11
Techniques of Differentiation
with Applications
Copyright © Cengage Learning. All rights reserved.
11.2
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.
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A First Application: Marginal Analysis
Marginal Cost
A cost function specifies the total cost C as a function of
the number of items x. In other words, C(x) is the total cost
of x items. The marginal cost function is the derivative
C(x) of the cost function C(x). It 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.
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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).
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Example 1 – Modeling Cost
Suppose that the cost in dollars to manufacture portable
CD players is given by
C(x) = 150,000 + 20x – 0.0001x2
where x is the number of CD players manufactured. Find
the marginal cost function C(x) and use it to estimate the
cost of manufacturing the 50,001st CD player.
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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
(CD players). Thus, C(x) is measured in dollars per CD
player.
The cost of the 50,001st CD player is the amount by which
the total cost would rise if we increased production from
50,000 CD players to 50,001.
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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 CD player.
In other words, we estimate that the 50,001st CD player will
cost approximately $10.
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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(x) and P(x) 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 x.
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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.
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Average Cost
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Example 4 – Average Cost
Suppose the cost in dollars to manufacture portable CD
players is given by
C(x) = 150,000 + 20x – 0.0001x2
where x is the number of CD players manufactured.
a. Find the average cost per CD player if 50,000 CD
players are manufactured.
b. Find a formula for the average cost per CD player if
x CD players are manufactured. This function of x is
called the average cost function,
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Example 4(a) – Solution
The total cost of manufacturing 50,000 CD players is given
by
C(50,000) = 150,000 + 20(50,000) – 0.0001(50,000)2
= $900,000.
Because 50,000 CD players cost a total of $900,000 to
manufacture, the average cost of manufacturing one CD
player is this total cost divided by 50,000:
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Example 4(a) – Solution
cont’d
= $18.00 per CD player.
Thus, if 50,000 CD players are manufactured, each CD
player costs the manufacturer an average of $18.00 to
manufacture.
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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 CD players:
Average cost function
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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.
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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.
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