1. No category

1.4 Linear Functions of Several Variables
Question 1: What is a linear function of several independent variables?
Question 2: What do the coefficients of the variables tell us?
Question 3: How do you find cost, revenue and profit functions with several variables?
In sections 1.1 through 1.3, we considered functions of one independent variable. The
cost, revenue, and profit, were each a function of some quantity Q. If we knew the
quantity Q produced and sold, we could use these functions to compute the
corresponding cost, revenue and profit at those production levels. As long as we deal
solely with one product and the cost, revenue, and profit involved with that product, a
function of one variable is adequate. But the total cost, revenue, and profit for a
business may come from many products. Each of these products may have its own
costs and prices. Using this information we can formulate cost, revenue and profit
functions of several variables. Each of these variables represents a quantity of a
different product produced and sold by the business.
In this section we’ll learn how to extend what we have learned about a function of a
single independent variable to functions of several independent variables.
1
Question 1: What is a linear function of several independent variables?
In section 1.1, we introduced a linear function of one variable, y  mx  b . There was
nothing special about the names of the variables, x and y, or the names of the
constants, m and b. Another possible form for a linear function of x and y is y  a0  a1 x .
In this format, a0 is the vertical intercept and a1 is the slope.
When several independent variables are introduced, it is prudent to use names for the
variables that make sense. If one variable is named x, we can extend this to n variables
using subscripts. Subscripts are numbers that appear to the right of the variable and
slightly lowered. The subscript is a part of the variable’s name and is useful to show
generically that there are many variables. For instance, if we wanted to define a function
with three independent variables that describe the quantities of three different products,
we might use Q1 , Q2 , and Q3 .
In general, let x1 , x2 , , xn be the names of n independent variables.
A linear function of n independent variables x1 , x2 , xn is any
equation that can be written in the form
y  a0  a1 x1  a2 x2  an xn
In this form, we say that y is a linear function of x1 , x2 , , xn . The
letters a0 , a1 , , an are real numbers corresponding to constants.
Function notation applies to functions of several independent variables as well as
functions of one independent variable. Recall that a linear function of one variable x
named f would be written as f ( x)  a0  a1 x . The independent variable for the function is
placed in parentheses after the name to distinguish the variables from the constants.
For a linear function of n independent variables, the n independent variables are placed
in the parentheses after the name to give
2
Example 1
f  x1 , x2 , , xn   a0  a1 x1  a2 x2    an xn Find Function Values
If f ( x1 , x2 , x3 )  10  2 x1  x2  3 x3 , find the value of f (6, 1, 2) .
Solution Substitute x1  6 , x2  1 and x3  2 into the function to yield
f (6, 1, 2)  10  2  6    1  3  2   3 3
Question 2: What do the coefficients of the variables tell us?
For a linear function of one independent variable, the coefficient on the variable is the
slope or rate of change. We can generalize this idea to linear functions of several
variables. Let’s consider the function g ( x1 , x2 )  2 x1  3 x2 . As we saw in Example 1, we
can substitute values for the variables to obtain a value for the function. For instance, if
we want to substitute 10 for x1 and 2 for x2 we get
g (10, 2)  2 10   3  2   14 Now compare this value to the value obtained when we increase the value for x1 by 1
unit,
g (11, 2)  2 11  3  2   16 The difference between these values is g 11, 2   g 10, 2   16  14  2 . Since the
coefficient on x1 is a 2, increasing the value for x1 increases the value from g by 2 units.
The coefficient on x1 tells us the amount the function will change when x1 increases by
1 unit.
In general, this applies for any value of x1 . If we fix x2 at some value and find the
difference between the function at x1 and at x1  1 we get
g  x1  1, 2   g  x1 , 2    2  x1  1  3  2     2 x1  3  2  
 2 x1  2  6  2 x1  6
2
We can also apply this reasoning to the coefficient of x2 . If we fix x1 at some value and
find the difference between the function at x2 and x2  1 we get
4
g 10, x2  1  g 10, x2    2 10   3  x2  1    2 10   3 x2 
 20  3 x2  3  20  3 x2
 3
The coefficient on x2 , -3, indicates the amount the function will change when x2 grows
by 1 unit.
This leads to the following generalization.
If f  x1 , x2 , , xn   a0  a1 x1  a2 x2    an xn is a linear function
of n independent variables, the coefficient ai for values of i
from 1 to n indicates the amount the function will change
when the variable xi is increased by one unit and all other
variables are fixed.
Example 2
Interpret Coefficients
The function
S ( A, E )  105 A  1412 E
describes the monthly sales,
in thousands of dollars, at a large dairy distributor with E employees and
A thousand dollars of monthly advertising costs.
a. What does the coefficient of E tell you about the monthly sales?
Solution The coefficient of E, 1412, relates the change in sales to the
variable E when it is increased by 1 unit. An increase of 1 unit in E
means an increase of 1 employee. This leads to an increase in sales of
1412 thousand dollars or $1,412,000.
b. What does the coefficient of A tell you about monthly sales?
Solution The coefficient of A, 105, relates the change in sales to the
variable A when it increases by 1 unit. An increase of 1 unit in A means
5
an increase of 1 thousand dollars in advertising. This leads to an
increase in sales of 105 thousand dollars or $105,000.
We can now interpret what coefficients a1 through an mean, but what about the
constant a0 ? The constant a0 is similar to the vertical intercept b for a linear function of
one independent variable f ( x)  mx  b . In that case, b is the value of the function when
the variable is set equal to zero, f  0   m  0   b  b . For a linear function of n
independent variables, the corresponding process is to set all independent variables
equal to zero. If f ( x1 , x2 , , xn )  a0  a1 x1    an xn , we can set each variable equal to
zero to obtain
f (0, 0, , 0)  a0  a1  0   a2  0     an  0   a0 The only term that remains is the term containing the constant a0 .
If f ( x1 , x2 , , xn )  a0  a1 x1    an xn is a linear function of n
independent variables, a0 is the value of the function when
all independent variables are set equal to zero.
Example 3
Fixed Costs for an Organic Dairy
The daily cost function (in dollars) for an organic dairy is given by the
linear function C (Q1 , Q2 , Q3 )  Q1  1.1Q2  1.25Q3  10, 000 where Q1 is the
number of gallons of whole milk produced, Q2 is the number of gallons
of 2% milk produced, and Q3 is the number of gallons of 1% milk
produced.
a. Find the fixed costs for the organic dairy based on this function.
Solution The fixed costs are costs incurred when nothing is produced.
We can find this cost by finding C (0, 0, 0) or by noting that the constant
6
gives the function’s value when all variables are zero. Either way, the
fixed costs are $10,000.
b. How much does each additional gallon of 2% milk cost to produce?
Solution The coefficient of Q2 gives the change in the cost when Q2 is
increased by 1 unit. In this case, the coefficient is 1.25 which tells us
that another gallon of 2% milk will cost an additional $1.25.
We can calculate the amount it would cost to change production in
Example 3 by any amount by adding the amount to the appropriate
independent variable and utilizing the cost function. If we want to find
the additional cost incurred from increasing the production of 1% milk
by 1000 gallons, calculate
C  Q1 , Q2 , Q3  1000   C  Q1 , Q2 , Q3    Q1  1.1Q2  1.25  Q3  1000     Q1  1.1Q2  1.25Q3 
 Q1  1.1Q2  1.25Q3  1250  Q1  1.1Q2  1.25Q3
 1250
The extra 1000 gallons cost $1250. This number is independent of the initial production
level. If we start with no 1% milk and increase production to 1000 gallons or increase
production of 1% milk from 10,000 gallons to 11,000 gallons, the additional cost will be
$1250. For linear functions like this one, the increase is always independent of the
production level. Another way of thinking about this is that the variable costs for 1% milk
are $1.25 per gallon. This is precisely the coefficient on the variable representing 1%
milk, Q3 .
7
Question 3: How do you find cost, revenue and profit functions with several variables?
In Example 3, we examined a cost function of several variables. This proved useful for
modeling the total cost in a business that produces several different goods like milk. If
the company receives revenue from several different products, we can utilize the fact
that the revenue from each individual product by multiplying the price per unit times the
number of units sold to get the revenue for that product. By summing the revenue for
each of the products, we can form a model for the total revenue.
In the next example, we’ll continue working with the organic dairy from Example 3 to
model the total revenue from selling three types of milk.
Example 4
Organic Dairy Revenue
The organic dairy in Example 3 charges $3.49 per gallon for whole milk,
$4.19 per gallon for 2% milk, and $4.59 per gallon for 1% milk. If the
variables Q1 , Q2 , and Q3 represent the number of gallons of whole, 2%
and 1% milk produced and sold respectively, find a linear function
R(Q1 , Q2 , Q3 ) to model the total revenue.
Solution We’ll find a linear function of the variables Q1 , Q2 , and Q3 .
The coefficients in this function correspond to the increase in revenue
from increasing the number of gallons of each type by 1 unit. These
coefficients are simply the prices the dairy charges for one gallon of
each type of milk. The linear function will have the form
R(Q1 , Q2 , Q3 )  p1Q1  p2Q2  p3Q3 where p1 , p2 , and p3 are the price per gallon for each type of milk.
Notice that there is no constant term in the sum since producing and
selling no milk leads to no revenue. If we had included a positive
constant, this would have been interpreted as revenue corresponding to
8
 Q1 , Q2 , Q3    0, 0, 0  . With the prices in place, we get a linear model for
total revenue in dollars,
R  Q1 , Q2 , Q3   3.49Q1  4.19Q2  4.59Q3 It is very common for cost and revenue functions to have many variables. Each variable
corresponds to a unique product or service that a business provides to consumers. The
coefficients of the cost function represent the variable costs per unit for each product or
service and the constant represents the fixed costs of production. The coefficients of the
revenue function represent the unit price for each product or service.
Example 5
Profit
The cost function for the organic dairy is
C (Q1 , Q2 , Q3 )  Q1  1.1Q2  1.25Q3  10, 000 and the revenue function is
R  Q1 , Q2 , Q3   3.49Q1  4.19Q2  4.59Q3 where Q1 is the number of gallons of whole milk, Q2 is the number of
gallons of 2% milk, Q3 is the number of gallons of 1% milk that the dairy
produces and sells.
a. Find the profit function Profit(Q1 , Q2 , Q3 ) .
Solution Profit is the difference between revenue and costs,
Profit(Q1 , Q2 , Q3 )   3.49Q1  4.19Q2  4.59Q3    Q1  1.1Q2  1.25Q3  10, 000 


 
R( Q1 ,Q2 ,Q3 )
C ( Q1 ,Q2 ,Q3 )
 2.49Q1  3.09Q2  3.29Q3  10, 000
9
b. Find the profit per unit for each type of milk.
Solution The coefficients on the profit function correspond to the
change in profit when the corresponding variable is increased by 1 unit.
Since the coefficient of Q1 is 2.49, the profit increases by $2.49 when 1
more gallon of whole milk is produced and sold. This is the profit per
gallon for whole milk. The same reasoning for the other types of milk
leads to a profit per unit for 2% milk of $3.09 per gallon and $3.29 per
gallon for 1% milk.
In real applications, the units are often modified to make the numbers more
manageable. In Example 2, we examined the sales of a business in thousands of
dollars given by S  A, E   105 A  1412 E where A is the amount spent on advertising in
thousands of dollars and E is the number of employees. If we were to find a value for
this function like
S (10,5)  105 10   1412  5   8110
We would interpret this as indicating that advertising expenses of $10,000 and using 5
employees would lead to sales of $8,110,000. The variable A is scaled in thousands of
dollars. Scaling a variable means that the actual amount is divided by an amount to
obtain the value for the variable. In this case, dividing $10,000 by $1000 yields A  10 .
The dependent variable, sales, is also scaled in thousands of dollars. If we are given a
value for this variable like 8110, we need to multiply by $1000 to get the actual amount.
The value 8110 thousand dollars is the same as 8110  $1000  or $8,110,000.
When independent variables are scaled, the cost and revenue functions change. In
Example 3, we found the revenue function for an organic dairy that charges $3.49 per
gallon for whole milk, $4.19 per gallon for 2% milk, and $4.59 per gallon for 1% milk.
The function describing the revenue was
R  Q1 , Q2 , Q3   3.49Q1  4.19Q2  4.59Q3 10
where Q1 is the number of gallons of whole milk, Q2 is the number of gallons of 2% milk,
Q3 is the number of gallons of 1% milk that the dairy produces and sells. Since an
organic dairy might produce and sell thousands of gallons of each week, we could
choose to scale the independent variables in thousands of gallons. Instead of inputting
values like 1000 into a variable, we would input 1 instead. This introduces subtle
changes into the revenue function since changes of 1 unit in the scaled variable results
in a change of 1000 in gallons of milk. Let’s look at how this will change the revenue
function.
Let R(Q1 , Q2 , Q3 )  p1Q1  p2Q2  p3Q3 represent the revenue for the organic dairy where
Q1 , Q2 and Q3 are the amounts of whole, 2%, and 1% milk in thousands of gallons
produced and sold respectively. As we found in Example 3, the coefficients p1 , p2 , and
p3 are the unit prices of each type of milk. However, when the variables are scaled the
meaning of one unit changes. Now the unit price corresponds to the price of 1 thousand
gallons of milk. A price of $3.49 per gallon is equivalent to $3490 per 1 thousand gallons
of milk. If we change each unit price to account for the scaling, we get
R  Q1 , Q2 , Q3   3490Q1  4190Q2  4590Q3 The value
R  0, 2, 0   3490  0   4190  2   4590  0   8380
represents the revenue from producing and selling 2000 gallons of 2% milk.
If we were to calculate the same revenue from the function in Example 3, we would get
R  0, 2000, 0   3.49  0   4.19  2000   4.59  0   8380 The key part of each calculation is middle term,
dollars
versus 4.19 dollars
4190 thousand
gallons  2 thousand gallons
gallon  2000 gallon
11
In each case, the units in red reduce to 1 leading to a function value of 8380 dollars. By
examining the units carefully in each term, we can deduce the units on the revenue. The
units on each term must be the same for the terms to add together properly.
Another possibility is to scale the variables in thousands of gallons but use the unit
prices in dollars per gallon. In this case the middle term would be
rs
4.19 dolla
gallon  2 thousand gallon
In this case, the numbers multiply to 8.38 and the units are thousands of dollars.
To define the function with the revenue in thousands of dollars we would write the
function as
R  Q1 , Q2 , Q3   3.49Q1  4.19Q2  4.59Q3 with each of the variables scaled in thousands of gallons. Now the revenue from 2000
gallons of 2% milk is calculated as
R  0, 2, 0   3.49  0   4.19  2   4.59  0   8.38 and the revenue is 8.38 thousand dollars or $8380.
Each of these models may be used to calculate the revenue for the organic dairy.
However, the units you choose for the independent and dependent variables in the
revenue and cost functions should match. If not, you might find yourself subtracting cost
in dollars from revenue in thousands of dollars. This would lead to a value for profit that
makes no sense.
12