3.3 Cost, Profit and Revenue
Functions
Learning Objective: to see how linear and quadratic functions are useful in
the business world.
Warm-up (IN)
Write what you know about these terms – cost,
demand, revenue, and profit.
Notes!
Cost (C)
If
Revenue (R)
break even
C=R
R>C
make a profit
R<C
loss of money
2 types of costs
Fixed Variable -
rent, insurance, etc.
Materials, wages, etc.
Dependent on # of items made or
hours worked
Cost Function - C= fixed costs + variable costs
C  a bx
Price-Demand Function -
p  m  nx
linear
m and n are constants
(depending on the problem)
x is the # of items that
can be sold at $p per item
linear
Revenue Function - # of items sold*price per item
R  x  m  nx  quadratic
P  R C
P  x  m  nx   a  bx 
R  xP
Profit Function or
or
quadratic
Ex 1 - Price-Demand data from a manufacturer of cameras:
P - Wholesale price per camera
x - millions of cameras sold
* note as price goes down, # sold goes up
a) Plot the data in the table and find the price-demand function. What is
the domain?
p  m  nx
p  x   94.75  4.96x
D  0  x  19
b) What is the company's Revenue function for this
camera? What is the domain of the function?
R  x  m  nx 
R  x  94.75  4.96x 
D  0  x  19
c) Complete the table, computing revenues to the nearest
million dollars
in
x
in
millions
1
3
6
9
12
15
R  x
millions
of $s
89.79
239.61
389.94
450.99
422.76
305.25
d) Graph the revenue function and change the window
appropriately. Sketch the function below.
What kind of graph is it?
e) What is the max revenue to the nearest $1,000? For
what output of cameras (nearest thousand)?
x=9.55
9,550,000 cameras
y=452.5
$452,500,000 revenue
f) What is the wholesale price per camera to nearest
dollar to produce the max revenue?
p  x   94.75  4.96x
 94.75  4.96  9.55
Use the # of cameras
that maximizes revenue
for x!
 $47.38 per camera
g) Given the cost data below, find the cost function for
manufacturing the cameras.
C  x   a  bx
C  x   156  19.65x
h) Find the company's profit function. Graph and find the
max profit and output.
P  x  m  nx   a  bx 
P  x  94.75  4.96x   156  19.65x 
P 94.75x 4.96x 156 19.65x
2
P  4.96x 75.1x 156
2
Max output
7.5 million cameras
Max profit
$128.27 million
i) Find the wholesale price for cameras to produce max
profit.
p  x   94.75  4.96x
 94.75  4.96  7.5
 $57.55
Use the # of cameras
that maximizes profit for
x!
per camera
j) Find where the company would break even, run at loss,
or have a profit.
break even P  x   0
loss
P  x  0
Profit
P  x 0
x  2.49 and x  12.66
0  x  2.49 and 12.66  x  19
2.49  x  12.66
Out –
Summary –
HW –
Don’t forget
about POW!!