Math 141 Lecture Notes for Section 1.3
Section 1.3 -
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Linear Functions and Mathematical Models
Function: A function f is a rule that assigns to each value of x one and only one value f (x) (used as
y for purposes of graphing). The value x is called the independent variable, and the value of f (x) is
called the dependent variable. The set of all possible values of x is called the domain, and the set of
all possible values of f (x) is called the range.
Linear Function: The function f defined by
f (x) = mx + b
(1)
where m and b are both constants is called a linear function.
Example 1.3.1 (Creating a Linear Model):
Suppose we wish to use a linear function to model the depreciation of an object. For example, suppose
you want to buy a new car with a sticker price of $27, 500. After 10 years, the value of the car will
drop to $2, 750. Assuming that the depreciation of the car is linear, let’s create a model to tell us what
the value of the car will be over time.
Step 1: Identifying Variables - First we need to figure out what our independent and dependent
variables are going to be in this particular model. In this case, since the variable we are going
to control is time, we’ll use t as our independent variable. Consequently, the value will be our
dependent variable, which we will note V (t).
Step 2: Start with a Model Equation - Since we are assuming that the car’s depreciation is linear,
we’re going to use a linear equation. From the definition we know that we will have the equation:
V (t) = mt + b.
This means that we need to find the constants m and b to finish our equation.
Step 3: Find Constants - From Step 2, we know that we are looking for the slope of the line and
the y-intercept. From the given information, we have two points (0, 27, 500) and (10, 2750)
corresponding to the value at purchase and the value after 10 years, respectively. First this allows
us to compute b since
27, 500 = V (0) = m(0) + b = b.
Second, we are able to compute m by using the definition of slope
m=
2750 − 27500
−24750
=
= −2475.
10 − 0
10
Step 4: Substitute Values for Constants - Now we have our values from Step 3, and we put them
into our equation from Step 2. Now we have the function of the cars value over time as:
V (t) = −2475t + 27500.
It is important to note here that the unit of time is in years, and the units of money is in the dollar.
Make sure you pay attention as sometimes you will be given a value in one unit and expected to
convert it into another.
Math 141 Lecture Notes for Section 1.3
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Cost, Revenue, and Profit Functions: Another common example of linear models is used in Cost-Profit
Analysis. These functions take on the following forms:
Cost: The linear cost function is
C(x) = cx + d.
where d and c are both constants. Here d represents the fixed cost of production and c represents
the cost of producing a single unit.
Revenue: The linear revenue function is
R(x) = sx.
where s is a constant representing the amount of money received from a sale of a single unit.
Profit: The linear profit function is
P (x) = R(x) − C(x) = (s − c)x − d.
where R(x) is the revenue function and C(x) is the cost function.
Example 1.3.2:
Suppose that you are operating a factory making computer memory chips. Your fixed cost for operating
the factory is $250, 000. The per unit cost for a memory chip is $50. Suppose that in a month where
you sell 10, 000 memory chips, your profit is $1, 250, 000. Let’s compute the cost, profit, and revenue
functions in that month.
Given the information we have, we start with a fixed cost of 250, 000 so we have our d value. We also
have our c value, the unit cost at 50. Therefore we can easily compute the cost function as
C(x) = 50x + 250, 000.
The only variable we don’t know is the value s, the unit revenue. From the definition of the profit
function, we have
P (x) = (s − 50)x − 250, 000.
and we know that P (10, 000) = 1, 250, 000. This allows us to set up the equation 1, 250, 000 =
P (10, 000) = (s − 50)(10, 000) − 250, 000. Then we can solve for s as follows
1, 250, 000 = 10, 000s − 500, 000 − 250, 000
1, 250, 000 = 10, 000s − 750, 000
2, 000, 000 = 10, 000s
200 = s.
Now we can express both the profit and revenue functions properly as
R(x) = sx = 200x
and
P (x) = (200 − 50)x − 250, 000 = 150x − 250, 000.
Math 141 Lecture Notes for Section 1.3
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Example 1.3.3:
Demand is another function we commonly use linear models for. Suppse that when a quantity of 3, 000
video game systems is demanded, the unit price is $250. Furthermore, suppose that when a quantity
of 1, 500 video game systems is demanded, the unit price is $400.
The choice of independent and dependent variables is tricky here. One would think that the independent
variable is price and the dependent variable is demand. Unfortunately, this is not the case, the actual
situation is the opposite. In the case of demand functions, the dependent variable is the price, and the
independent variable is the quantity demanded. Therefore we treat the variable x as the number of
units demanded, and the price p(x) as the dependent variable. Since the demand model is linear we
have the equation
p(x) = mx + b.
Where m and b are both constants. Now we compute the slope from the two points given
m=
150
1
400 − 250
=
=−
1500 − 3000
−1500
10
Using the point-slope equation of a line, we have
1
(x − 1500)
10
1
p(x) − 400 = − x + 150
10
1
p(x) = − x + 550
10
p(x) − 400 = −
And now we have a function relating price and demand.
Example 1.3.4:
Since we have used demand as a linear function, now we can move forward to supply. Again we have
a linear function of the form
p = mx + b.
With x being the number of units supplied, m and b are both constants, and p(x) is the unit price
dependent on the number of units supplied. Suppose we’re given a supply equation for video game
consoles in the form
6p − 4x − 504 = 0
and we want to rewrite this as a standard linear function. First we need to find the p-intercept, which
we can do by setting x to 0.
6p − 504 = 0
6p = 504
p = 504/6 = 84
Now we have that b = 84. We need to find the slope m for the linear equation. By the notes for
Section 1.2 we know the slope for this equation is − 64 . Thus we can write the suppy equation as
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p(x) = − x + 84.
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Math 141 Lecture Notes for Section 1.3
Suggested Homework Problems: 11, 13, 15, 21, 23, 35, 39, 47.
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