1.2 Applications of Linear Functions
Question 1: What are the pieces of a linear function?
Question 2: How do you graph a linear function?
Question 3: How do you find a linear function through two points?
Question 4: How do you find cost and revenue functions?
Question 5: What are demand and supply?
Linear functions are prevalent throughout business and economics. They provide a
simple way to model economic quantities. Even if a linear function is not the best type of
function to model a quantity, often they are used as an initial model to help understand
a situation and all of its complexities. In this section we’ll look at a number of different
situations in which a linear model might be appropriate. In later chapters we’ll revisit
these applications and modify them as needed to account for properties which linear
models don’t provide.
Linear functions are composed of two pieces separated by addition or subtraction. Each
of these pieces is called a term. We’ll start this section by examining how we can
determine what these terms tell us about the function. Once we understand how to find
the terms, we’ll look at some specific examples of linear functions in business. The first
linear function we’ll examine is a linear cost function. In section 1.1 we looked at some
examples of linear cost functions to help us understand variables and function notation.
In this section we’ll look at the individual pieces of a linear cost function to see what they
represent in business. Then we’ll conclude this section with several more examples of
linear functions. Each of these examples will help to model milk production at a dairy.
1
Question 1:
What are the pieces of a linear function?
In section 1.1, we defined a linear function in the variables x and y,
y  mx  b When the function is written this way, we say “y is a linear function of x”. This means the
variable x is the independent variable and the variable y is the dependent variable. The
other letters in this equation, m and b, correspond to numbers. For instance, in the
equation y  2 x  3 the value of the constant m is 3 and the value of b is 3. We
distinguish between variables and constants since in any problem the constants will be
given (or can easily be calculated) and the variables will vary as needed.
In every linear function of one independent variable, the right side of the function
contains two terms. The first term, 2 x , contains the constant m  2 and the independent
variable x. Since this term contains the variable, it is often called the variable term. As
different values of x are put into the function, the product of the constant and the
variable will change.
The second term 3 contains the constant b. Since this term never changes as values
are put into the function, it is often called the fixed term.
In the case of y  2 x  3 , we can try values of x in the function and see how y depends
on these values. Suppose we pick several values of x like x  3, 2, , 3. If we think of
this equation as a function f ( x)  2 x  3 , we can calculate the corresponding values of y:
f (3)  2  3  3  3
f (2)  2  2   3  1

f (3)  2  3  3  9
Let’s examine this behavior of the variable term and the constant term in a table.
2
Table 1
x
-3
-2
-1
0
1
2
3
variable term
-6
-4
-2
0
2
4
6
constant term
3
3
3
3
3
3
3
y  f ( x)
-3
-1
1
3
5
7
9
For each x value, the constant term does not change. However, as the x value
increases by 1 unit, the variable term increases by 2 units. This increase is to the value
of m, 2, multiplying the variable x. The y value in the last row is the sum of the variable
and constant terms. Since the variable term increases by 2 units when the x value
increases by 1 unit, so does the y value.
3
Question 2: How do you graph a linear function?
Consider the x value and the y value on an equation y  2 x  3 as an ordered pair,  x, y  .
The values in can be graphed as shown below.
y
x
Figure 1 – The ordered pairs from Table 1and the line
passing through the ordered pairs.
Each of the ordered pairs we just found lie along a straight line. This line is related to
the constants m and b in the linear function.
Suppose f ( x)  mx  b is a linear function in the independent
variable x. The linear function intercepts the vertical axis at the
ordered pair (0, b) .
For the linear function f ( x)  2 x  3 , the value of the constant b is b  3 . This indicates
that the function crosses the y axis at the ordered pair  0,3 .
4
y
x
Figure 2 – A closer look at y = 2x + 3 showing the y intercept at
(0, 3) and the changes in the variables at ordered pairs along
the line.
The constant m in a linear function is related to how the variables change with respect
to each other as you move from point to point along the line. In Figure 2, three points
are shown on the line. The horizontal change between each ordered pair is 1 unit. The
vertical change between each ordered pair is 2 units. The ratio of these changes
describes how the variables change with respect to each other and is equal to
Vertical Change
2 units

Horizontal Change 1 unit
This ratio is called the rate of change of the linear function.
Suppose f ( x)  mx  b is a linear function in the independent
variable x. The rate of change of the linear function is the constant m.
When viewed on a graph, this rate is called the slope of the line. Knowing the slope and
the vertical intercept of the line corresponding to a linear function allows us to graph the
linear function quickly.
5
Example 1
Graph a Linear Function
For each part below, graph the line corresponding to the linear function.
a.
f ( x)  3x  1
Solution To graph this function, we’ll identify the values of the constants
and interpret them to find points on the line. To match the form
f ( x)  mx  b , rewrite the function as f ( x)  3x   1 . This tells us that
b  1 and m  3 . Geometrically we know that the vertical intercept is at
 0, 1
and the slope is 3. Place a point on the graph at the vertically
intercept and then draw a second with respect to that point using
Vertical Change
3
3
 or
1
Horizontal Change 1
Each of these ratios is equal to 3. We could also use any other ratio
equal to 3, but these are the simplest to visualize.
y
x
Figure 3 – From the intercept at (0, -1), move either left or right by 1 unit and up or
down by 3 units to generate a second point on the line. Once two points are graphed,
the graph of the linear function can be drawn through the two points.
6
b.
1
g ( x)   x  4
2
Solution For this function, b  4 and m   12 . This means the vertical
intercept is located at  0, 4  and the ratio of vertical change to horizontal
change is
1
2
or
1
2
. Either ratio gives the same slope. The line passing
through the corresponding points is graphed in Figure 4.
Figure 4 – By moving left or right 2 units and then up or down 1 unit gives two more
points on the line.
c.
h( x)  2
7
Solution Write the function in the form h( x)  0 x   2  . For this
function, b  2 and m  0 . Since the slope is zero, the line must have
no vertical change and is a horizontal line.
Figure 5 – A constant linear function is a horizontal line through the vertical intercept.
The different parts of Example 1 illustrate the role that the constant m plays on the
graph of a linear function. A slope that is positive leads to a graph that rises as you
move from left to right. A graph with this type of behavior is said to be increasing. A
slope that is negative leads to a graph that falls as move from the left to right. A graph
with this type of behavior is said to be decreasing. A graphs with zero slope stay at the
same level as you move from left to right on the graph. A graph with this type of
behavior is said to be constant.
Figure 6 – Examples of (a) an increasing linear function, (b) a decreasing linear function, and
(c) a constant linear function.
8
Question 3: How do you find a linear function through two points?
The examples above demonstrate how to create points with a given slope between
them. This is useful if we are given the function to begin with. However, we may also be
given at least two points on the graph of the function and want to find the function’s
formula. In this case we need to find the slope of the line passing through the points.
Suppose you are given two points,  x0 , y0  and  x1 , y1  , on a line. The
slope of the line is found by calculating
m
y y
Vertical Change
 1 0.
Horizontal Change x1  x0
When you use this formula be careful to subtract the y values in the numerator (the
vertical change) and the x values in the denominator (the horizontal change). Also that
you subtract in the proper order.
Example 2
Find the Slope Between Two Points
For each part, find the slope of the line passing through the two points.
a.
 3,1
and  5,8 
Solution Let  x0 , y0    3,1 and  x1 , y1    5,8  . Substitute these ordered
pairs into the slope formula to yield
m
b.
 3, 2 
8 1 7
 53 2
and  4, 1
Solution Let  x0 , y0    3, 2  and  x1 , y1    4, 1 . Substitute these
ordered pairs into the slope formula to yield
9
m
1  2
3
 4   3
7
Once we have the slope between a pair of points, we can use the linear function form
f ( x)  mx  b to find the formula for the line.
Example 3
Find the Function Passing Through Two Points
Find the function for the line shown in the graph below.
Solution To find the formula of the function f ( x)  mx  b , we‘ll need to
first find the slope between  x0 , y0    2,3 and  x1 , y1    5, 2  . The slope
is
m
23
1
 52
3
With this value for the slope, the linear function becomes f ( x)   13 x  b .
To find b, we must substitute one of the given points on the line into the
function. If we input x  2 into the function, the output should equal
y  3:
f (2)   13  2   b  3 10
We can solve for b by solving  13  2   b  3 . The first term simplifies to
 23 . Adding
2
3
to both sides leads to b  113 . The function passing
through the two points is
f ( x)   13 x  113 To check this function, substitute each point into the function to insure
each x value yields the correct y value:
f (2)   13  2   113  3
TRUE
f (5)   13  5   113  2
TRUE
11
Question 4: How do you find cost and revenue functions?
All businesses produce some kind of product. That product may be something that you
can hold in your hand like an MP3 player or it may be a service like cleaning an office.
The outputs from the business take the form of goods and services. To produce these
outputs, the business requires inputs such as workers, machines and supplies. A
business takes the inputs and converts them to outputs.
The process of converting inputs to outputs requires technology. This is more than our
everyday meaning where we think of technology as a cool new product or technique. In
economics, technology is any process a business uses to convert inputs into outputs. It
depends on many factors. For instance, in a dairy the factors might include the capacity
of the milking machinery, the skill of the laborers who operate the machinery and the
quality of the feed the cows are fed.
Technological change may occur if the businesses inputs and outputs change. This can
occur in two basic ways. First, there may be a change in the outputs for the same
number of inputs. If greater output occurs for the same level of inputs, positive
technological change has occurred. If lower output occurs for the same level of input,
negative technological change has taken place. For instance, if a dairy produces more
milk with the same number of cows, positive technological change has taken place.
Technological change can also take place if the outputs are held constant and the
inputs change. Positive technological change occurs if the same output occurs, but from
a lower level of inputs. On the other hand, negative technological change occurs if the
same outputs occur for an increased level of inputs. If a dairy produces the same
amount of milk from fewer cows, positive change has taken place. If it takes more cows
to produce the same level of milk, negative change has taken case.
Businesses are interested in knowing the relationship between the inputs and outputs
with respect to the costs they incur to produce the outputs. Since there can be many
different inputs that affect their costs, at least one of the businesses input are held
constant. This period of time is called the short run. The period of time for which any of
12
the businesses inputs can vary or it can change its technology or capacity is called the
long run.
Let’s consider several inputs for a dairy that could lead to the output of milk. A dairy
could change the number of cows producing milk, the number of breeding cows, the
number of laborers, the capacity of the milking machines, the type of feed or the number
of acres used to graze the cows on. The long run is the period of time over which the
dairy can vary all of these inputs. The short run is the period of time over which at least
one of the inputs is fixed. If the number of cows producing milk is varied over a year, but
all other inputs are fixed, the short run is a year. Depending on what is fixed and what is
left to vary influences the short run.
To examine the total cost of producing outputs, it is often easiest to look at the short run
and to fix as many of the inputs as possible. If only one variable changes, the total cost
function will contain only one variable. A total cost function with two variables means
that two inputs are varied and so on.
Costs typically fall into two categories, variable costs and fixed costs. To determine
which category a cost falls, we need to examine how the cost changes as the outputs
change. If a cost varies as the output changes, the cost is a variable cost. A cost that
does not change as the output changes is a fixed cost. For a dairy, the labor costs,
veterinary costs, feed costs, and utility costs change as the output level of milk changes.
To produce more milk, more people need to feed and milk the cows. More milk also
means that it will cost more to maintain the health of the herd as well as to operate the
milking machines. Costs like real estate taxes, interest on short term loans, and
depreciation on equipment are fixed as milk production changes and would be
considered fixed costs. All costs are either variable or fixed.
The total cost of operating the business is the sum of the variable costs and the fixed
costs,
Total Cost  Fixed Cost  Variable Cost
13
For a cost function in which all inputs are fixed except one, a single variable is used and
the cost function is called C (Q) or Cost  Q  The variable Q is the input that is varied and
needs to be defined so that the function makes sense.
Example 4
Dairy Costs
Suppose we are interested in finding a function that describes the total
annual costs C (Q) of running a dairy as a function of the number of
cows Q producing milk. By saying “annual”, we are assuming that the
short run is a period of one year. Since the function C (Q) has only one
input Q, all others are being held constant in the short run.
The table below describes the fixed, variable and total costs as the
number of dairy cows is increased.
Table 1
Number of
Dairy Cows
Fixed Annual Cost
(Dollars)
Variable Annual Cost
(Dollars)
Total Annual Costs
(Dollars)
0
68,688
0
68,688
10
68,688
28,908
97,596
20
68,688
57,816,
126,504
30
68,688
86,724
155,412
40
68,688
115,632
184,320
50
68,688
144,540
213,228
Notice that in each row of Table 1 the total cost is the sum fixed cost
and the variable cost.
Let’s get a feel for C (Q) by graphing the pertinent columns of this table
in a scatter plot. The variable Q represents the number of dairy cows so
14
this will be the input or independent variable on the scatter plot. The
dependent variable represents the total cost. This means that we’ll
graph the first and fourth columns of the table as ordered pairs,
Number of
Dairy Cows
Total Annual Costs
(Dollars)
0
68,688
10
97,596
20
126,504
30
155,412
40
184,320
50
213,228
The scatter plot of the ordered pairs is shown below.
Figure 7 – Scatter plot of data in Table 1.
The ordered pairs appear to follow a straight line. To check, let’s
calculate the slope between adjacent points on the graph.
15
Table 2
Adjacent Ordered Pairs
Slope
(0, 68688) and (10, 97596)
97596  68688
 2890.8
10  0
(10, 97596) and (20, 126504)
126504  97596
 2890.8
20  10
(20, 126504) and (30, 155412)
155412  126504
 2890.8
30  20
(30, 155412) and (40, 184320)
184320  155412
 2890.8
40  30
(40, 184320) and (50, 213228)
213228  184320
 2890.8
50  40
Adjacent sets of points have the same slope, 2890.8, so all points lie
along a line.
The slope of the line is 2890.8 and the vertical intercept is 68688. Since
the graph corresponds to a linear function, we can write the total cost
function C (Q) as
C (Q)  2890.8Q  68688
where Q is the number of dairy cows. The term 68688 is fixed as Q
changes and corresponds to the fixed costs.
16
Figure 8 – The linear function corresponding to the data in Figure 7.
The slope indicates the rate at which costs are changing. In this case,
we can think of the slope as
Change in Costs
28908 dollars 2890.8 dollars


Change in Dairy Cows
10 cows
1
cows
This tells us that increasing the number of cows by one increases the
total cost by 2890.8 dollars. Thus the variable term 2890.8x corresponds
to the variable cost since the term changes as x changes.
The slope is also called the marginal cost since it describes how the
costs change when the input changes by 1 unit. In general, the term
marginal describes how one quantity changes when the input changes
by 1 unit.
As discussed before, businesses convert inputs to outputs. This process costs
businesses money which is described by the total cost function C (Q) . To compensate
for these costs, businesses receive money for selling goods or services. The amount
received depends on the inputs to the businesses and is called revenue. The revenue
function is denoted by the name R or the name Revenue and is a function of the
17
company’s inputs. Typically revenue is described in the short term where at least one of
the inputs is fixed. If all but one of the inputs are fixed, the revenue can be denoted by a
function of one variable R(Q) or Revenue  Q 
Example 5
Dairy Revenue
A dairy can increase its milk production and thus its revenue in several
ways. By changing the feed or administering certain hormones, the
amount of milk produced can be increased and lead to higher total
revenue. Revenue can also be changed by increasing or decreasing the
number of dairy cows. More cows can produce more milk thus leading
to higher revenue. We are interested in varying the number of cows Q
and seeing how the total annual revenue R(Q) changes. All other inputs
to the business will be held constant in the short run.
The table below describes the level of total annual revenue for various
numbers of dairy cows.
Table 3
Number of
Dairy Cows
Total Annual Revenue
(Dollars)
0
0
10
35,470
20
70,940
30
106,410
40
141,880
50
177,350
Find the revenue function R (Q ) .
18
Solution This is a function of the number of dairy cows. On a graph, we
know that the number of dairy cows will be graphed horizontally and the
total annual cost will be graphed vertically.
Figure 9 – Scatter plot of data in Table 3.
Like the total annual cost, the total annual revenue appears to follow a
straight line. Let’s calculate the slope between adjacent points.
Table 4
Adjacent Ordered Pairs
Slope
(0, 0) and (10, 35470)
35470  0
 3547
10  0
(10, 35470) and (20, 70940)
70940  35470
 3547
20  10
(20, 70940) and (30, 106410)
106410  70940
 3547
30  20
(30, 106410) and (40, 141880)
141880  106410
 3547
40  30
(40, 141880) and (50, 177350)
177350  141880
 3547
50  40
19
The slope between any pair of points on the line is 3547. This indicates
that the ordered pairs all lie along a line that has a slope of 3547. Since
the vertical intercept of this line is the point (0,0), we can use the slopeintercept form of a line to write
R(Q)  3547Q
The graph of this function mirrors the trend of the data in the scatter plot
above.
Figure 10 – The linear function corresponding to the data in Figure 9.
We can think of the slope of the revenue function as a ratio,
Change in Revenue
35470 dollars 3547 dollars


Change in Dairy Cows
10 cows
1 cows .
An increase in the number of dairy cows by 10 leads to an increase in
revenue of $35,470. This means that each additional dairy cow adds
$3547 to the revenue. The amount of revenue added when the input
increases by one unit is called the marginal revenue.
20
Question 5: What are demand and supply?
An important application of linear functions is the connection between the price of a
good or service and the quantity of a good or service. If we are interested in the
relationship between the price and the quantity demanded by consumers at that price,
this connection is described by the demand function. The relationship between price of
a good or service and the quantity supplied by manufacturers at that price is described
by the supply function. These functions describe how consumers and suppliers behave
as price is changed.
The consumer’s behavior is given by a table called a demand schedule. This table
reflects the relationship between the price of some good or service and the quantity of
this good or service demanded at this price. The table below is a typical demand
schedule in some market.
Table 5
Average Price of a Gallon of Milk
(dollars per gallon)
Quantity of Milk Sold Per Week
(thousands of gallons)
1.50
125
2.00
115
2.50
105
3.00
95
3.50
85
In one column of the demand schedule are prices of some good. In the other column is
the corresponding numbers of that good sold to consumers. From this demand
schedule for dairies in a competitive market, we can see that as the price per gallon for
milk increases the quantity of milk sold per week decreases. The law of demand states
that if everything else is held constant, as the price of the good increases, the quantity
demanded drops. This pattern is visible in the demand curve shown below.
21
The demand function is traditionally graphed with the quantity as the independent
variable and the price as the dependent variable. Because of this, quantity is graphed
on the horizontal axis and price is graphed on the vertical axis. This means that the
information indicating that 125,000 gallons of milk is sold per week at a price of $1.50 per
gallon corresponds to the ordered pair (125, 1.50). The demand function takes an input
of quantity and outputs price,
P  D(Q) .
Demand curves can have many shapes, but the simplest curve is a demand curve
follows a straight line. In the case of the demand function for dairies in a market shown
here, we’ll find a linear demand function of the form
D(Q)  mQ  b .
This is the same form defined earlier, f ( x)  mx  b , with the variable changed to Q from
x and the name changed to D from f.
Figure 11 – The demand D(Q) for milk.
Example 6
Find the Demand Function for Milk
Find the equation of the demand curve graphed in Figure 11.
22
Solution The graph above gives us a hint that the demand function is a
linear function. To make sure the ordered pairs all lie along a straight
line, we need to find the slope between adjacent ordered pairs.
Table 6
Adjacent Ordered Pairs
Slope
(125, 1.50) and (115, 2.00)
2.00  1.50
 0.05
115  125
(115, 2.00) and (105, 2.50)
2.50  2.00
 0.05
105  115
(105, 2.50) and (95, 3.00)
3.00  2.50
 0.05
95  105
(95, 3.00) and (85, 3.50)
3.50  3.00
 0.05
85  95
Since the slope between any pair of adjacent points is -0.05, all of the
points lie along a line with slope -0.05.
Insert the slope into the form for a linear demand function to yield
D(Q)  0.05Q  b
All that remains is to find the vertical intercept b in the function. None of
the points are the vertical intercept. To find b, substitute any of the
points into the function. Let’s try the ordered pair (125, 1.50):
D(125)  0.05 125  b  1.50
To find the value of b, simplify the equation to give
6.25  b  1.50
23
Adding 6.25 to both sides leads to b equal to 7.75. The function for
demand is
D(Q)  0.05Q  7.75
The supply function also relates the price of a good or service to a quantity of the good
or service. Unlike the demand function which describes the consumer’s willingness and
ability to buy a good or service, the supply function describes the willingness and ability
of a business to supply a good or service at a given price. If all other factors are held
constant, as the price rises the quantity supplied will also increase. A higher price leads
to more profit for a business encouraging higher production.
A supply schedule is a table that shows the relationship between the price of a good or
service and the quantity supplied by a firm.
Table 7
Average Price of a
Gallon of Milk
(dollars)
Quantity of Milk Supplied
Per Week
(thousands of gallons)
0
0
2.40
76
3.00
95
3.60
114
4.20
133
In this supply schedule, an increase in the average price of a gallon of milk corresponds
to an increase in the quantity of milk supplied by dairies. A graph of the data shows a
straight line with a positive slope.
24
Figure 12 - Data from the supply schedule and the corresponding supply curve.
Keep in mind that while this is a “typical” supply schedule, supply functions may have
many different shapes. A supply function may be curved concave down or up. It may
even change its concavity.
Figure 13 - Three examples of supply functions. In a, the supply curve is concave down. In b,
the supply curve is concave up. In c, the supply curve changes from concave down to
concave up.
A linear function often models the data adequately and simplifies later calculations. A
linear supply function has the form
S (Q)  mQ  b
where m is the slope of the line and b is the vertical intercept.
25
Example 7
Find the Supply Function for Milk
Find the linear supply function S  Q  , where Q is the quantity of milk
supplied, passing through the ordered pairs in Table 7.
Solution Since the function will be a function of Q, interchange the
columns of the table. This change is necessary since the independent
variable is typically listed in the first column.
Quantity of Milk Supplied
Per Week
(thousands of gallons)
Average Price of a
Gallon of Milk
(dollars)
0
0
76
2.40
95
3.00
114
3.60
133
4.20
Let’s check the slope between adjacent points in the Table 7.
Table 8
Adjacent Ordered Pairs
Slope
(0, 0) and (76, 2.40)
2.40  0 2.4

 0.032
76  0
76
(76, 2.40) and (95, 3.00)
3.00  2.40 0.6

 0.032
95  76
19
(95, 3.00) and (114, 3.60)
3.60  3.00 0.6

 0.032
114  95
19
(114, 3.60) and (132, 4.20)
4.20  3.60 0.6

 0.032
132  114
19
26
The fractions
2.4
76
and
0.6
19
simplify to
3
95
. Since all of the slopes are the
same at about 0.032, we know the points lie along a straight line.
Unlike the demand function, we know that the vertical intercept is (0, 0)
so the equation of the line is
S (Q) 
3
Q
95
Notice that the fraction has been used for the slope. If the rounded
decimal had been used instead of the exact fraction, the line would go
very close to the data in the supply schedule, but not through exactly.
Decimals are acceptable when they are exact decimals and not
rounded or where approximations are acceptable.
Example 8
Find the Surplus at a Fixed Price
At a price of $3.36 per gallon, dairies will be willing to supply more milk
than consumers are willing to buy. Find the surplus amount of milk at
this higher price.
Solution To find the amount of the surplus, we must find the difference
between the quantity supplied and the quantity demanded. These
quantities are found by setting the outputs of the supply and demand
functions equal to $3.36.
To find the quantity supplied, set S(Q) equal to 3.36 and solve for the
quantity Q :
3
Q  3.36
95
Q  3.36 
95
 106.4
3
Multiply by reciprocal of 3/95
27
At a price of $3.36 per gallon, the dairies would be willing to supply
106.4 thousand gallons or 106,400 gallons of milk per week.
To find the quantity demanded by consumers, set D(Q) equal to 3.36
and solve for Q:
0.05Q  7.75  3.36
0.05Q  4.39
Q  87.8
Subtract 7.75 from both sides
Divide both sides by -0.05
The demand at a price level of $3.36 per gallon is 87,800 gallons of milk
per week.
The surplus of milk is the difference between the two quantities or
Surplus  106, 400  87,800  18,600 gallons per week
28