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4 - 9 R ates of Change in Economics

Rates of Change in Economics
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and the Sciences
and the Sciences

It has often been said that mathematics is the language of nature. Today, the concepts of
P1: OSO/OVY P2: OSO/OVY QC: OSO/OVY calculus are being applied in virtually every field of human endeavor. The applications
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in this section represent but a small sampling of some elementary uses of the derivative.
Recall that the derivative of a function gives the instantaneous rate of change of
that function. So, when you see the word rate, you should be thinking derivative. You
can hardly pick up a newspaper without finding reference to some rates (e.g., inflation
rate, interest rate, etc.). These can be thought of as derivatives. There are also many
familiar quantities that you might not recognize as rates of change. Our first example,
286 CHAPTER 4 • • Applications of Differentiation 4-76
which comes from economics, is of this type.
In economics, the term marginal is used to indicate a rate. Thus, marginal cost
In economics, the term marginal is used to indicate a rate. Thus, marginal cost is is
the derivative of the cost function, marginal profit is the derivative of the profit function
the derivative of the cost function, marginal profit is the derivative of the profit function
and so on.
and so on.
Suppose that you are manufacturing an item, where your start-up costs are AED
Suppose that you are manufacturing an item, where your start-up costs are AED4000
4000 and production costs are AED 2 per item. The total cost of producing x items
and production costs are AED2 per item. The total cost of producing x items would then be
would then be 4000 + 2x. Of course, the assumption that the cost per item is constant
4000 + 2x. Of course, the assumption that the cost per item is constant is unrealistic. Ef-
ficient mass-production techniques could reduce the cost per item, but machine main-
is unrealistic. Efficient mass-production techniques could reduce the cost per item, but
tenance, labor, plant expansion and other factors could drive costs up as production (x) as
machine maintenance, labor, plant expansion and other factors could drive costs up
increases. In example 9.1, a quadratic cost function is used to take into account some
production (x) increases. In example 9.1, a quadratic cost function is used to take into
account some of these
of these extra factors. extra factors.
Whenthecostperitemisnotconstant,animportantquestionformanagerstoanswer
is how much it will cost to increase production. This is the idea behind marginal cost.
EXAMPLE 9.1 Analyzing the Marginal Cost of Producing
a Commercial Product
Suppose that
2
C(x) = 0.02x + 2x + 4000
is the total cost (in dollars) for a company to produce x units of a certain product.
Compute the marginal cost at x = 100 and compare this to the actual cost of
producing the 100th unit.

Solution The marginal cost function is the derivative of the cost function:
′
C (x) = 0.04x + 2
′
and so, the marginal cost at x = 100 is C (100) = 4 + 2 = 6 dollars per unit. On the
other hand, the actual cost of producing item number 100 would be C(100) − C(99).
(Why?) We have
C(100) − C(99) = 200 + 200 + 4000 − (196.02 + 198 + 4000)
= 4400 − 4394.02 = 5.98 dollars.
Note that this is very close to the marginal cost of AED 6. Also notice that the
marginal cost is easier to compute.
Another quantity that businesses use to analyze production is average cost. You
can easily remember the formula for average cost by thinking of an example. If it costs a
( )
total of $120 to produce 12 items, then the average cost would be $10 $ 120 per item.
12
In general, the total cost is given by C(x) and the number of items by x, so average cost is Copyright © McGraw-Hill Education
defined by
C(x)
C(x) = .
x

306 Business managers want to know the level of production that minimizes average cost.
306 | Lesson 4-9 | Rates of Change in Economics and the Sciences
EXAMPLE 9.2 Minimizing the Average Cost of Producing
a Commercial Product
Suppose that
2
C(x) = 0.02x + 2x + 4000

is the total cost (in dollars) for a company to produce x units of a certain product.
Find the production level x that minimizes the average cost.
Solution The average cost function is given by

2
0.02x + 2x + 4000
−1
C(x) = = 0.02x + 2 + 4000x .
x