sale of one more item.

Average Cost

Example 4 – Average Cost Suppose the cost in dollars to manufacture portable music players is given by C(x) = 150,000 + 20x – 0.0001x2 where x is the number of music players manufactured. a. Find the average cost per music player if 50,000 music players are manufactured. b. Find a formula for the average cost per music player if x music players are manufactured. This function of x is called the average cost function,

Example 4(a) – Solution The total cost of manufacturing 50,000 music players is given by C(50,000) = 150,000 + 20(50,000) – 0.0001(50,000)2 = $900,000. Because 50,000 music players cost a total of $900,000 to manufacture, the average cost of manufacturing one music player is this total cost divided by 50,000:

Example 4(a) – Solution = $18.00 per music player. cont’d = $18.00 per music player. Thus, if 50,000 music players are manufactured, each music player costs the manufacturer an average of $18.00 to manufacture.

Example 4(b) – Solution cont’d If we replace 50,000 by x, we get the general formula for the average cost of manufacturing x music players: Average cost function

Average Cost Average Cost Given a cost function C, the average cost of the first x items is given by The average cost is distinct from the marginal cost C(x), which tells us the approximate cost of the next item.

Average Cost Quick Example For the cost function C(x) = 20x + 100 dollars Marginal Cost = C(x) = $20 per additional item. Average Cost = = $(20 + 100/x) per item.