Project 1
Published and Distributed by
The
Mathematical Association of America
Marketing
Computer Drives
Part 2: Calculus & Optimization
Video introduction by
Title
Prof. Christopher
Lamoureux
Department of Finance
University of Arizona.
For
7
e 200
c
i
f
f
O
Mathematics for
Business Decisions
Part 2
Release 2, 2009
Graphing Functions
Trend Lines
Demand, Revenue, Cost, and Profit
Differentiation
Using Solver
Integration
Marketing Example
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© 2009 by The Arizona Board of Regents for The University of Arizona. All rights reserved.
Graphing Functions. Plotting Points
Graphing Functions, Plotting Points
1. PLOTTING POINTS
The ability to produce informative, attractively formatted plots is of
importance in almost all areas of business. Graphing the functions that are
used in marketing, or probability densities and cumulative distributions for
continuous random variables requires some special computer skills. In
particular, many points are needed for Excel to accurately graph these, and
other “smooth” functions.
Example 1. We will start by plotting a simple function which could
be the p.d.f. of a continuous random variable that takes values in the interval
[1, 3].
  0 . 75  x 2  3  x  2 . 25 if 1  x  3
f (x)  
 0 otherwise
To graph f over the interval [0, 4], we subdivide [0, 4] with a large
number of points. For this example, we will use 501 points x0, x1, , x500 that
are evenly spaced from 0 to 4, with x0 = 0 and x500 = 4.
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Graphing
Functions. Plotting Points: page 2
Graphing, Points
With this arrangement, the distance between any two adjacent
points is (4  0)/500 = 0.008. This means that x0 = 0.000, x1 = 0.008, x2 =
0.016, , x500 = 4.000.
In the sheet Graph 1 of the Excel file Graph Examples.xlsx we
have entered 0 in Cell B9 and 0.008 in Cell B10. The pattern is extended by
dragging these two cells down to Cell B509.
The formula for f is entered in Cell C9, using the IF function.
Recall that IF is found in the Logical list under Paste Function. It takes the
form
IF(statement, rule 1, rule 2)
If the statement is true, IF uses rule 1. When the statement is false,
IF uses rule 2. We will use -0.75*B9^2+3*B9-2.25 for rule 1 and 0 for rule
2. With these rules, we want our statement to be 1  x  3, that is, 1  B9 
3.
Graph Examples.xlsx
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Graphing Functions. Plotting Points:
page 3
Graphing, Points
Excel does not recognize a
continued inequality such as 1  B9 
3. We must realize that this means 1
 B9 and B9  3. These are entered
using the AND function, found in the
Logical list of the Insert Function
Dialog Box.
Graph Examples.xlsx
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Graphing Functions. Plotting Graphing,
Points:Points
page 4
The And function is used as the statement in the
IF function, completing the definition of f in Cell C9.
Cell C9 in now dragged down to Cell C509.
Graph Examples.xlsx
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Graphing
Functions.
Graphing,
Points
Plotting Points: page 5
The best graph of our
501 points is made using an XY
(Scatter) plot, with smoothed
lines and no markers. To find
this, click Insert, then on the
Create Chart box
We select this type of
plot and enter our data range.
Create Chart
Graph Examples.xlsx
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Graphing
Functions.
Plotting Points: page 6
Graphing,
Points
Right click in the plot region, then left
click on Select Data. Once the data is entered, we
can format the plot in any way that we like.
Graph Examples.xlsx
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Graphing
Functions.
Graphing, Points
Plotting Points: page 7
The best effect is usually obtained by
clicking on the plotted points, clicking on Format/
Selected Data Series, and then selecting a medium
line width with no markers.
Graph Examples.xlsx
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Graphing
Functions.
Plotting Points: page 8
Graphing,
Points
Here is our final graph, copied from Graph Examples.xlsx.
0.8
0.6
f(x) 0.4
0.2
0.0
0.0
0.5
1.0
1.5
2.0
x
2.5
3.0
3.5
4.0
Graph Examples.xlsx
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Graphing Functions.
Graphing, Points Plotting Points: page 9
Start with a blank Excel file and use 701 evenly
spaced points to plot the graph of f(t) = 2,000e0.06t over the interval [12,
12]. Recall that Excel uses EXP(x) for ex.
This graph shows you the value of $2,000, invested at 6%,
compounded continuously for t years. When t is negative, f(t) gives the past
value. When t is positive, f(t) gives the future value of the $2,000.
Use the plot that you made in Exercise 1 to estimate
the length of time that it will take for $2,000 to grow to $3,000, at 6%
compounded continuously.
Start with a blank Excel file and use 501 evenly
spaced points to plot the graph of
 0 if x  0
g (x)  
 x / 1 .5
if 0  x
1  e
over the
interval [2, 10]. Note that this function is the c.d.f. for an exponential
random variable with parameter  = 1.5.
Excel
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Graphing Functions. Plotting
page 10
Graphing, Points:
Points
Plot the graph of f(x) = 0.000025x2 + 100 over the
interval [0, 2,000].
Start with a blank Excel file and use 501 evenly
spaced points to plot the graph of f ( x ) 
1
2 
e
 0 . 5 x
2
over the
interval [4, 4]. To enter the constant  in Excel, type PI(). In Excel, the
square root function is entered as SQRT.
(i) Have Excel display vertical gridlines, spaced 0.5
units apart, in the graph that you made in Exercise 5. Add horizontal
gridlines, spaced 0.1 units apart, so that the area of each gridline rectangle
is 0.05 square units. (ii) Use the rectangles to estimate the total area of the
region between the x-axis and the graph of f, over the interval [4, 4].
Excel
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Graphing Functions. Several Functions
Graphing Functions, Several Functions
2. SEVERAL FUNCTIONS
Graphing two or more functions in the same plot can often help us
understand the connections between the quantities that are represented by the
functions.
Example 2. Here are two functions that are typical of those which
we will encounter in our study of marketing.
3
f ( x )   0 . 000025  x  100  x
 50  x  15 , 000
g(x)  
 25  x  35 , 000
if x  800
if 800  x
We will plot both f and g over the interval [0, 2,000], using 401
equally spaced points x0, x1, , x400, with x0 = 0 and x400 = 2,000. The
distance between any two consecutive points is 2,000/400 = 5. This means
that x0 = 0, x1 = 5, x2 = 10, , x400 = 2,000.
In the sheet Graph 2 of Graph Examples.xlsx we have entered 0 in
Cell B9 and 5 in Cell B10. The pattern is extended by dragging these two cells
down to Cell B409.
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Graphing Functions.
Several
Functions: page 2
Graphing,
Several
The formulas for f and
g are entered in Cells C9 and
D9, using the IF function for g.
We select an XY (Scatter) plot,
with smoothed lines and no
markers. Right click in the plot
region, then left click on Select
Data. Enter the indicated data
range.
Graph Examples.xlsx
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Graphing Functions. Several Functions:
page 3
Graphing, Several
Click on Add under
Legend Entries (Series). We
name Series 1 as f and enter its
data ranges. Series 2 is named
g, and its data is entered.
Finally, the plot is formatted as
desired.
Graph Examples.xlsx
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Graphing Functions. Several Functions:
page 4
Graphing, Several
80,000
f
g
values
60,000
40,000
20,000
0
0
500
1,000
x
1,500
Use the plot of
f and g from Example 2 to estimate the
two values of x at which f(x) = g(x).
Graph Examples.xlsx
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2,000
Look at the graph in the
sheet Graph 2 of Graph
Examples.xlsx to see how the
final formatting was done. The
graph is copied on this page.
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Graphing
Functions.
Several Functions: page 5
Graphing,
Several
Show the graphs of both f(x) = 60x + 20,000 and g(x)
= 0.2x2 + 200x in the same plot, over the interval [0, 1,000].
Let f ( x )  x and g ( x )  0 . 25  x  1 . (i) Show the
graphs of both f and g in the same plot, over the interval [0, 8]. Recall that
in Excel, the square root function is entered as SQRT. (ii) Use your plot to
estimate the coordinates of the point at which f(x) = g(x). (iii) How would
you describe the relationship between the graphs of f and g at the point
where they intersect?
The c.d.f. for an exponential random variable X,
 0 if
x0
. Show three
with parameter  is given by F ( x )  
x /
if 0  x
1  e
graphs of F, with  = 2, 3, and 4, in a single plot, over the interval [5, 15].
Excel
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Graphing Functions.
A Graphing Utility
Graphing, A Graphing Utility
Click here for information on
running the macro in Graphing.xlsm.
3. A GRAPHING UTILITY
It is important that you know how to create your own graphs. This
will be necessary for special plotting needs. Understanding the process of
computer graphing also helps you know what the computer is doing and assists
you in your interpretation of the results.
However, in routine situations the creation of plots can be time
consuming and could distract us from our main goal of understanding the
business applications of a function. For this reason, we are supplying an Excel
file, Graphing.xlsm, that serves as a graphing utility.
To use Graphing.xlsm, open the file and follow the directions in the
blue box. Be sure to note that you are to enter a formula for f(x) in terms of
the letter x, not a cell reference. x is the only letter that can be used as a
variable in Graphing.xlsm.
Graphing.xlsm
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Graphing Graphing,
Functions.
A Graphing Utility: page 2
Utility
Before running Graphing.xlsm, be sure that Excel is set for
Automatic Calculation.
When entering a function in Graphing.xlsm, you can use the
letters s, t, u, v, or w as constants. If this is done, you must enter values for
any letters that you use.
For example, to plot f(x) = x s, where s is a number that you enter in
Cell L17, enter the following function in Cell C18.
= x^s
Accept the function and then click on the Graph button at the right
of the plot. Entering a new number in Cell L17 will automatically change
the function and redraw the plot.
If values are entered in Cells L17:L21; then constants s, t, u, v, or
w may also be used to describe the range, [a, b], of the plot. If these
constant names are used, they must be entered in Cells H18:I18 preceded
with "=" signs. For example, =s.
Graphing.xlsm
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Graphing Functions.Graphing,
A Graphing
Utility: page 3
Utility
Use Graphing.xlsm to plot the graph of f(x) =
2,000e0.06t over the interval [12, 12].
Use Graphing.xlsm to plot the graph of F(x) over
the
interval [5, 15].
 0 if x  0
FX ( x)  
.
 x / 3 .5
if 0  x
1  e
Note that this function
is the c.d.f. for an exponential random variable with parameter  = 3.5.
Use Graphing.xlsm to plot the graph of fX over the
interval [5, 15]. f X
 0 if x  0
(x)   1
.
 x / 3 .5

e
if
0

x

 3 .5
Note that this function
is the p.d.f. for an exponential random variable with parameter  = 3.5.
Graphing.xlsm
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Graphing
A Graphing Utility: page 4
Graphing,Functions.
Utility
Redo Exercise 13, but leave the parameter  as a
constant. Changing  and recomputing the sheet should change the plot
correspondingly. Hint: use the built-in constant, s, as a replacement for .
Let  = 1, and use Graphing.xlsm to plot the
graph of f ( x ) 
1
2 
e
 0 . 5 ( x   )
2
over the interval [4, 8]. To enter the
constant  in Excel, type PI(). In Excel, the square root function is entered
as SQRT.
(i) Redo Exercise 15, but leave the
parameter  as a constant. Changing  and recomputing the
sheet should change the plot correspondingly. Hint: use the
built-in constant, s, as a replacement for . (ii) Display graphs
for  = 0, 1, 2, and 4.
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Graphing.xlsm
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Trend Lines,
Fitting Models Fitting Models
Trend
Lines.
1. FITTING MODELS
Mathematical analysis of business situations often requires formulas
for the functions that are used as models. Unfortunately, the real world almost
never provides us with formulas. What we do find in business are data points
that provide approximate values for the models. Trend lines allow us to use
data points to generate formulas which can be used for business planning.
Example 1. (An Exponential Model) A county Child Protective
Services, CPS, agency has records showing the number of cases that it has
handled during the years from 1998 through 2006.
Y ear
1998
1999
2000
2001
2002
2003
2004
2005
2006
C ase L oad
2,373 2,825 2,806 3,534 3,808 4,312 5,067 5,368 6,445
Noting that the case load is increasing rather rapidly, the agency
would like to predict how many cases it is likely to handle in the year 2010.
The Excel file Case Load.xlsx shows a plot of the nine data points.
For convenience in later work, we have rescaled time as the number of years
after 1998.
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Case Load.xlsx
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TrendTrend
Lines.
Fitting Models: page 2
Lines, Models
We would like to find a formula for the function that represents the
underlying pattern of the case load growth. To do this, we must first use our
judgment to select the type of functions that we will consider. For the CPS
data we note the following. (i) The data points appear to increase and curve
slightly upward. (ii) Many changing populations have patterns of exponential
growth. (iii) Experience in public administration around the country suggests
that social service case loads increase exponentially.
For these reasons we will assume that the number of cases handled in
a year that is t years after 1998, is given by an exponential function of the form
f(t) = uevt, where u and v are constants. Recall that e is a constant that is the
base for the natural exponential function. It can be shown that e  2.71828.
The exponential function is defined for all values of x, and is entered in Excel
with upper case letters as EXP().
Once we have selected a form for f(t), Excel can find values for the
constants u and v that produce the function of the given type, that best fits
the data points. Such a function is called a trend line.
Case Load.xlsx
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Trend
Lines.
Trend Lines,
ModelsFitting Models: page 3
We will select an exponential trend line for the CPS data. Open
Case Load.xlsx to see a plot of the nine original data points.
To add a trend line:
1. Left click on any data point.
2. Right click to see the pull-down menu.
3. Click on Add Trendline.
C A SE LO A D
N um ber of C ases
12,000
10,000
8,000
6,000
4,000
2,000
0
1
2
3
4
5
6
7
8
9
10 11 12
Years A fter 1998
98
Case Load.xlsx
00
02
04
06
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08
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Trend
Fitting Models: page 4
Trend Lines.
Lines, Models
4. Click on
Exponential.
5. Click on
Trendline Options.
Case Load.xlsx
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Trend
Lines.
Trend Lines,
Models Fitting Models: page 5
6. To move to 2010,
set Forward to 4.
8. Click on Close.
7. Click on Display
Equation on chart.
Case Load.xlsx
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N um ber of C ases
Trend
Lines.
Trend Lines,
Models Fitting Models: page 6
C A S E L O A D y = 2373.6435 e 0 .1 2 1 7 x
12,000
10,000
8,000
6,000
4,000
2,000
0
1
2
3
4
5
6
7
8
9
10 11 12
Years A fter 1998
98
00
02
04
06
08
10
The equation of the exponential trend line is displayed on the plot.
We see that the coefficients are u = 2373.6435 and v = 0.1217. To use these
numbers in the Excel file, you must either copy them manually, or with
copy/paste.
Case Load.xlsx
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Trend
Lines.
Fitting Models: page 7
Trend
Lines, Models
Let f(t) = 2,373.6435e0.1217t. Since we are using t as the number of
years since 1998, the expected case load in a year y is f(y  1998). Case
Load.xlsx shows that the agency can expect 10,225 cases in the year 2010.
Their load has been growing at a yearly rate of approximately 12%.
The extension of a trend line beyond the range of the known data is
called extrapolation. This must be done with great caution. It is not
unreasonable to expect the current pattern of growth to continue for the next
four years. However, it would be absurd to expect the same circumstances to
continue for the next 100 years. As Case Load.xlsx shows, f(2106  1998) =
1,212,295,365. Obviously, we should not assume that CPS will handle over
one billion cases in the year 2106.
The choice of a basic form for a trend line is of
major importance. Further training in business will help you recognize the
appropriate types of functions to use in common situations.
Case Load.xlsx
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Trend
Lines.
Fitting Models: page 8
Trend Lines,
Models
Example 2. (A Linear Model) We will illustrate the effect of
different types of trend lines by fitting a linear function to the CPS data
points. This means looking for the best fitting curve of the form
f(t) = at + b.
This is done in exactly the same way that we used for the exponential
model, except that we select the Linear box on Excel’s menu.
Once the coefficients of
the trend line are computed, we
can click on the equation box and
format the numbers. For the
linear trend line, we have
displayed a and b as integers.
Case Load.xlsx
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Trend
Lines.
Trend Lines,
Models Fitting Models: page 9
N um ber of C ases
C A SE LO A D
y = 487 x + 2112
12,000
10,000
8,000
6,000
4,000
2,000
0
1
2
3
4
5
6
7
8
9
10 11 12
Years A fter 1998
98
00
02
04
06
08
10
We see that the coefficients are a = 487 and b = 2,112. To use these
numbers in the Excel file, you must either copy them manually, or with
copy/paste.
Case Load.xlsx
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Trend
Lines.
Fitting Models: page 10
Trend
Lines, Models
Let f(t) = 487t + 2,112. The expected case load in a year y is f(y 
1998). Case Load.xlsx shows that the agency can expect 7,956 cases in the
year 2010. This is approximately 22% fewer cases than were predicted with
the exponential model. Will growth be exponential, linear, or neither? Only
time can tell with any certainty.
The difference in choice of models is particularly noticeable over
long periods of extrapolation. As computed in Case Load.xlsx, the linear
model predicts a load of 54,708 cases in the year 2106. This is approximately
0.005% of the 1,212,295,365 cases predicted by the
exponential model!
How do I know
what type of trend line
to use?
Case Load.xlsx
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Trend Lines. Fitting
TrendModels:
Lines, Models page 11
(i) Use the exponential model in the CPS example to
predict the case load in the year 2011. (ii) Repeat Part i, using the linear
model.
(i) Use the exponential model in the CPS example
to predict the case load in the year 2013. (ii) Repeat Part i, using the linear
model.
(i) Fit a 3rd degree polynomial trend line through the
data points in the CPS example. (ii) What case load does this model predict
for the year 2010?
(i) Fit a 6th degree polynomial trend line through the
data points in the CPS example. (ii) What case load does this model predict
for the year 2010? (iii) Does increasing the degree of a polynomial trend line
seem to improve its predictive value?
Case Load.xlsx
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Trend
Fitting Models: page 12
TrendLines.
Lines, Models
During the first 6 days of a special promotion, a
small business records the following sales information. The dollar amount
listed for each day is the total cumulative sales from the start of the
promotion.
Day
1
2
3
4
5
6
Total Sales $29,390 $37,358 $47,116 $50,604 $57,066 $60,666
(i) Fit a Power trend line to the data and use it to estimate the total
sales during the first 9 days of the promotion. (ii) Repeat Part i, using a
Linear trend line. (iii) Which model do you think is more realistic?
Fit an exponential trend line to the following data
and use the line’s formula to predict the highest closing price during July.
Highest Monthly Close of MONEY .com Stock Shares
Month Name
January February March
April
May
Month Number
1
2
3
4
5
Highest Close
$65.35
$65.95
$67.71
$68.62
$71.24
Excel
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Trend Lines. Fitting
TrendModels:
Lines, Modelspage 13
(i) Fit a linear trend line to the data
in Exercise 6, and use this model to predict the highest
closing price during July. (ii) Fit a 3rd degree polynomial
trend line to the data in Exercise 6, and use this model to
predict the highest closing price during July. (iii) Do you
think that trend lines can provide a reliable way to predict
stock prices?
Your company, which
manufactures and distributes beach wear, has kept sales
records for the months of March, April, May, June, and
July of the current year. These indicate that sales volume
is increasing. Which, if any, of the types of trend lines
that are available in Excel might be used to predict sales
volume for the rest of the year? Explain your answer.
Excel
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Trend Lines. Fitting
TrendModels:
Lines, Modelspage 14
The Flimsy Plastic company produces toys for
children in the age group from 3 to 6 years. A regional sales office has kept
records on the number of children in this range that live in its territory.
3-6 Year Olds in Sales Territory
Year
1997 1998 1999 2000 2001
Population (K’s) 390 410 440 460 470
Year
2002 2003 2004 2005 2006
Population (K’s) 510 540 570 580 620
(i) Experiment with several different types of trend lines to
determine which model best fits the data. (ii) Use the type of trend line that
you select to predict the number of 3-6 year olds that will be in the sales
territory in the year 2011.
Excel
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Demand, Revenue,
Cost, & Profit,
Demand, Revenue,
Cost,
&Income
Profit. Income
1. INCOME
The amount of revenue that a producer receives from the sale of a
good depends on the number of units that are sold and on the price per unit that
is paid. In a given market, some quantity q of the product can be sold at a
price p per unit. These are related by the demand function D(q). Specifically,
p = D(q) is the price at which q units of the good can be sold.
Example 1. A regional restaurant chain plans to introduce a new
buffalo steak dinner. Managers tested various prices in their establishments
and arrived at the following estimates for weekly sales.
Price
$14.95 $19.95 $24.95 $29.95
Number sold per week 2,800 2,300 1,600
300
For example, they believe that D(2,800) is approximately $14.95. In
addition, their experience suggests a quadratic demand function.
In the sheet Income of the Excel file Dinners.xlsm we have plotted
the four given points and fitted a 2nd degree polynomial trend line to the data.
Dinners.xlsm
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Demand,
Revenue, Cost, & Profit. Income: page 2
D, R, C, & P, Income
Demand Function
y = -0.0000018x2 - 0.0002953x + 30.19
$32
D(q)
$24
$16
$8
$0
0
1,000
2,000
q
3,000
4,000
Define the demand function to be
D(q) = aq2 + bq + c, where a = 0.0000018,
b = 0.0002953, and c = 30.19.
Dinners.xlsm
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Demand,
Revenue,
Cost, & Profit. Income: page 3
D, R, C, & P,
Income
The amount of money that a producer receives from the sale of q
units of its product is called revenue, and is denoted by the revenue function
R(q). If D(q) is the demand function, and revenue will come from selling q
units at a price of D(q) dollars per unit, then R(q) = qD(q).
In the case of the restaurants, the amount of money that diners pay for
q of the buffalo steak dinners per week is weekly revenue. The following
graph is copied from the sheet Income in Dinners.xlsm.
Setting a high price
for the dinners will result in
very few orders, and a small
revenue. Setting a low price
for the dinners will also bring
in low revenue.
Revenue Function
$50,000
R(q)
$40,000
$30,000
$20,000
$10,000
Dinners.xlsm
$0
0
1000
2000
q
3000
4000
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Demand,
Revenue,
Cost, & Profit. Income: page 4
D, R, C, &
P, Income
It appears from the graph that the maximum revenue would result
from selling approximately 2,300 buffalo steak dinners. Looking back at the
graph of the demand function, we see that a price of around $20 can be
expected to result in the sale of 2,300 dinners. Open the sheet Income in
Dinners.xlsm and explore this with the Computation cells.
(i) Find the price, per
dinner, that will result in the sale of 2,600 buffalo
steak dinners. (ii) What revenue can be expected
from 2,600 dinners?
Dinners.xlsm
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D, R, C, & P, Income
Demand,
Revenue, Cost, & Profit. Income: page 5
(i) Experiment with different
values in Cell B19 of the sheet Income in Dinners.xlsm to
find the number of dinners that would be sold at a price of
$19.95. (ii) What revenue can be expected if dinners are
priced at $19.95?
Suppose that the demand function
for a good is given by D(q) = 0.1q + 150. (i) Use
Graphing.xlsm to plot D(q) and the revenue function, R(q).
(ii) Estimate the price that will yield the greatest revenue.
Suppose that D(q) = 0.00006q2
+ 250 is the demand function for a certain model of audio
speaker. (i) Use Graphing.xlsm to plot D(q) and the
revenue function, R(q). (ii) Estimate the price that will
yield the greatest revenue.
Graphing.xlsm
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D, R, C, & P, Income
Demand,
Revenue, Cost, & Profit. Income: page 6
DEMAND FUNCTION
500
400
D (q ) in $
The
demand function for a new type
of car alarm system is shown in
the adjacent plot. Note that the
quantities sold are given in
thousands. Estimate revenue,
in dollars, from selling 300,000
alarms.
300
200
100
0
0
100
200
300 400
q in K's
500
600
(i) Use your team’s data and Trend Lines to find the
formula for a quadratic demand function. (ii) Plot D(q) and R(q) for your
team’s product. Use the same units as in the work on the Class Project in the
sheet Functions of Marketing Focus.xlsm.
Excel
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D, R, C, Revenue,
& P, Expenses & Profit Cost, & Profit. Expenses & Profit
Demand,
2. EXPENSES AND PROFIT
Every producer incurs costs in the production of its product. These
usually include fixed costs, which do not depend upon the amount of a good
that is produced, and variable costs. Fixed costs might include such things as
plant overhead, minimal labor costs, and debt service. Variable (production)
costs cover such items as materials, labor, and distribution expenses. We will
denote fixed cost by C0, and denote the variable cost for q units of a good by
VC(q).
A producer’s total cost function, C(q), for the production of q units
is given by C(q) = C0 + VC(q).
Example 2. We will continue Example 1, and develop a cost function
for the restaurant chain’s new buffalo steak dinners. Managers estimate that
the new menu item will have to support $9,000 out of the complete weekly
fixed cost of operating the chain. Hence, C0 = $9,000. The head chef makes
the following estimates for the costs of preparing various numbers of dinners.
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Demand,
Revenue,
D, R, C,
& P, Expenses Cost,
& Profit & Profit. Expenses & Profit: page 2
Number of dinners, q
1,000
Variable cost, VC (q ) $14,000
2,000
$22,000
3,000
$28,000
Notice that there is some economy of scale. The cost of preparing the
first thousand dinners is $14,000, of preparing the second thousand dinners is
an additional $8,000, and of preparing the third thousand dinners is an
additional $6,000. In order to develop a formula for the variable cost function,
we will have Excel fit a trend line through the three estimated data points. The
managers know that, in former menu offerings, food preparation costs have
usually turned out to follow power function models. That is, VC(q) = uqv, for
some constants u and v.
In the sheet Profit of the file Dinners.xlsm, we have plotted the data
points and added a power trend line. Excel finds that u = 177 and v = 0.633.
Thus, VC(q) = 177q0.663, and the total weekly cost function for the buffalo
steak dinners is
C(q) = C0 + VC(q) = 9,000 + 177q0.633.
Dinners.xlsm
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Demand,
Cost, & Profit.
D, R, C, & Revenue,
P, Expenses & Profit
Expenses & Profit: page 3
0.633
Variable Costs Function
y = 177x
$50,000
VC(q)
$40,000
$30,000
$20,000
$10,000
$0
0
1,000
2,000
q
3,000
4,000
Cost Function
$50,000
C(q)
$40,000
$30,000
$20,000
$10,000
Dinners.xlsm
$0
0
1000
2000
q
3000
4000
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D, R, C, & P, Cost,
Expenses &
& Profit
Demand, Revenue,
Profit.
Expenses & Profit: page 4
The producer of any good is interested
in profit. We let P(q) be the profit
obtained from producing and selling q units of a good at the price D(q).
Profit = Revenue  Cost
P(q) = R(q)  C(q)
The sheet Profit of Dinners.xlsm shows a
plot of both the revenue and cost functions on a single
set of axes. As the graph shows, revenue appears to
exceed cost in a range of approximately 700 to 3,100
dinners per week. The sheet Profit also plots the
profit function, P(q). A maximum profit of roughly
$14,000 seems to result from the preparation and sale
of approximately 2,000 dinners per week. Note that
this is somewhat less than our estimate of 2,300
dinners per week that would produce the maximum
revenue for the chain. Open the sheet Profit and
explore this with the Computation cells.
Dinners.xlsm
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How can I recognize the graphs of
demand, cost, and profit
functions?
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D, R, C, &Revenue,
P, Expenses & Profit
Demand,
Cost, & Profit. Expenses & Profit: page 5
Cost
Revenue
Revenue and Cost Function
$50,000
Dollars
$40,000
$30,000
$20,000
$10,000
$0
0
1000
2000
q
3000
4000
Profit Function
$15,000
P(q)
$10,000
$5,000
$0
-$5,000
0
1000
2000
3000
4000
-$10,000
q
Dinners.xlsm
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D, R, C, & P,Revenue,
Expenses & Profit
Demand,
Cost, & Profit. Expenses & Profit: page 6
(i) Find the cost of preparing 2,300 buffalo steak
dinners. (ii) What revenue can be expected from preparing and selling 2,300
dinners? (iii) What weekly profit can be expected from preparing and selling
2,300 dinners?
(i) Find the cost of preparing 1,500 buffalo steak
dinners. (ii) What revenue can be expected from preparing and selling 1,500
dinners? (iii) What weekly profit can be expected from preparing and selling
1,500 dinners?
(i) Experiment with computation in the sheet
Profit of Dinners.xlsm to find the number of dinners that could be sold at a
price of $18. (ii) Find the cost of preparing these dinners. (iii) What profit
can be expected from preparing and selling buffalo steak dinners at $18?
Dinners.xlsm
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Demand,
Revenue,
& Profit. Expenses & Profit: page 7
D, R, C, &
P, Expenses & Cost,
Profit
Returning to our plot of the demand function, D(q), we can estimate
that a price of approximately $22 would produce sales of 2,000 dinners. In
summary, the restaurant chain should prepare around 2,000 buffalo steak
dinners per week and price them at approximately $22. If this is done, they
can expect a weekly profit of around $14,000 from the new menu item.
The graph of P(q) shows us how sensitive the weekly profit is to
deviation from the optimal price per dinner of around $22. For example, using
rough estimates from the plot of the demand function, it appears that raising
the price of a dinner to $25 would reduce the demand to around 1,600 dinners
per week. Estimating from the plot of the profit function, we see that this
would reduce the chain’s weekly profit to around $13,000. That is a 7% drop
from the maximum profit of approximately $14,000.
Dinners.xlsm
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Demand,
Cost, & Profit. Expenses & Profit: page 8
D, R, C, & P,Revenue,
Expenses & Profit
Estimate the percentage drop in
weekly profit that would result from dropping the price of
buffalo steak dinners from $22 to $20.
Use the graphs in this section to
estimate the range of dinner prices that would yield weekly
profits of at least $10,000.
Use the graphs in this section to
estimate the range of dinner prices that would yield weekly
revenues of at least $35,000.
Dinners.xlsm
Use the graphs in this section to
estimate the weekly profit that would result from spending
$30,000 to prepare buffalo steak dinners.
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Demand,
Revenue,
D, R, C, & P, Expenses
& ProfitCost, & Profit. Expenses & Profit: page 9
Refer to the demand function given in Exercise 3.
Suppose that the fixed cost for producing this good is C0 = $12,000 and that
the variable costs are given by VC ( q )  1,300  q . (i) Use Graphing.xlsm
to plot D(q) and P(q). (ii) Use your graphs to estimate the number of units
that should be produced, and how they should be priced in order to attain the
maximum profit. (iii) Approximately what maximum profit might be
expected?
Refer to the demand function given in Exercise 4.
Suppose that the fixed cost for producing the speakers is C0 = $60,000 and
that it costs $110 to produce each speaker. (i) Use Graphing.xlsm to plot
D(q) and P(q). (ii) Use your graphs to estimate the number of speakers that
should be produced, and how they should be priced in order to attain the
maximum profit. (iii) Approximately what maximum profit might be
expected?
Graphing.xlsm
Excel
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COST FUNCTION
C (q ) in millions of $
Refer to
the demand function whose graph is
shown in Exercise 5. The cost
function for the alarms is plotted in
the adjacent graph. Note that the
quantities of alarms are given in
thousands and the total costs are
given in millions of dollars. (i)
Estimate the number of alarms that
would be sold at a price of $250
each. (ii) Estimate the revenue, in
dollars, that would result from the
sale of alarms priced at $250. (iii)
Estimate the cost, in dollars, of
producing the alarms that would be
sold at $250. (iv) Estimate the
profit, in dollars, that would result
from the sale of alarms at $250.
Demand,
Revenue,
& Profit.
D, R, C, &
P, Expenses & Cost,
Profit
Expenses & Profit: page 10
140
120
100
80
60
40
20
0
0
100
200
300 400
q in K's
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Demand,
Cost, & Profit. Expenses & Profit: page 11
D, R, C, & P,Revenue,
Expenses & Profit
(i) Use your team’s
data to find a formula for the cost function of
your product. (ii) Plot C(q) and P(q) for your
team’s data. Use the same units as in the work
on the Class Project in the sheet Functions of
Marketing Focus.xlsm. Study the Focus pages
to see how to use the custom programmed
function COST.
Excel
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Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
Drives
on the project
How can demand, revenue,
cost, and profit functions help us price
12-GB drives?
Obviously, Card Tech would like to
price its new drives in such a way that profit is
maximized. To compute values of their profit
function, they must first develop demand,
revenue, and cost functions.
Marketing Focus.xlsm
D, R, C, & P, Focus
Class Project
(material
continues)
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Our information on test markets and variableMarketing
costs is summarized
Calculus, Mathematics,
Homework,
in the sheet Data of the Excel file Marketing
Focus.xlsm. Tests,
Since
we areComputers
Computer
dealing with quite large numbers, we will adopt some conventions for
on the project
Drives
units.
 Prices for individual drives are given in dollars.
 Revenues from sales in the national market are given in
millions of dollars.
 Quantities of drives in the test markets are actual numbers of
drives.
 Quantities of drives in the national market are given in
thousands of drives.
At the top of the sheet Functions in Marketing Focus.xlsm, these
units are applied to our data, and the projected yearly sales in the national
market are computed for each test market price.
Marketing Focus.xlsm
D, R, C, & P, Focus
Class Project
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continues)
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 test market
on the project

sales



 size of 
 test market 

Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
 national 
Drives
 sales (K' s) 


size of


 national market (K' s) 
Since it was assumed that the demand function is quadratic, the
data points for national sales are plotted and fitted with a second degree
polynomial trend line. Excel finds that
D(q) = 0.00005349q2  0.03440302q + 414.53444491
where D(q) is the price, in dollars, at which q thousand drives can be sold.
We formatted the trend line equation so that its coefficients are
displayed with 8 decimal places. Further precision would result in only very
minor changes in computed values. This apparent increase in accuracy
would be spurious, since it exceeds the accuracy of our market data.
(material
D,
R,
C,
&
P,
Focus
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Marketing Focus.xlsm Class Project
continues)
D e m a n d F un ctio n
$ 5 00
D (q )
$ 4 00
on the project
Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
Drives
$ 3 00
$ 2 00
$ 1 00
$0
0
400
80 0
1 ,2 0 0
1 ,60 0
2,0 00
2,4 0 0
2 ,80 0
q (K 's)
In our units, the revenue function R(q) is to give the revenue, in
millions of dollars from selling q thousand drives. D(q) gives the price, in
dollars per drive, at q thousand drives. Hence, D(q)q1,000 gives the
revenue, in dollars, from selling q thousand drives. To express this revenue
in millions of dollars, we divide by 1,000,000.
R(q) = (D(q)q1,000)/1,000,000 = D(q)q/1,000
Marketing Focus.xlsm
D, R, C, & P, Focus
Class Project
(material
continues)
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R even u e F u n ction
R (q ) (M 's)
$400
on the project
$300
$200
$100
Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
Drives
$0
0
400
800
1,200
1,600
2,000
2,400
2,800
q (K 's)
It appears that we might obtain the maximum revenue, if we sold
approximately 1,400 thousand drives. Cells C91:E91 in the sheet
Functions of Marketing Focus.xlsm allow us to compute D(q) and R(q) for
any value of q. We find that D(1,400) = $261.53 and R(1,400) = 366.142
million dollars.
We want the total cost function, C(q), to give the cost, in millions
of dollars, of producing q thousand drives.
C(q) = fixed cost + variable cost for q thousand drives
Marketing Focus.xlsm
D, R, C, & P, Focus
Class Project
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continues)
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Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Let marginal cost denote the price, in dollarsComputer
per drive, for
production
atproject
a given number of drives. Our cost information can now be
on
the
Drives
summarized in the following table. The Batch Size column displays the
numbers of drives, in K’s, that will be produced in first, second, and all
further production lots.
Fixed Cost
(M's)
$135.0
Variable Costs (M's)
1
2
3
Batch Size (K's)
First
800
Second
400
Further
Marginal Cost
$160.00
$128.00
$72.00
C(q) depends upon 7 numbers: q (Quantity), Fixed Cost, Batch
Size 1, Batch Size 2, Marginal Cost 1, Marginal Cost 2, and Marginal Cost
3.
Marketing Focus.xlsm
D, R, C, & P, Focus
Class Project
(material
continues)
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Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
on the
Forproject
example, suppose that we produce 1,400 thousand
Drivesdrives. The
fixed cost is 135 million dollars. The first 800 thousand drives cost $160
per drive, for a total cost of 8001,000160/1,000,000 = 800160/1,000 =
128 million dollars. The next 400 thousand drives cost $128 per drive, for a
total cost of 4001,000128/1,000,000 = 400128/1,000 = 51.2 million
dollars. The remaining 200 thousand drives cost $72 per drive, for a total
cost of 2001,00072/1,000,000 = 20072/1,000 = 14.4 million dollars.
Adding these, we find that C(1,400) = 135.0 + 128.0 + 51.2 + 14.4
= 328.6 million dollars.
It is possible, though very tedious, to use nested IF functions in
Excel to produce a formula for C(q). This is demonstrated in Cell D125 of
the sheet Functions in Marketing Focus.xlsm.
Marketing Focus.xlsm
D, R, C, & P, Focus
Class Project
(material
continues)
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When it is difficult or very complicated to compute with built-in
Excel functions, we can seek help from custom programming.
In particular,
Marketing
Calculus,
Mathematics,user defined
Tests, Homework,
Computers
the Visual Basic language can be used
to construct
functions
in
Computer
any Excel file. These are accessed via pull-down menus, in the same way
on the
the
project
Drives
that
built-in
functions are used.
Given the complexity of our cost function, it is desirable to custom
program the formula in Visual Basic and attach it to the Excel file. This has
been done in Marketing Focus.xlsm, with the resulting function, COST,
being placed in the list of functions under User Defined. Look in the
formula bar for Cell E125 in the sheet Functions of Marketing Focus.xlsm
to see how the function is used.
To view the Visual Basic code for COST in Office 2007, click on
the Developer tab, then on Visual Basic. In earlier versions, click on
Tools/Macro/Visual Basic Editor. In the side bar, select VBAProjects
(Marketing Focus.xlsm), double click on Modules, then double click on
Module1.
Marketing Focus.xlsm
D, R, C, & P, Focus
Class Project
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continues)
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on the project
Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
Drives
You are not expected to do any Visual Basic programming in
Mathematics for Business Decisions. However, you will probably find it
convenient to use the function COST while working on your team’s project.
To transfer the Cost function from Marketing Focus.xlsm into another
Excel file, open both files. Open Module1 (as directed above), click on its
icon in the side bar under VBAProject (Marketing Focus.xlsm) and drag
the icon into the VBAProject side bar area of the new file.
You can also move modules containing macros or custom functions
with export/import. To do this, open the module, click on File/Export File,
then select a folder to contain a copy of the module. To attach the module to
another Excel file, open the Visual Basic Editor in that file, click on
File/Import File, and open the module.
Marketing Focus.xlsm
D, R, C, & P, Focus
Class Project
(material
continues)
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R even u e & C ost F u n ction s
$500
(M 's)
$400
on the project
Marketing R evenue
Calculus, Mathematics,
Tests, Homework, Computers
Computer C ost
Drives
$300
$200
$100
$0
0
400
800
1,200
1,600
2,000
2,400
2,800
q (K 's)
The Cost function can be added to Graphing.xlsm and plotted
with that utility, or it can be plotted as in the sheet Functions of Marketing
Focus.xlsm. It is most interesting to display the graphs of both R(q) and
C(q) on the same set of axes.
We will make a profit when R(q) > C(q) and will operate at a loss
where R(q) < C(q). It appears that we can make a profit by selling between
approximately 650 and 1,650 thousand drives. Verify this with numerical
experimentation in Cells B170:F170 in the sheet Functions of Marketing
Focus.xlsm.
Marketing Focus.xlsm
D, R, C, & P, Focus
Class Project
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continues)
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Our main interest is in profit, which is given Marketing
by revenue minus cost.
Let P(q) be the profit, in millions of dollars,
from selling q thousand
drives.
Calculus, Mathematics,
Tests, Homework,
Computers
on the project
P(q) = R(q)  C(q)
Computer
Drives
A plot of the profit function has been copied from Marketing
Focus.xlsm into the next Focus page. Each of the two “humps” represent a
local maximum profit. It appears that the local maximum that occurs on the
right, at approximately 1,280 thousand drives, is slightly higher than the
local maximum on the left. Hence, the largest possible profit, or global
maximum, is approximately P(1,280). This is close to 42 million dollars.
Looking at the plot of the demand function, we estimate that the 1,200
thousand drives should be priced at approximately $280 each. Verify this
with numerical experimentation in Cells B170:F170 in the sheet Functions
of Marketing Focus.xlsm.
What we need now is some way to replace graphical estimates with
more precise computations.
Marketing Focus.xlsm
D, R, C, & P, Focus
Class Project
(material
continues)
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P rofit F u n ction
$50
Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
Drives
P (q ) (M 's)
$40
$30
$20
on the project
$10
$0
-$10 0
400
800
1,200
1,600
2,000
-$20
q (K 's)
WHAT SHOULD YOU DO?
Each team should now find formulas for, and make plots of, its
demand, revenue, cost, and profit functions. Work with the same units and
coefficient precision as in the Class Project. Use your plots to estimate the
production range that would yield a profit, and to estimate the maximum
possible profit.
Marketing Focus.xlsm
D, R, C, & P, Focus
Class Project
(material
continues)
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Differentiation, Marginal
Analysis
Differentiation.
Marginal
Analysis
1. MARGINAL ANALYSIS
When working with the cost function for a product it is often helpful
to determine the instantaneous rate of change in total cost per unit of the good.
This cost per unit at a given level of production is called the marginal cost.
Example 1. We consider the cost function C(q) = C0 + VC(q) =
9,000 + 177q0.633 that was developed in the Expenses and Profit section of
Demand, Revenue, Cost, and Profit. Recall that a restaurant chain is planning
to introduce a new buffalo steak dinner. C(q) is the cost, in dollars, of
preparing q dinners per week.
We let MC(q) be the marginal cost at q dinners. That is, MC(q) is the
cost for an additional dinner, when q dinners are being prepared. For the
moment, we will think of this as the cost of preparing the next, or (q + 1)st,
dinner. Since C(q + 1) is the cost of preparing q + 1 dinners, and C(q) is the
cost of preparing q dinners, the (q + 1)st dinner costs C(q + 1)  C(q) dollars.
This gives a First Plan formula for the marginal cost.
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Differentiation.
Marginal Analysis: page 2
Differentiation, Marginal
MC ( q )  C ( q  1)  C ( q )

  9 ,000  177  q 0 .663 
0 . 633
0 . 633

 177  ( q  1)
q
 9 , 000  177  ( q  1)
0 . 633
We can use a calculator or Excel to compute values of MC(q). For
example,
0 . 633
0 . 633


 1, 000
0 . 633
0 . 633

 177  (1, 001 )
 1, 000
MC (1, 000 )  177  (1, 000  1)
 8 . 877636 .
Thinking in terms of money, the marginal cost at the level of 1,000
dinners, is approximately $8.88 per dinner. Similar computations show that
MC(2,000)  $6.88 and MC(3,000)  $5.93.
Since the marginal cost per dinner depends upon the number of
dinners currently being prepared, it is helpful to look at a plot of MC(q) against
q. This is created in the sheet M Cost of the Excel file Dinners.xlsm.
Excel
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Differentiation.
Differentiation, Marginal Marginal Analysis: page 3
Marginal Cost Function
MC(q) $ /dinner
$25
$20
$15
$10
$5
$0
0
1,000
2,000
3,000
q dinners
4,000
Visiting that file, we
see that the First Plan
marginal cost in Column D
is very large, when only a
small number of dinners are
prepared. MC(0) =
$177.00/dinner. However,
the cost per dinner drops
rapidly. MC(100) is down to
$20.63/dinner, and
MC(3,999) = $5.34/dinner.
When we talk about the cost per dinner, when q dinners are being
prepared, it is not clear whether we should consider the cost of the qth or the (q
+ 1)st dinner. Rather than arbitrarily choosing to move ahead to the (q + 1)st
dinner, as in the First Plan, we get a better indication of the marginal cost at q
by averaging the change in both directions from q. We will call this the
Second Plan for computing marginal cost.
Dinners.xlsm
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Differentiation.
Differentiation, Marginal
Marginal Analysis: page 4
MC ( q ) 
C ( q  1)  C ( q )   C ( q )  C ( q  1) 
2

C ( q  1)  C ( q )  C ( q )  C ( q  1)
2

C ( q  1)  C ( q  1)
2
The marginal costs, computed with the Second Plan, are shown in
Column E of M Cost in the Excel file Dinners.xlsm.
What can we learn from MC(q)? The restaurant managers might
want to know how many dinners would need to be prepared per week in order
to get the price per dinner for further dinners down to $8 or less. Glancing
down Column E in the sheet M Cost shows that MC(q)  $8 for q  1,327.
One further refinement is necessary in our computation of marginal
cost. MC(q) is best defined as the instantaneous rate of change in total cost,
per unit. The formula
C ( q  1)  C ( q  1)
is only an average change over an
2
interval of length 2.
Dinners.xlsm
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Differentiation.
Marginal Analysis: page 5
Differentiation, Marginal
To improve the computation, we let mathematics do something that
cannot be done in the real world. We consider a non-integer change in the
number of dinners that are prepared. If h is any small number then
C (q  h)  C (q  h)
2h
is the average change over an interval of length 2h. Since the numerator is a
difference of two numbers and the entire expression is a quotient, it is usually
called a difference quotient. As h is assigned smaller and smaller values, the
difference quotient gives better and better approximations for MC(q).
Marginal Cost at q = 1,000
h
(C (1,000+h ) C (1,000h ))/2h
10
$13.58196
1
$13.58151
0.1
$13.58151
0.01
$13.58151
0.001
$13.58151
0.0001
$13.58151
0.00001
$13.58151
Dinners.xlsm
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We indicate that MC(q)
is an instantaneous rate of
change, by writing
MC ( q )  lim
C ( q  h)  C ( q  h)
2 h
h0
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Differentiation.
Marginal Analysis: page 6
Differentiation, Marginal
In practice, we will compute marginal cost by evaluating the difference
quotient
C (q  h)  C (q  h)
2h
at a small value of h. Since excessively small
values may cause numerical problems, we will often use h = 0.00001.
This method is the Final Plan, that is used in Column F of M Cost in
the Excel file Dinners.xlsm.
Explain, in terms of real world dinners and dollars,
why it is plausible that, for buffalo steak dinners, MC(q) is very large for
small values of q, and that it is always positive.
Excel
Dinners.xlsm
Use Excel, with
h = 0.00001, to compute MC(1,000) for
buffalo dinners, rounded to 2 decimal
places.
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Differentiation.
Marginal Analysis: page 7
Differentiation, Marginal
Use Excel, with h = 0.00001, to compute
MC(2,500) for buffalo dinners, rounded to 4 decimal places.
Marginal analysis is also used with revenue, demand, and profit
functions. For example, the marginal revenue, MR(q), at q dinners is the
instantaneous rate of change in revenue per dinner, when q dinners are being
prepared and ordered at the demand price per dinner. This is defined by
MR( q )  lim
R( q  h)  R( q  h)
h0
2 h
and is computed by evaluating the difference quotient
,
R (q  h)  R (q  h)
2h
for a
small value of h. Similar definitions apply to the marginal demand, MD(q), and
marginal profit, MP(q), which are defined by the following limits.
MD( q )  lim
h0
Dinners.xlsm
D( q  h)  D( q  h)
2 h
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MP ( q )  lim
P ( q  h)  P ( q  h)
2 h
h0
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Differentiation, MarginalMarginal Analysis: page 8
Differentiation.
Values for all of our marginal functions are computed in the sheets M Cost and M Profit of the Excel file
Dinners.xlsm. The graphs of MD(q), MR(q), and MP(q) are also displayed in
those sheets.
Marginal Demand Function
Demand Function
$0.000
$32
MD(q) $/dinner
0
D(q)
$24
$16
$8
1,000
2,000
3,000
4,000
-$0.005
-$0.010
-$0.015
$0
0
1000
2000
q
3000
4000
-$0.020
q
Many aspects of the demand function are reflected in properties of the
difference quotients for marginal demand, and in the marginal demand
function. D(q) is always decreasing. Hence, all difference quotients for
marginal demand are negative, and MD(q) is always negative. The more
rapidly D(q) drops, the more negative are the difference quotients, and the
further negative is MD(q).
Dinners.xlsm
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Differentiation, Marginal Marginal Analysis: page 9
Differentiation.
Revenue Function
$50,000
Marginal Revnue Function
$40
$30,000
$20,000
$10,000
$0
0
1000
2000
q
3000
4000
MR(q) $/dinner
R(q)
$40,000
$20
$0
0
1,000
2,000
3,000
4,000
-$20
Where the revenue function
-$40
R(q) is increasing, the difference
q
quotients for marginal revenue are
positive, and MR(q) is positive. For example, MR(1,300) is approximately
$20. Thus, when 1,300 dinners are prepared and sold, the restaurant chain
takes in $20 more for each extra dinner. Likewise, where R(q) is decreasing,
MR(q) is negative. This shows that the maximum revenue will occur at the
value of q where the marginal revenue is equal to 0. Computations in the
sheet M Profit show that MR(2,309) = $0.01 and MR(2,310) = $0.01. Hence,
the maximum revenue occurs at either 2,309 or 2,310 dinners. Direct
computation shows that the maximum revenue is R(2,310) = $45,975.65.
Dinners.xlsm
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Differentiation, MarginalMarginal Analysis: page 10
Differentiation.
Cost
Revenue
Revenue and Cost Function
$50,000
Profit Function
$15,000
$30,000
$10,000
$20,000
$5,000
P(q)
Dollars
$40,000
$10,000
$0
$0
0
1000
2000
q
$/dinner
Marginal Cost & Marginal Revnue
Functions
$40
3000
4000
2,000
1000
2000
3000
4000
q
M Cost
M Revenue
Marginal analysis can tell us
a great deal about the profit function.
Refer back to these plots while
reading the next pages.
$0
1,000
0
-$10,000
$20
0
-$5,000
3,000
4,000
-$20
-$40
q
Dinners.xlsm
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Differentiation.
Differentiation, MarginalMarginal Analysis: page 11
Since P(q) = R(q)  C(q), profit will increase with more dinners if the
increase in revenue per dinner is greater than the increase in cost per dinner.
This happens where MR(q) > MC(q). Similarly, profit will decrease with
more dinners if the change in revenue per dinner (positive or negative) is less
than the increase in cost per dinner. This happens where MR(q) < MC(q).
Profit stops increasing, and starts to decrease at its maximum value.
Hence, the maximum profit must occur where MR(q) = MC(q). From the
plot of MR(q) and MC(q), it appears that the two graphs cross at a point where
q is slightly greater than 2,000. The computations in the sheet M Profit of
Dinners.xlsm show that MR(2,025) = $6.84 = MC(2,025). Hence, the
maximum profit will occur at q = 2,025 dinners. Direct computation shows
that D(2,025) = $22.21 and that P(2,025) = $14,052.
Considering only the mathematics of our analysis, the restaurant
chain should expect to sell 2,025 buffalo steak dinners per week, priced at
$22.21 per dinner, for a total profit of $14,052. Note that all of this
information is consistent with the estimates that we made in the section
Demand, Revenue, Cost, and Profit.
Dinners.xlsm
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Differentiation, Marginal
Differentiation.
Marginal Analysis: page 12
Marginal Profit Function
MP(q) $/dinner
$20
$10
$0
0
1,000
2,000
3,000
-$10
-$20
q
4,000
Marginal analysis of
profit offers another way to
determine the maximum
profit. Where profit, P(q), is
increasing, marginal profit,
MP(q), is positive. Where
P(q) is decreasing, MP(q) is
negative. The change from
increasing to decreasing
profit occurs at the maximum
profit.
This must be where MP(q) changes from positive to negative. Thus,
the maximum profit occurs when the marginal profit is zero, MP(q) = 0.
The computations in Column F of M Profit in Dinners.xlsm show
that MP(2,025) = $0.00. This agrees with the value found from the marginal
analysis of revenue and cost.
Dinners.xlsm
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Differentiation.
Differentiation, MarginalMarginal Analysis: page 13
Exercises 4, 5, and 6 refer to the situation that was discussed in
Exercises 3 and 14 of Demand, Revenue, Cost, and Profit. The demand and
cost functions for a good were given by D(q) = 0.1q + 150 and
C ( q )  12 , 000  1, 300  q , respectively.
Use the methods of Marginal Analysis to plot the
marginal cost function, MC(q), and the marginal revenue function, MR(q), on
the same set of axes.
(i) Use the methods of Marginal Analysis to plot
MP(q). (ii) Use your graph to estimate the value of q that maximizes profit.
(iii) What is the maximum possible profit? (iv) At what price should the
good be sold, in order to realize the maximum profit?
(i) Use your graphs from Exercise 4 to estimate the
value of q that makes MR(q) = MC(q). (ii) How does this value relate to your
work in Exercise 5?
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Differentiation.
Differentiation, Marginal Marginal Analysis: page 14
Exercises 7, 8, and 9 refer to the situation that was discussed in
Exercises 4 and 15 of Demand, Revenue, Cost, and Profit. The demand and
cost functions for one model of audio speaker were given by D(q) =
0.00006q2 + 250 and C(q) = 60,000 + 110q, respectively.
Use the methods of Marginal Analysis to plot the
marginal cost function, MC(q), and the marginal revenue function, MR(q), on
the same set of axes.
(i) Use the methods of Marginal Analysis to plot
MP(q). (ii) Use your graph to estimate the value of q that maximizes profit.
(iii) What is the maximum possible profit? (iv) At what price should the
speakers be sold, in order to realize the maximum profit?
(i) Use your graphs from Exercise 7 to estimate the
value of q that makes MR(q) = MC(q). (ii) How does this value relate to your
work in Exercise 8?
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Differentiation. Marginal
Analysis:
Differentiation,
Marginal page 15
Exercises 10 and 11 refer to the demand and cost functions for car
alarm systems that are plotted in Exercises 5 and 16 of Demand, Revenue,
Cost, and Profit.
Use your conceptual
understanding of marginal demand and the plot of the
demand function to sketch a rough graph of MD(q).
Use your conceptual
understanding of marginal cost and the plot of the cost
function to sketch a rough graph of MC(q).
Sketch what you believe might be a typical plot of a
marginal revenue function. Explain the reasoning behind your sketch.
Excel
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Differentiation, Derivatives
Differentiation.
Derivatives
2. DERIVATIVES
Difference quotients are used in many business situations, other than
in marginal analysis. For this reason, it is convenient to have a standard name
and common notation for limits of such quotients.
Let f be a function which is defined on an open interval containing a
real number x. Suppose that
f (x  h)  f (x  h)
2h
approaches a number m as h
is taken to be smaller and smaller. We write
m 
lim f ( x  h )  f ( x  h )
,
2h
h 0
and call m the derivative of f at x.
To show that m depends on both the function, f, and the number, x, we
denote the derivative by f (x).
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f ' ( x )  lim
Differentiation.
Differentiation, Derivatives
Derivatives: page 2
f ( x  h)  f ( x  h)
2 h
h0
This is read as “the derivative of f, at x, is f
prime of x.” The process of computing f (x) is called
differentiation and f (x) is interpreted as the
instantaneous rate of change in f(x) with respect to x.
Click here for
technical information
about our definition.
In the new terminology, marginal demand, MD(q), is the derivative
of the demand function, D(q). Hence, MD(q) = D(q). Likewise, MC(q) =
C(q), MR(q) = R(q), and MP(q) = P(q). In this notation, the units of the
marginal functions will reflect the units of the original functions.
We can compute the approximate value of a derivative, f (x), by
having Excel compute
f (x  h)  f (x  h)
2h
for a very small value of h.
Example 2. Evaluate f (2), if f(x) = x3 + 6. We let h = 0.00001
and approximate f (2) with the difference quotient
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f (x  h)  f (x  h)
2h
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Differentiation.
Differentiation, DerivativesDerivatives: page 3
The following computation can be done with a calculator or with
Excel. If it is likely that we will need to evaluate f (x) for several different
values of x, then Excel is a more efficient tool.
f (x  h)  f (x  h)
2h


f ( 2  0 . 00001 )  f ( 2  0 . 00001 )
2  0 . 00001
f ( 2 . 00001 )  f (1 . 99999 )
0 . 00002
3

2 . 00001  6   1 . 99999

3
6

0 . 00002

14 . 00012  13 . 99988
0 . 00002
 12 . 000
Example 3. Let f(x) = x3 + 6. Use Excel to plot both f(x) and f (x)
over the interval from 3 to 3. The work is shown in the file Example 3.xlsx
and the resulting graph is shown on the following page.
Example 3.xlsx
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E x m p le 3 : f(x ) a n d f'(x )
Differentiation.
Derivatives: page 4
Differentiation, Derivatives
40
30
fu n ctio n
20
va lu es
d eriva tive
10
0
-3
-2
-1
-1 0
0
1
2
3
-2 0
x
Let f(x) = 1  ex/2. Use
Excel and the method of Example 3 to plot both
f(x) and f (x) over the interval from 0 to 10.
Let f(x) = 2x. (i) Use Excel and the method of
Example 3 to plot the graph of f  over the interval [0, 4]. (ii) Use the
interpretation of the derivative as an instantaneous rate of change to explain
the form of your graph.
Example 3.xlsx
Graphing.xlsm
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Differentiation.
Derivatives:
page 5
Differentiation,
Derivatives
The evaluation of difference quotients provides a good understanding
of differentiation, but it is rather tedious. Carrying out this much work, while
in the middle of a business application, could easily distract us from the basic
business problem.
What we are doing is called numerical differentiation. Many
software packages, but (unfortunately) not Excel, have built-in functions that
perform numerical differentiation. To meet this need, we have created an
Excel file Differentiating.xlsm, that serves as a numerical differentiation
utility.
To use Differentiating.xlsm, open the file and follow the instructions
in the blue box. In particular, be sure that Excel is set for Automatic
Calculation. The file works in much the same way as Graphing.xlsm. Once
you have entered the formula for f(x), you can get numerical values for f(x) and
f (x), and can see the graphs of these functions over any interval.
Differentiating.xlsm
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Differentiation.
Differentiation, Derivatives Derivatives: page 6
From now on, unless otherwise indicated, we will assume that all
differentiation is performed with Differentiating.xlsm.
When entering a function in Differentiating.xlsm, you can use the
letters s, t, u, v, or w as constants. If this is done, you must enter values in
Cells O23:O27 for any letters that you use.
For example, suppose that you want to differentiate f(x) = x s, where s
is a number that is in Cell O23. You would enter = x^s in Cell C24.
Accept the function and then click on the Differentiate button at the
right of the plots. Since Excel is set for automatic calculation, entering a new
number in Cell O23 will automatically change the function, redraw the plots,
and recompute the derivative.
Click here for
information on running the
macro in Differentiating.xlsm.
Differentiating.xlsm
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Differentiation,Derivatives:
Derivatives
Differentiation.
page 7
If values are entered in Cells O23:O27; then constants s, t, u, v, or w
may also be used to describe the range, [a, b], of the plot. If these constant
names are used, they must be entered in Cells I24:J24 preceded with "="
signs. For example, =s.
If f '(x) is constant, the displayed plot
in Differentiating.xlsm will be distorted. To correct this,
format the y-axis to have realistic fixed minimum and
maximum values.
Example 4. Use Differentiating.xlsm to redo Examples 1 and 2. Open
Differentiating.xlsm and enter =x^3+6 in Cell C24. Click on  to accept the
function, and then click on the Enter button on the Differentiating toolbar. Enter
2, 3, and 3 in Cells E24, I24, and J24, respectively. The resulting computation
and plot of both the function and its derivative are shown on the following page.
Differentiating.xlsm
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Differentiation.
Derivatives: page 8
function
derivative
FUNCTION & DERIVATIVE
40
Differentiation, Derivatives
30
20
10
f(x)
f '(x)
0
-4
-3
-2
-1
0
1
2
3
4
-10
x
2
-20
-30
Computation
f (x )
f ' (x )
14
12.000
x
Use Differentiating.xlsm
to redo Part i of Exercise 5.
Use Differentiating.xlsm
to redo Part i of Exercise 8.
Differentiating.xlsm
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Differentiation.
Differentiation, DerivativesDerivatives: page 9
Exercises 17, 18, and 19 refer to the situation that was discussed in
Exercises 3 and 14 of Demand, Revenue, Cost, and Profit. The demand and
cost functions for a good were given by D(q) = 0.1q + 150 and
C ( q )  12 , 000  1, 300  q , respectively.
Use Differentiating.xlsm to plot marginal cost, C,
over the interval from 0 to 1,500.
(i) Use Differentiating.xlsm to plot marginal
profit, P over the interval from 0 to 1,500. (ii) Experiment with the
Computation boxes to find a value for q that is greater than 100, and at
which P(q) = 0. (Hint: This value of q might not be an integer.)
(i) Use Differentiating.xlsm to plot marginal
revenue, R over the interval from 0 to 1,500. (ii) Experiment with the
Computation boxes to find a value for q at which R(q) = 0.
Differentiating.xlsm
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Differentiation.
Derivatives: page 10
Differentiation, Derivatives
Exercises 20, 21, and 22 refer to the situation that was discussed in
Exercises 4 and 15 of Demand, Revenue, Cost, and Profit. The demand and
cost functions for audio speakers were given by D(q) = 0.00006q2 + 250
and C(q) = 60,000 + 110q, respectively.
Use Differentiating.xlsm to plot marginal revenue,
R, over the interval from 0 to 2,100.
(i) Use Differentiating.xlsm to plot marginal
profit, P, over the interval from 0 to 2,100. (ii) Experiment with the
Computation boxes to find a value for q at which P(q) = 0. (Hint: This
value of q might not be an integer.)
Use Differentiating.xlsm to plot marginal cost,
C, over the interval from 0 to 2,100. Hint: remember the caution about
using Differentiating.xlsm, when the graph of f  is distorted.
Differentiating.xlsm
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Differentiation, Properties
& Applications
Differentiation.
Properties
and Applications
3. PROPERTIES AND APPLICATIONS
A few basic properties of derivatives can save us time while working
on business problems. Since a derivative represents the instantaneous rate of
change in a function, it is not surprising that multiplying a function, f, by a
constant, a, also multiplies its derivative by the same constant.
If g(x) = af(x), then g(x) = af (x).
Thus, for example, doubling the production costs for a product will
double the marginal cost.
The derivative of the sum or difference of two functions is the sum or
difference of their derivatives.
If h(x) = f(x)  g(x), then h (x) = f (x)  g(x).
This property is often used with marginal profit. If R, C, and P, are
the revenue, cost, and profit functions for a good, then P(q) = R(q)  C(q).
Thus, P(q) = R(q)  C(q).
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Differentiation, P. & A.
Differentiation.
Properties & Applications: page 2
Since the marginal profit is equal to the marginal revenue minus the
marginal cost, the marginal profit is equal to zero exactly when the marginal
revenue is equal to the marginal cost.
Some useful properties of differentiation are clear from the definition
of a derivative. A function, f, is constant on an interval, if f(x) is the same
number for all values of x in the interval. For example, we might have f(x) =
100 for all numbers, x, with 0  x  10. Since the function does not change,
the derivative of a constant function is zero.
Approximately constant demand functions are encountered in the case
of some commodities, such as wheat or corn, where a change in production
level by a single farmer has no effect on the price per bushel for his or her
crop. In such a case the marginal demand is zero.
A function of the form f(x) = mx + b, where m and b are constants, is
called linear. We know that the derivative of f is the sum of the derivatives of
mx and b. Since b is a constant, its derivative is zero. The graph of y = mx is
a straight line, with a slope of m. Thus, the instantaneous rate of change for
mx is always equal to m.
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Differentiation.
Properties & Applications: page 3
Differentiation, P. & A.
Putting these facts together, we see that, if f(x) = mx + b, then f (x)
= m, for all values of x.
Linear functions are often used as models for parts of demand and
cost functions. In this case, there is no need for numerical differentiation. For
example, if the cost function for a good is given by C(q) = 120q + 5,000, then
its marginal cost is C(q) = 120, for all quantities, q.
If f(x) = 0.75x + 1.94, find a formula for f (x).
What is f (5)?
Let f(x) = 3g(x) + 12 and suppose that g(5) = 2.
When 12,000 mountain bicycles are being
produced and sold the marginal revenue is $895 and the marginal cost is
$787. (i) What is the marginal profit at this production level? (ii) What
does this number mean in terms of bicycles and dollars?
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Differentiation.
Properties
& Applications: page 4
Differentiation,
P. & A.
A demand function, D(q) = 0.8q + 200, gives
the price, in dollars, at which q items can be sold. (i) Find a formula for the
marginal demand. (ii) Relate your formula to dollars and items.
The demand function D(q) = 0.00006q2 + 250
for audio speakers has been considered in previous exercises. The
corresponding revenue function is given by R(q) = qD(q) = 0.00006q3 +
250q. (i) Use Differentiating.xlsm to plot R(q) and use Graphing.xlsm to
plot the function g(q) = 0.00018q2 + 250. (ii) What do your plots suggest
about the form of the marginal revenue function?
Use the
formulas for differentiation to explain
what happened in the solution of
Exercise 23.
Since this stuff takes a lot of
effort, you might want to have a
backup plan. Have you considered
OMPHALOSKEPSIS?
Differentiating.xlsm
Graphing.xlsm
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Differentiation.
& Applications: page 5
Differentiation, P. & Properties
A.
So far we have used differentiation for simple numerical problems
and as new terminology for marginal analysis. Many other types of business
problems can be solved with differentiation. As an illustration, we will apply
differentiation to a situation in inventory control.
Example 5. A large computer manufacturer uses 50,000 CD drives at
a uniform rate throughout the year. These are outsourced, and must be ordered
in batches from the supplier. Drives cost $20 each, and there is an additional
$1,200 expense per order for processing, shipping, receiving and handling the
drives. Thus, it is cheaper to order in large batches.
The drives must be kept in inventory from the time that they arrive
until they are used in computers. The estimated cost of keeping each drive
stored is one cent per day.
The cost of maintaining the inventory suggests placing smaller orders,
that will be used up more quickly. Assuming that orders can be placed so that
the drives arrive just in time to be used, how often should the company order a
batch of drives, how many should be ordered, and how much will it cost them,
per year, to order and store the CD drives?
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Differentiation, P. &Properties
A.
Differentiation.
& Applications: page 6
Suppose that C(t) is the total yearly cost if drives are ordered every t
days, starting from the first of the year. Our goal is to find the value of t that
minimizes C(t).
If orders are placed every t days, then there will be 365/t orders per
year, and each order will be for
50 , 000
50 , 000  t
drives. In order to

365
365
t
understand the problem, let It(x) be the number of drives in the company’s
inventory at x days after the arrival of an order. As an example, we will plot
It(x), assuming that the company orders every 30 days, that is for t = 30. In
this case, each order will be for
50 , 000  30
 4 ,110
drives.
365
Note that, on average, there will be one-half the number of parts
that are ordered in storage.
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Differentiation, P.Properties
& A.
Differentiation.
& Applications: page 7
inventory
average
INVENTORY
Drives In Stock
5,000
4,000
3,000
2,000
1,000
0
0
30
60
Day
90
120
In general, if the company orders every t days, the mean number of
drives in the inventory will be
50 , 000  t
365  2
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
25 , 000  t
.
365
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Differentiation.
Properties & Applications: page 8
Differentiation, P. & A.
At one cent per day per drive for storage, it will cost $0.01365 =
$3.65 per year to store a single drive. We can now set up a formula for C(t).
 Cost of   Cost of 
C (t )  


orders
  storage 

 Cost per   Number of   Mean Number of   Cost per 





stored
drives
orders
order
  drive to store 
 
 

50,000  t
  365   25,000  t 

 20   
  1,200 
   3.65 

365
365

  t  


1, 200  365
 50 , 000  20 
25 , 000  3 . 65
365
t

438 , 000
t
 1, 000 , 000  250  t
t
Differentiating.xlsm
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Differentiation. Properties
& Applications:
page 9
Differentiation,
P. & A.
This expression for C(t) is entered into Differentiating.xlsm as f(x).
The resulting plots are shown below. Large scale plots show that f(x) has
only one possible minimum value, and that the minimum seems to occur
when x is close to 40.
DERIVATIVE
FUNCTION
1028000
200
1027000
1026000
0
-200
1025000
f(x) 1024000
1023000
1022000
f '(x) -400
1021000
1020000
-800
0
20
40
60
80
-600
0
20
40
60
80
-1000
x
Differentiating.xlsm
x
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Differentiation, P. &Properties
A.
Differentiation.
& Applications: page 10
Where f(x) is decreasing, its difference quotients are negative, and it
derivative is negative. Where f(x) is increasing, f (x) is positive. Hence, the
minimum value of f(x) occurs where f (x) = 0. This is exactly what is shown
in the graphs.
By experimenting with the values of x in Cell E24 of Differentiating
.xlsm, we find that f (41.8569) = 0.000 and that f(41.8569) = 1,020,928.
Computation shows that 41.8569 days is equal to 41 days, 20 hours, 33
minutes, and 56 seconds. While this is very precise, it is not a highly practical
time interval at which to place orders! The natural plan would be to round this
time to 42 days. Putting 42 into Cell E24 of Differentiating.xlsm, shows that
f(42) = 1,020,929. Since this is only $1 more than the minimum possible value
of f(x), it does no harm to order every 42 days.
Returning to the language of our problem, we have found that the
most economical plan is to order (50,000/365)t  5,753 drives every 42 days.
This will result in a total cost of approximately $1,021,000 per year for
procuring and storing the drives.
Differentiating.xlsm
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Differentiation,
P. & A.11
Differentiation. Properties & Applications:
page
Redo Example 5, assuming that the fixed cost of
an order is $1,000 and that it costs 1.5 cents per day to store a CD drive.
Redo Example 5, assuming that the fixed cost of
an order is $1,500 and that it costs 0.8 cents per day to store a CD drive.
(i) Use your team’s data to plot MR, MC, and
MP for its product. (ii) Prepare computational cells, as in the sheet Marginal
of Marketing Focus.xlsm and use these to answer your team’s Questions 1-4
for its product. Use the same units as in the work on the Class Project in
Marketing Focus.xlsm. Study of the Focus pages will help with this work.
Differentiating.xlsm
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Differentiation, Tangents & Slopes
Differentiation.
Tangents & Slopes
4. TANGENTS & SLOPES
The subject of differentiation is much more sophisticated than we
have indicated in our presentation. However, the numerical methods that we
have introduced provide good intuition, and are adequate for a wide range of
simple business problems. Differentiation is one of the main components of
the mathematical subject, calculus.
There is a geometric interpretation of the derivative that often helps in
the understanding of business graphs. Recall that the slope of a straight line is
defined to be the change in height divided by the horizontal change between
any two points on the line. We would like to extend this notion to define
“slope” for the graph of a function f at a single point (x, f(x)).
Example 6. Let f ( x )  x and consider the point (1, f(1)) = (1, 1).
Moving an increment h units to the left and to the right of (1, 1), locates the
points (1  h, f(1  h)) and (1 + h, f(1 + h)) on the graph of f. The straight line
connecting these points is the secant line determined by the increment h.
* This section is not needed for the study of
other material in Mathematics for Business Decisions.
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Differentiation,
T. & S. & Slopes: page 2
Differentiation.
Tangents
Notice that the slope, mh, of the blue secant line is given by the
difference quotient
mh 
f (1  h )  f (1  h )
2h
SECANT & TANGENT LINES
1.5
f(x)
Tangent
Secant
1.0
(1, f(1))
yy
0.5
2h
0.0
.
The red line
that we will call tangent
to the graph of f at the
point (1, f(1)) has a slope
which is the limit of the
slopes of the secant
lines, as the increment,
h, approaches 0. That is,
lim f (1  h )  f (1  h ) .
2h
h 0
f(1 + h)  f(1  h)
0.0
0.5
1.0
1.5
2.0
x (material continues)
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Differentiation,
T. & S.
Differentiation.
Tangents
& Slopes: page 3
f (1  h )  f (1  h )
, we see that the derivative,
Since f ' (1)  lim
h 0
2h
f (1), is the slope of the tangent line at (1, f(1)).
The Excel file Slope.xlsx plots both the tangent and secant lines for
selected values of h. Open that file and experiment with different increments.
For relatively small values of h, the blue secant line appears to coincide with
the red tangent line.
What we have just learned about the tangent line for f ( x )  x at the
point (1, f(1)) is also true for any differentiable function. If f(a) exists, then
the slope of the tangent line to the graph of f at the point (a, f(a)) is defined
to be f(a). For this reason, f(a) is called the slope of the graph of f at the
point (a, f(a)). This is often shortened to “the slope of f at a.”
In practical situations, the slope, f (a), of f at (a, f(a)) is interpreted as
the instantaneous rate of change of f(x) at a. This follows from the fact that
the slope mh of a secant line is the average rate of change in f(x) over the
interval from a  h to a + h. The limit of this average rate, as the increment, h,
approaches 0, is the instantaneous rate of change.
Slope.xlsx
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Differentiation,
T. & S.
Differentiation.
Tangents
& Slopes: page 4
Since we know the slope of the tangent line and have one point on
that line, we can determine its equation. If f (a) exists, then the tangent line
to the graph of f at the point (a, f(a)) has the equation
y = f (a)(x  a) + f(a).
When using the equation for a tangent line it is
important to distinguish between a and x. a is the first coordinate of the point
at which the line is tangent. x is simply the independent variable in the
equation of the tangent line.
Example 7. Find the slope of the graph of f(x) = x2  4x + 4 at the
point (3, f(3)), and find an equation for the tangent line at that point.
Differentiating.xlsm shows that f (3) = 2. Hence, the slope of f is 2
at x = 3. This means that f(x) is increasing at an instantaneous rate of 2 units
per unit increase in x, when x = 3. In the tangent line equation, we have a = 3,
and f(a) = f(3) = 32  43 + 4 = 1.
Differentiating.xlsm
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Differentiation,
T. & S.
Differentiation. Tangents
& Slopes:
page 5
TANGENT LINE
4
Tangent Line
y  f ( a )  ( x  a )  f ( a )
function
tangent line
3
 2  ( x  3)  1
 2x 5
2
yy
1
0
0
1
2
3
4
5
-1
x
(i) Find the slope of the graph of f(x) = 2x at the
point (1, f(1)), and (ii) find an equation for the tangent line at that point. (iii)
Use Excel to show both the graph of f and the graph of the tangent line in a
single plot. Plot both functions over the interval [0, 3].
Excel
Differentiating.xlsm
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Differentiation, T. Tangents
& S.
Differentiation.
& Slopes: page 6
(i) Find the slope of the graph of the natural
logarithm function, f(x) = ln(x) at the point (2, f(2)), and (ii) find an equation
for the tangent line at that point. (iii) Use Excel to show both the graph of f
and the graph of the tangent line in a single plot. Plot both functions over the
interval [1, 5].
Let f(x) = 1 + x/2. (i) Find the slope of the graph
of f at the point (4, f(4)), and (ii) find an equation for the tangent line at that
point. (iii) Use Excel to show both the graph of f and the graph of the
tangent line in a single plot over the interval [0, 6]. (iv) Use the formulas for
differentiation to explain what happened in Part iii.
Many formulas have been developed which
allow us to find the derivative of a function symbolically, without the need
for numerical approximation. These will not be needed in our work with
differentiation, but they might be of interest to you. To see a few examples,
click on symbolic.
Excel
Differentiating.xlsm
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Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
Drives
on the project
How can differentiation
help us price 12-GB drives?
We must consider the effect of
the units that we are using. Recall; from the
sheet Functions in Marketing Focus.xlsm and
from the Focus pages of the section Demand,
Revenue, Cost, and Profit; that C(q) gives the
cost, in millions of dollars for producing q
thousand drives. C'(q) gives the rate of change
in C(q), with respect to q.
Differentiation,
Focus
Marketing
Focus.xlsm
Class Project
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continues)
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Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
on the
project
Drives
Hence, C'(q) gives marginal cost in millions of dollars per
thousand drives, at a production level of q thousand drives.
From our variable cost data, we see that the marginal cost for q
thousand drives is $162 per drive if q is less than or equal to 800; $128 per
drive if q is greater than 800, but less than or equal to 1,200; and $72 per
drive for q greater than 1,200. These values give marginal cost in dollars
per drive at a production level of q thousand drives.
How does this relate to the derivative? 1,000,000C'(q) is in dollars
per thousand drives, and (1,000,000C'(q))/1,000 is in dollars per drive. The
last quantity simplifies to 1,000C'(q). Hence, where C is differentiable,
1,000C'(q) gives the marginal cost, MC(q), in dollars per drive, at a
production level of q thousand drives.
Differentiation,
Focus
Marketing
Focus.xlsm
Class Project
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continues)
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Marketing
Calculus, Mathematics,
Tests, Homework, Computers
For 12-GB drives, R(q) gives the revenue, in
millions of dollars,
Computer
from
saleproject
of q thousand drives, and R'(q) gives the rate
of change in
on the
the
Drives
R(q), with respect to q. Hence, R'(q) gives marginal revenue in millions of
dollars per thousand drives, at a production level of q thousand drives.
Since our data for marginal profit occurs naturally in dollars per
drive, we will us these units for all 12-GB drive marginal functions.
(1,000,000R'(q))/1,000 is in dollars per drive. This simplifies to
1,000R'(q). MR(q) = 1,000R'(q) is the marginal revenue, MR(q), in
dollars per drive, when q thousand drives are being sold.
The functions MR(q), and MC(q) are plotted in the sheet Marginal
of Marketing Focus.xlsm. These graphs are also shown on the next page.
Note that the marginal revenue is positive where the revenue function is
increasing and negative where revenue is decreasing. Since cost increases
with quantity, marginal cost is always positive.
Differentiation,
Focus
Marketing
Focus.xlsm
Class Project
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continues)
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Revenue & Cost Functions
$500
Revenue
Cost
Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
Drives
$300
$200
$100
$0
Margin al R even u e an d C ost
$600
M R(q)
on the project
0
400
800
1,200
1,600
$400
2,000
$ P er D rive
(M's)
$400
2,400
q (K's)
M C (q)
$200
2,800
$0
- $200
0
400
800
1,200
1,600
2,000
2,400
2,800
- $400
- $600
q (K 's)
P(q) = R(q)  C(q) gives the profit, in millions of dollars, from the
sale of q thousand drives. Since P(q) = R(q)  C(q), we know that, in
dollars per drive,
MP(q) = MR(q)  MC(q)
= 1,000R'(q)  1,000C'(q)
= 1,000(R'(q)  C'(q))
= 1,000P'(q).
Differentiation,
Focus
Marketing
Focus.xlsm
Class Project
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P rofit F u n ction
$70
$60
Margin al P rofit
$400
$40
$30
$20
$10
$0
- $10 0
- $20
on the project
400
800
1,200
1,600
M P (q ) $ P er D rive
P (q ) (M 's)
$50
Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
Drives
$200
$0
0
2,000
- $200
400
800
1,200
1,600
2,000
q (K 's)
- $400
q (K 's)
Since, in dollars per drive, MP(q) = 1,000P'(q), we see that MP(q)
and P'(q) always have the same sign, and are equal to zero at exactly the
same values of q. Hence, MP(q) = 0 if and only if R (q) = C (q).
The above graphs are copied from the sheet Marginal in Marketing
Focus.xlsm. The changes in cost per drive at q = 800 and q = 1,200
thousand drives cause drops in the plot of MC(q) and produce corresponding
jumps upward in MP(q). Due to the second gap, marginal profit crosses the
axis in two places. These reflect the two local maxima, or “humps”, in the
profit function.
Differentiation,
Focus
Marketing
Focus.xlsm
Class Project
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continues)
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From the plot of marginal profit, it appears that MP(q) = 0 when q
is approximately equal to 1,150 or 1,250. Looking at the plot of profit, it
seems likely that profit has local maxima at the zeros Marketing
of MP.
Calculus, Mathematics,
Tests, Homework, Computers
Computer
Cells B107:I107 of the sheet Marginal in Marketing
Focus.xlsm
compute
values
for all of our functions, when a value of Drives
q is entered in
on the
project
B107. Experimentation with different values of q produces the following
results.
q (K 's)
D (q )
1138.96
$305.96
q (K 's)
D (q )
1262.27
$285.88
R (q ) (M 's) M R (q ) $/d rive C (q ) (M 's) M C (q ) $/d rive P (q ) (M 's) M P (q ) $/d rive
$348.478
$128.00
$306.387
$128.00
$42.092
$0.00
R (q ) (M 's) M R (q ) $/d rive C (q ) (M 's) M C (q ) $/d rive P (q ) (M 's) M P (q ) $/d rive
$360.860
$72.00
$318.683
$72.00
$42.176
$0.00
As we have guessed from the graphs, the local maximum at the
larger value of q is greater than at the smaller value of q.
According to our model, Card Tech would earn a maximum profit
of $42,176,000 from selling 1,262,270 12-GB drives, priced at $285.88.
This answers Questions 1, 2, and 3 from the original discussion of our Class
Project.
Differentiation,
Focus
Marketing
Focus.xlsm
Class Project
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continues)
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Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
We will answer Question 4 for our Class Project, by investigating
on
the
project
Drives
the sensitivity of the maximum profit to changes from the
optimal price.
Numerical experimentation in Cells B107:I107 of the sheet Marginal in
Marketing Focus.xlsm shows that raising the price per drive by $2 to
$287.88 would lower the number of drives sold to 1,250,431, and decrease
the profit to 42.143 million dollars. This is a drop of only 0.08% from the
maximum possible profit.
Similar work shows a 0.08% drop in profit if the price per drive is
lowered to $283.88. Clearly, profit is not very sensitive to $2 changes in
price.
On the other hand, a $10 change in price would cause a noticeable
effect on profits. Raising the price per drive to $295.88, drops the profit by
2.01% and lowering the price to $275.88, drops the profit by 1.91%.
Differentiation,
Focus
Marketing
Focus.xlsm
Class Project
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continues)
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on the project
Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
Drives
Since profit is not very sensitive to small changes in price, Card
Tech would probably adjust the “optimum” price of $285.88 to a more
common value, such as $284.99 or $285.99. Experimentation in the sheet
Marginal of Marketing Focus.xlsm, shows that these could be expected to
reduce the profit by only 0.02%.
WHAT SHOULD YOU DO?
Each team should now plot MR, MC, and MP for its product and
prepare computational cells, as in the sheet Marketing. Use these to answer
your team’s Questions 1-4 for its product. Use the same units as we have
used in the Class Project.
Differentiation,
Focus
Marketing
Focus.xlsm
Class Project
(material
continues)
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Using Solver, Excel’s
Solver
Using
Solver.
Excel’s Solver
1. EXCEL’S SOLVER
The utility Solver is one of Excel’s most useful tools for business
analysis. This allows us to maximize, minimize, or find a predetermined value
for the contents of a given cell by changing the values in other cells.
Moreover, this can be done in such a way that it satisfies extra constraints that
we might wish to impose.
Example 1. The size limitations on boxes shipped by your plant are
as follows. (i) Their circumference is at most 100 inches. (ii) The sum of their
dimensions is at most 120 inches. You would like to know the dimensions of
such a box that has the largest possible volume. Let H, W, and L be the
height, width, and length of a box; respectively; measured in inches. We wish
to maximize the volume of the box, V = HWL, subject to the limitations that
the circumference C = 2H + 2W  100 and the sum S = H + W + L  120.
This problem is set up in the Excel file Shipping.xlsx. We will
outline its solution with screen captures and directions. First, enter any
reasonable values for the dimensions of the box in Cells B7:D7.
Shipping.xlsx
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Using
Solver, Solver
Using Solver.
Excel’s
Solver: page 2
FRAGILE
Crush slowly
H
L
To use Solver, click on Data, then
Solver in the Analysis box. In older
versions of Excel select Tools in the main
Excel menu, then click on Solver.
Enter cell that computes volume.
Select Max.
Enter cells that contain
dimensions
Click on Add.
Computer Problem?
Shipping.xlsx
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Using Solver,
Solver Solver: page 3
Using Solver.
Excel’s
The requirement that the circumference be at most 100 inches is
called a constraint. We want to have the contents of Cell E7 be at most 100.
Enter cell that computes circumference.
Select <=.
Click on OK.
Enter the limiting number.
Repeat the above process to add the constraint F7 <= 120, then click
on Solve.
Shipping.xlsx
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Using Solver,
SolverSolver: page 4
Using Solver.
Excel’s
Click on Solve.
Click on Keep Solver Solution.
Click on OK.
Shipping.xlsx
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Using Using
Solver.
Excel’s Solver: page 5
Solver, Solver
The dimensions that maximize
volume are now shown in Cells B8:D8. The maximum volume, the value of
the circumference and the sum of the dimensions are now displayed. For a
maximum volume of 43,750 cubic inches, the box should be 25 inches high,
25 inches wide, and 70 inches long.
B
6
7
C
D
E
H (height) W (width) L (length) C (circumference)
25
25
70
100
F
G
S (sum)
120
V (volume)
43750
In rare cases; such as very large or small initial values of H, W, or L;
you may need to add the constraints B7 >= 0, C7 >= 0 and D7 >= 0.
Suppose that the shipping company in Example 1
requires that each dimension of a box must be at least 3. Note that the
2
2
2
longest item which can be shipped in a box has a length of H  W  L
(i) Modify Shipping.xlsx and use Solver to find the dimensions of an
allowable box that will accept the longest possible item. (ii) What is the
maximum length of such an item?
Shipping.xlsx
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Using Using
Solver.
Excel’s Solver: page 6
Solver, Solver
Big Box shipping company limits the
circumference of its boxes to at most 80 inches with the sum of their
dimensions to be at most 150 inches. (i) Modify Shipping.xlsx and use
Solver to find the dimensions of the box with maximum volume that will be
accepted by Big Box. (ii) What is the maximum volume of a box which
Big Box will ship?
Rush! shipping company limits the size of the
boxes that it accepts by limiting their volume to at most 16 cubic feet
(27,648 cubic inches). For it to ship a box, each dimension must be between
3 and 54 inches. (i) Modify Shipping.xlsx and use Solver to find the
dimensions of a Rush! box which will accept the longest possible item.
Hint: Use different initial values for each dimension. (ii) What is the
maximum length of such an item? Note that the longest item which can be
shipped in a box has a length of
H
Shipping.xlsx
2
W
2
2
L .
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Using Solver.
Excel’s
Solver: page 7
Using Solver,
Solver
Example 2. Return to the restaurant management example that has
been studied in Demand, Revenue, Cost and Profit; and Differentiation. We
will consider the demand, profit, and marginal profit functions; D(q), P(q), and
MP(q); that were developed in those sections. Recall that a restaurant chain is
planning to introduce a new buffalo steak dinner. D(q) is the price, in dollars,
at which q dinners per week would be ordered. P(q) is the profit from selling
q dinners priced at D(q) dollars and MP(q) =P(q) is the marginal profit.
What is the minimum number of dinners that the restaurants would
have to sell in order to make a profit?
In the sheet Solver of Dinners.xlsm we have plotted D(q), P(q), and
P(q); and have built computation boxes that evaluate the functions at q.
Referring to the graph in that sheet we see that P(q) is first positive when q is
approximately equal to 700. Solver is now used to find a value close to 700, in
Cell G11 that makes Cell I11 equal to 0.
Dinners.xlsm
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Using Solver,page
Solver 8
Using Solver. Excel’s Solver:
G
Dinners.xlsm
q
700
H
I
Computation
D (q )
P (q )
$29.10
$178.58
MP (q )
$17.0095
q
689.51027
Computation
D (q )
P (q )
$29.13
$0.00
MP (q )
$17.0382
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Using Solver,page
Solver 9
Using Solver. Excel’s Solver:
Solver finds that 689.51027 dinners would be needed to produce a
non-negative profit. Since this makes no sense in terms of actual meals, the
restaurants would need to serve at least 690 buffalo steak dinners in order to
make a profit. Recomputing
Computation
the Excel sheet shows that this
q
D (q )
P (q )
MP (q )
would still require a price of at
690
$29.13
$8.34
$17.0369
least $29.13 per dinner.
Different initial values for q may cause Solver to
produce slightly different values in the target cell.
Three questions are of great interest to the restaurant managers. (i)
How should the buffalo steak dinners be priced in order to obtain the
maximum profit? (ii) How many dinners per week could they expect to sell at
that price? (iii) What maximum profit can they expect?
We can use Solver and the computation boxes in the sheet Solver of
Dinners.xlsm to answer all of these questions. We start with a guess value of
2,100 for q, and then maximize P(q).
Dinners.xlsm
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Solver, Solver
UsingUsing
Solver.
Excel’s Solver: page 10
C om p u tation
D (q )
P (q )
q
2,024.8609
Dinners.xlsm
$22.21
$14,052.02
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M P (q )
$0.0000
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Using
Solver, Solver
Using Solver. Excel’s
Solver:
page 11
P(q) has a maximum value when q = 2024.8609. As expected, we
note that the marginal profit, MP(q), is equal to zero when P(q) is at a
maximum.
Customers do not order 0.8609 dinners! Hence, q = 2024.8609 is not
a practical answer. If we add the constraint that Cell G11 is to be an integer,
then Solver can find the whole number of dinners which will maximize profit.
q
2025
Dinners.xlsm
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Computation
D (q )
P (q )
$22.21
$14,052.02
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MP (q )
-$0.0029
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Solver, Solver
Using Solver.Using
Excel’s
Solver: page 12
This completes our work with the restaurant
example. When we first considered the problem in
Demand, Revenue, Cost, and Profit; we used graphs
to estimate that the restaurant chain should prepare
around 2,000 buffalo steak dinners per week and price
If Solver can
them at approximately $22, expecting a weekly profit
maximize or minimize
of around $14,000. In Differentiation, we used
any function, why
marginal analysis to refine these estimates,
should I use derivatives
concluding that they should expect to sell 2,025
for optimization?
buffalo steak dinners per week, priced at $22.21, for a
total profit of $14,052. Finally, Solver has given us a
much easier way to obtain the same information.
Although such precision is clearly unnecessary, we could use Solver
to find the expected profit from more realistic pricing. 2,059 dinners, priced at
$21.95 would yield an expected weekly profit of $14,040.
Entrée
Buffalo Steak Dinner . . . only $21.95
Dinners.xlsm
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Ha, ha, ha!!!
Solver, Solver
UsingUsing
Solver.
Excel’s Solver: page 13
Use Solver to find the maximum
number of dinners that the restaurant chain in our example
could sell and still make a non-negative profit.
Oh, Oh.
Use Solver to find the expected
maximum profit if the buffalo dinners are priced at (i)
$19.95 or (ii) at $24.95.
(i) Use Solver to determine the optimum number of
buffalo dinners for the restaurant chain, by finding where MP(q) = 0. (ii)
Why do you think that Solver returns a non-integer value for q, even if a
constraint is added, requesting only integer values?
Dinners.xlsm
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Using
Solver.
Using Solver,
Solver Excel’s Solver: page 14
(i) Use Solver to find the number qrev of buffalo
dinners that will produce the maximum revenue in the restaurant example.
(ii) What price should be put on the dinners in order to sell qrev of them? (iii)
What maximum revenue can be expected?
Consider the situation that was discussed in
Demand, Revenue, Cost, and Profit. The demand and cost functions for a
good were given by D(q) = 0.1q + 150 and C ( q )  12 ,000  1, 300  q ,
respectively. (i) Use Differentiating.xlsm to plot profit, P, and marginal
profit, P, over the interval from 0 to 1,500. (ii) Use Solver to find a value
for q that is greater than 100, and maximizes profit. (iii) Use Solver to find a
value for q that is greater than 100, and at which P(q) = 0.
Refer to the good whose demand function is given
in Exercise 8. (i) Use Differentiating.xlsm to plot revenue, R, and marginal
revenue, R, over the interval from 0 to 1,500. (ii) Use Solver to find a value
for q that maximizes revenue. (iii) Use Solver to find a value for q at which
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R(q) = 0.
Dinners.xlsm
Differentiating.xlsm
Excel
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Using Solver,
Solver Excel’s Solver: page 15
Using
Solver.
Consider the situation that was discussed in
Demand, Revenue, Cost, and Profit. The demand and cost functions for a
certain model of audio speaker are D(q) = 0.00006q2 + 250 and C(q) =
60,000 + 110q, respectively. (i) Use Differentiating.xlsm to plot profit, P,
and marginal profit, P. (ii) Use Solver to find a value for q that maximizes
profit. (iii) Use Solver to find a value for q at which P(q) = 0.
Refer to the audio speakers whose demand
function is given in Exercise 10. (i) Use Differentiating.xlsm to plot revenue
and marginal revenue. (ii) Use Solver to find a value for q that maximizes
revenue. (iii) Use Solver to find a value for q at which R(q) = 0.
In Differentiation, we considered the function C(t)
that gives the total yearly expense of purchasing and storing CD drives, if
drives are ordered every t days, starting from the first of the year.
C (t ) 
438 , 000
 1, 000 , 000  250  t
t
Use Solver to find the value of t that minimizes C(t).
Differentiating.xlsm
Excel
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Using Solver.
Excel’s
Using Solver,
Solver Solver: page 16
Consider the Class
Project on pricing 12-GB computer drives. (i)
Use Solver to find the number q0 of drives, in
thousands, such that MR(q0) = 0. (ii) What price
should be put on the drives in order to sell q0
drives? (ii) What profit would result from selling
q0 drives?
(i) Use Solver and your
team’s data to check your earlier numerical
results. (ii) Modify your project files and use
Solver to answer your team’s Questions 6-9.
Study of the Focus pages will help with this
work.
Marketing Focus.xlsm
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Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
Drives
on the project
How can Solver help us
price 12-GB drives?
We will apply Solver to the
computation table in Rows 106 and 107 of the
sheet Marginal in the Excel file Marketing
Focus.xlsm. Starting with initial values of q =
600 and 1,700, and finding values of q that
make P(q) = 0, we see that a positive profit
will result from selling between 641.462
thousand and 1,666.152 thousand drives.
Using Solver, Focus
Marketing Focus.xlsm
Class Project
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Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
With an initial value of 1,300 for q, Solver finds a maximum value
ofon
$42,176,000
for P(q). Changing the cell formatting toDrives
display three
the project
decimal places shows that maximum profit results from the production and
sale of 1,262,120 drives. Since D(1,262.120) = $285.91, the theoretical
optimum price for the drives is $285.91.
It is interesting to note that the lack of sensitivity of profit to price
results in rather unstable results when Solver maximizes. The prices and
quantities, but not the maximum profit, often depend upon the initial value of
q.
Marginal analysis provides more accurate results. The
computation table in the sheet Marginal shows that MP(1,262.120) = $0.07
per drive. Since there is a positive marginal profit at 1,262.120 thousand
drives, the maximum profit will actually occur at a slightly higher production
level.
Using Solver, Focus
Marketing Focus.xlsm
Class Project
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Taking an initial value of 1,300 for q and using Solver to find q such
that MP(q) = 0, we find a sales level of 1,262.274 thousand drives. These
should be priced at $285.88, and their sale will produce
a profit of 42.176
Marketing
million dollars. We will use 1,262,274
drives
as our optimum
marketing
Calculus,
Mathematics,
Tests,
Homework, Computers
Computer
level.
on the
project
Drives
In order to duplicate these results, you may need to click on
Options in the Solver window, and reset Precision to 0.000001.
The values obtained with Solver are in very close agreement with
the results that we found by numerical experimentation in the Focus section
of Differentiation. The use of Solver can increase accuracy and save a
tremendous amount of time by eliminating the need for graphical estimation
and numerical experimentation.
Differentiation and Solver do have limitations, and cannot be used
to replace thought and understanding. For example, the change in marginal
cost at a production level of 1,200 thousand drives gives P(q) a local
minimum at exactly 1,200. However, the graph of P(q) is not smooth at
1,200. Since it has an angled ”corner,” P(1,200) does not exist. If Solver is
constrained to look near 1,200, it cannot find a value of q such that P(q) =0.
Using Solver, Focus.xlsm
Focus
Marketing
Class Project
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Due to the lack of sensitivity of profit to production level, Solver
can only approximate the level that produces the local minimum 1,200. If it
Marketing
is constrained to look only between 1,190
and
1,210,
given
an initial value of
Calculus, Mathematics,
Tests, Homework, Computers
Computer
1,201, and asked to find a value of q that minimizes P(q);
Solver returns
1,199.993.
on the project
Drives
Question 6 in the description of the Class Project asks us to find the
profit that Card Tech could expect, if they price the drives at $299.99. We
use Solver in the computation table at the bottom of the sheet Marginal.
This shows that D(q) = $299.99, when q = 1,176.693, and that the profit from
selling 1,176,693 drives will be $41,779,000.
Questions 7 and 8 from the Class Project ask how much Card Tech
should pay for an advertising campaign that would increase demand for the
12-GB drives by 10% at all price levels. We go to the sheet Functions in
Marketing Focus.xlsm. Setting the Demand Factor in Cell F11 at 1.10
increases the sales in all of the test markets by 10%. When the sheet is
recalculated, new coefficients for the trend line are displayed in the plot of
the demand data. These must by copied into Cells H28:H30.
Using Solver, Focus
Marketing Focus.xlsm
Class Project
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We now return to the computation table at the bottom of the sheet
Marginal and use Solver to find a value for q such that
MP(q) = 0. Starting
Marketing
Mathematics,
Computers
with an initial value of q = 1,390, theCalculus,
optimum
strategy is toTests,
sellHomework,
1,388,590
Computer
drives, priced at $285.88 per drive. This will result in a maximum profit of
on the project
Drives
$$69,192,000.
The increase in profit due to the 10% increase
in demand is
$69,192,000  $42,176,000. These values are rounded to the nearest
thousands of dollars. Computation in Excel with unrounded values shows
that the difference is $27,015,000, rounded to thousands. Hence, Card Tech
should pay no more than $27,015,000 for the advertising campaign. It is
interesting to note that the optimum selling price per drive remained
unchanged by the change in demand.
Question 9 in the description of the Class Project asks if it would
be wise for Card Tech to put $15,000,000 into training and streamlining
which would reduce its non-fixed, variable costs by 7%. To answer this, we
open the sheet Functions of Marketing Focus.xlsm and enter 0.93 as the
Marginal Cost Factor in Cell G102. When the sheet is recalculated, this
reduces all of the marginal costs by 7%.
Using Solver, Focus
Marketing Focus.xlsm
Class Project
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Jump start your business career!
Click
on
Marketing
Calculus, Mathematics,LEVEL
Tests,
Homework, Computers
EXECUTIVE
THINKING.
Computer
Solver
can now be used with MP in the computation
table of the
on the
project
Drives
sheet Marginal. With q = 1,280 as an initial value, we find a maximum
profit of $$55,061,00. Since this is up $12,885,000 from the present
maximum profit of $42,176,000, Card Tech would not expect to recover the
$15,000,000 cost for training and streamlining. This makes the investment
look unwise.
WHAT SHOULD YOU DO?
Study the file Marketing Focus.xlsm and all of the earlier Focus
sections. When you have a clear understanding of the marketing process,
perform similar computations with your team’s data. In particular, each team
should now use Solver to answer Questions 1 - 4 and 6 - 9 in its team
project.
Using Solver, Focus
Marketing Focus.xlsm
Class Project
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Integration, Area Under A Curve
Integration.
Area Under A Curve
1. AREA UNDER A CURVE
There is a very convenient way to picture the revenue that results
from selling q items with a demand function D for a good. Since R(q) is the
product of q and D(q), it is the area of a rectangle that is q units wide and D(q)
dollars high. It is easy to picture such a rectangle under the graph of the
demand function.
Demand Function
D(q)
Revenue
D(q)
q
-1.2
-10
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Integration.
Integration,Area
Area Under A Curve: page 2
Since q is a number of items and D(q) is the price in dollars per item,
the area of any region under the graph of the demand function represents
income from sales. The unit of such an area is (items)(dollars/items) =
dollars. In particular, the area of the entire region under the graph of the
demand function and over the horizontal axis represents the total possible
revenue for the good.
The total possible revenue is the money that the producer would
receive if everyone who wanted the good, bought it at the maximum price
that he or she was willing to pay. This is the greatest possible revenue that a
seller or producer could obtain
when operating with a given
Demand Function
demand function.
Total Possible
Revenue
-1.2
-8
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Integration.
Integration,Area
Area Under A Curve: page 3
We have seen that our revenue from selling a good at a single price
that must be paid by all buyers is represented by the area of a rectangle that is
part of the region under the graph of the demand function. The remaining
space under the graph of the demand function is partitioned into two other
regions. These represent income that is lost from selling at a single price.
Some buyers would have been willing pay a higher price for the good
than we charged. The total extra amount of money that people who bought the
good would have paid is called the consumer surplus. It is represented by the
area of the region under
the graph of the demand
Demand Function
Consumer
function and above the
Surplus
revenue rectangle.
D(q)
Revenue
Not
Sold
-1.2
-8
q
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Integration.
Integration,Area
Area Under A Curve: page 4
Fixing a single price means that some people can buy the good for
less money that they would have been willing to pay. It also means that some
potential customers do not buy the good, because they feel that the price is too
high. The total amount of this lost income, which we will call not sold, is
represented by the area of the region under the graph of the demand function to
the right of the revenue rectangle.
We learned how to find the single price that maximizes the revenue
that we receive from selling a good. This is an important tool. However, the
skilled marketer is always looking for creative ways to claim some of the lost
income that is represented by the other regions under the graph of the demand
function. The first step in this direction is computing the dollar values of the
total possible revenue, the consumer surplus, and the not sold lost income.
Example 1. We revisit the problem of pricing buffalo steak dinners
that was considered in several earlier sections. The restaurants’ demand
function is given by D(q) = 0.0000018q2  0.0002953q + 30.19. Taking q =
2,300 dinners as an example, we find that D(2,300) = $19.99.
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Integration.
Integration,Area
Area Under A Curve: page 5
If the restaurants priced buffalo steak dinners at $19.99, they could
expect to sell 2,300 dinners per week, and take in revenue of approximately
($19.99)(2,300) = $45,977.
$32
$24
$19.99
D(q) = 0.0000018q2  0.0002953q + 30.19
Consumer
Surplus
Demand Function
$16
q = 2,300
$8
D(2,300) = $19.99
Revenue
$45,977
Not
Sold
$0
0
1000
2000
3000
4000
5000
2,300
It appears that we will realize only a little more than half of the total
possible revenue, if we sell at the single price of $19.99.
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Integration,
Area
Integration. Area Under
A Curve:
page 6
Watch the effect of different
prices on the various regions. Click
anywhere in the plot to run the
animation. See Using The Computer
Text for information on controlling a
media animation.
It appears that the
maximum revenue occurs
when approximately 2,328
buffalo steak dinners are
sold, priced at $19.75. This
gives rough graphical
confirmation of our earlier
computation.
Computer Problem?
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Integration.
Area
Integration,
Area Under A Curve: page 7
Experimentation with the
animation on the preceding page finds that the approximate maximum
revenue from buffalo steak dinners is $45,972. This comes from the sale of
2,328 dinners. At that quantity, the consumer surplus is shown as $15, 945
and the lost revenue from dinners not sold is $18,082. Adding the areas of
these three regions, we find that the total possible revenue is $45,972 + $15,
945 + $18,082 = $79,999. Notice that, with single price selling, we can claim
only $45,972/$79,999  57% of the total possible revenue.
In Differentiation we found that the maximum profit
from the sale of buffalo steak dinners resulted from the sale of 2,025 dinners,
priced at $22.21 per dinner. (i) Experiment with the animation on the
preceding page to approximate the consumer surplus, revenue, and lost
income from not sold dinners. Hint: the quantity 2,007 is closest to 2,025 in
the animation. (ii) Use your results from Part i to approximate the total
possible revenue. (iii) What percentages of the total possible revenue are
accounted for by consumer surplus, revenue, and lost income from not sold
dinners?
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Integration.
Integration,Area
Area Under A Curve: page 8
In order to deal effectively with marketing problems we will have to
learn how to compute the areas of regions under the graphs of demand
functions. This is only one of many practical business applications that
involve the computation of areas.
As a second example, recall from your work in Part 1 of
Mathematics for Business Decisions that probability information about a
continuous random variable, X, is carried by its probability density function,
fX. Specifically, P(a  X  b) is given by the area under the graph of fX and
over the interval [a, b].
fX
Area of region A
= P(a  X  b)
A
a
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Integration.
Integration,Area
Area Under A Curve: page 9
Computing probabilities for continuous random variables (such as
time and money), defining and computing the mean of a continuous random
variable, computing the value of an income stream, and computing consumer
surpluses; all involve finding the areas of regions that are bounded by the
graphs of functions. Integration is the mathematical tool that helps us make
these computations.
The development of integration is a major undertaking, which will
require time and effort. However, there is no way to acquire the quantitative
information necessary for many business decisions without this important tool.
We will start with a very simple problem. What is the area of the
region R that is enclosed between the x-axis and the graph of
f(x) = 2x  x2/2,
for x between 1 and 4?
The first step is to graph the function. Try this with Graphing.xlsm
and compare your plot with the following.
Graphing.xlsm
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Integration.
Under A Curve: page 10
Integration, Area
Area
3
2
2
1
2
R
f(4) = 24  42/2 = 8  8 = 0
1.5
1
f(1) = 21  12/2 = 2  1/2 = 1.5
4
We should never start a computation without having some idea of the
expected results. The most common of all area formulas states that the area of
a rectangle is the product of its height times its width.
Note that the region R is completely contained in a rectangle of height
2 and width 3. Hence, the area of R is less than 6.
R completely contains a rectangle of height 1.5 and width 2. Hence,
the area of R is greater than 3.
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Integration.
Area
Integration,
Area Under A Curve: page 11
We can improve on our crude estimates by using rectangles in a
different way. The interval from 1 to 4 can be divided into 6 subintervals, each
of width 0.5. Next, we form rectangles over the subintervals. Since we want
the rectangles to approximate the area of R, it seems reasonable to cut each one
off at the place where the graph of f crosses the middle of its width.
2
1
1
1.5 2 2.5 3 3.5 4
As is always the case,
the introduction of mathematical
notation simplifies our
discussion. The points 1, 1.5, 2,
2.5, 3, 3.5, and 4 will be named
x0, x1, x2, x3, x4, x5, and x6. Let x
= 0.5, which is the width of each
subinterval.
Denote the midpoints of the 6 subintervals by m1, m2, m3, m4, m5, and
m6. Since m1 is the midpoint of [1, 1.5], we have m1 = (1 + 1.5)/2 = 1.25.
Likewise, m2 = 1.75, m3 = 2.25, m4 = 2.75, m5 = 3.25, and m6 = 3.75.
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Integration, Area Area Under A Curve: page 12
Integration.
2
f (m1)
f (x)  2  x 
x
2
2
1
f ( m1 )  2  m1 
{
4
1 m1 m2 m3 m4 m5 m6
x0 x1 x2 x3 x4 x5 x6
m1
2
2
 2  1 . 25 
1 . 25
2
2
 2 . 5  0 . 78125
x
 1 . 71875
Similar calculations allow us to compute f(mi) for i = 2, 3, , 6.
i
1
2
3
4
5
6
mi
1.25
1.75
2.25
2.75
3.25
3.75
f (m i )
1.71875
1.96875
1.96875
1.71875
1.21875
0.46875
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Integration.
Area Under A Curve: page 13
Integration, Area
The height of the ith rectangle is f(mi) and its width is x = 0.5.
Hence, the area of the ith rectangle is f(mi)x.
6
f ( mi )   x
i
1
2
3
4
5
6
th
mi
f (m i ) Area of i rectangle
1.25 1.71875
0.859375
1.75 1.96875
0.984375
2.25 1.96875
0.984375
2.75 1.71875
0.859375
3.25 1.21875
0.609375
3.75 0.46875
0.234375
4.531250
Sum of areas:
Sum of areas 

f (mi )  x
i 1
 4 . 531250
4.531250 is our
approximation for the area of R.
The 6 rectangles do not
completely fill R and, in some
places, they extend outside of R.
Thus, it is not clear whether our
estimate is too big or too small.
If we used more than 6 rectangles, it is likely that our approximation
for the area of R would improve. For example, consider 20 rectangles, each
with a height that is the value of f at the midpoint of its base. These are shown
on the next page. The shaded region inside the rectangles appears to be quite
similar to the area under the graph of f.
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Integration.
Area Under A Curve: page 14
Integration, Area
We need notation that
will allow us to compute the
sums of areas for an arbitrary
number, n, of rectangles. Let x
= (4  1)/n, and let xi = 1 + ix
for i = 0, 1, , n. The midpoint
of [xi-1, xi] is
2
1
1
4
n
Sn 
 f ( m i )  x
i 1
mi = (xi-1 + xi)/2
for i = 1, 2, , n. We will let Sn
denote the midpoint sum of the
areas of the n rectangles.
Excel can help us do the arithmetic that is needed to compute
midpoint sums for large values of n. The sheet n = 20 of the file Area
Example.xlsx shows that S20 = 4.502813.
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Integration, Area
Integration.
Area Under A Curve: page 15
The sheet Any n in Area Example.xlsx allows us to compute Sn for
any value of n between 1 and 2000. The sums are shown rounded to 6 decimal
places. Experiment with different values of n in this sheet.
We discover that Sn gets closer and closer to 4.5 as n is increased.
When n is up to 1501, S1501 = 4.5, rounded to 6 decimal places. Further
increases in n do not change this value. Based on this evidence, we believe
that the limit of Sn, as n increases without bound, is 4.5. This is denoted by
lim S n  4 . 5 .
n 
We have now found that the area of the region R is 4.5 square units.
Experiment with Area Example.xlsx, to find the
smallest value of n for which Sn differs from 4.5 by at most 0.01.
Experiment with Area Example.xlsx, to find the
smallest value of n for which Sn differs from 4.5 by at most 0.001.
Area Example.xlsx
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Integration. Area Under AIntegration,
Curve:Areapage 16
Experiment with Area Example.xlsx, to find the
smallest value of n for which Sn differs from 4.5 by at most 0.0001.
The same plan that we have just used in our example can be applied
to the problem of computing the area between the horizontal axis and the
graph of any function, over an arbitrary interval [a, b].
We let x = (b  a)/n, and let xi = a + ix for i = 0, 1, , n. The
midpoint of [xi-1, xi] is mi = (xi-1 + xi)/2 for i = 1, 2, , n. We let
n
S n ( f , [a , b]) 
 f ( m i )  x
i 1
denote the midpoint sum of the areas of the n rectangles.
If these midpoint sums approach a number A, as n is made larger and
larger, we write lim S n ( f , [ a , b ])  A , and consider A to be the area between
n 
the axis and the graph of f, over [a, b].
Area Example.xlsx
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Integration.
Integration, AreaArea Under A Curve: page 17 Click anywhere in the plot to see the
effect of n on the midpoint sums for a typical
probability density function. See Using The
Computer Text for information on controlling
a media animation.
Computer Problem?
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Integration.
Area
Integration,
AreaUnder A Curve: page 18
Modify sheet n = 20 in Area Example.xlsx, so that it
computes S20(f, [0, 5]) for f(x) = 2x  x2/2.
Area Example.xlsx
(i) Use Graphing.xlsm to plot f(x) = e-x over the
interval [0, 10] and use this plot to estimate the area under the graph of f,
above the x-axis, and over [0, 10].
(ii) Create a new Excel file that computes S1000(f, [0, 10]).
(iii) Based upon this computation, what do you think is the area
under the graph of f, above the x-axis, and over [0, 10]?
(iv) Since f gives the non-zero part of an exponential probability
density function, what would you expect to be true about this area?
Excel
Graphing.xlsm
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Integration.
Area
Integration,
AreaUnder A Curve: page 19
Modify sheet n = 20 in Area Example.xlsx, so that it
computes the sum S100(f, [0, 4]), with 100 subintervals, for f(x) = 2x  x2/2.
Area Example.xlsx
(i) Create a new Excel file that computes S1000(f, [5,
5]), where the function f is given by the following.
f (x) 
1
2 
e
 0 . 5 x
2
(ii) Based upon this computation, what do you think is the area
under the graph of f, above the x-axis, and over [5, 5]?
(iii) Do you think that it is likely, or unlikely, that f is a probability
density function? Justify your answer.
Excel
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Integration.
Area
Integration,
AreaUnder A Curve: page 20
(i) Use Graphing.xlsm to plot
f(x) = (1/6)x3  x2 + (3/2)x
over the interval [0, 3] and use this plot to estimate the area under
the graph of f, above the x-axis, and over [0, 3].
(ii) Create a new Excel file that computes S1000(f, [0, 3]).
(iii) Based upon this computation, what do you think is
the area under the graph of f, above the x-axis, and over [0, 3]?
(iv) Do you think that it is likely, or unlikely, that f is the
non-zero portion of a probability density function? Justify your
answer.
Excel
Graphing.xlsm
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Integration. Integrals
Integration, Integrals
2. INTEGRALS
What would happen if we computed midpoint sums for a function
which might assume negative values in the interval [a, b]?
+
a
Where f(mi) < 0, the product
f(mi)x is also negative. Thus, the
midpoint sums
n
S n ( f , [a , b]) 

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b
 f ( m i )  x
i 1
will approximate the “signed area”
of the region between the x-axis and
the graph of f, over [a, b]. This is the
algebraic sum of the area above the
axis, minus the area below the axis.
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Integration.
Integration, IntegralsIntegrals: page 2
For most of the functions, f, that occur in simple business
applications, as n increases, the midpoint sums Sn(f, [a, b]) over any reasonable
interval will approach a limiting value. This value is called the integral of f
b
over [a, b] and is denoted by the expression  f ( x ) dx . The function, f, is
a
called the integrand and the symbol  is called an integral sign. Historically,
this was derived from a stretched out letter S, to remind people that an integral
is a limit of sums. The dx at the end of the integral expression is placed there
to tell us that the variable of integration is x, and to indicate the end of the
integrand.
To summarize our main result, the integral of f over [a, b] is
b
n
f ( m i )  x
 f ( x ) dx  nlim
 
a
i 1
and it represents the algebraic sum of the signed areas of the regions
between the horizontal axis and the graph of f, over [a, b].
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Integration.
Integrals: page 3
integration, Integrals
The Excel file Midpoint Sums.xlsm lets you experiment with midpoint sums. You can enter any
function, any interval, and any number of subdivisions, n (with n  500).
The program will then plot the function and the approximating rectangles. It
will also display x and the value of the midpoint sum.
Click here for information on
running the macro in Midpoint Sums.xlsm.
(i) Use Midpoint Sums.xlsm to plot f(x) = 3x2 over
the interval [0, 2], and compute Sn(f, [0, 2]) for n = 5, 10, 100, and 500.
2
(ii) Use your results from Part i. to guess the exact value of

2
3  x dx .
0
(iii) Experiment with the interval [0, v] for several values of v, and use the
v
method of Parts i and ii to guess the exact value of

2
3  x dx .
0
Midpoint Sums.xlsm
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Integration.
Integration, IntegralsIntegrals: page 4
(i) Use Midpoint Sums.xlsm to plot f(x) = x3  2x
over the interval [2, 2], and compute Sn(f, [2, 2]) for n = 3, 10, 40, and 99.
(ii) Use the idea of signed areas to explain what you found in Part i.
2
(i) Use Midpoint Sums.xlsm to plotf ( x )  1  x
over the interval [1, 1], and compute Sn(f, [1, 1]) for n = 10 and 480. (ii)
Noting that the graph of f is the upper half of a circle with radius r = 1,
explain the sum that you found for n = 480. Hint: The area of a disk with
radius one is  square units.
Excel
interprets x2 as (x)2. You must enter the
expression as (x2). However, Excel interprets 1
 x2 correctly. No parentheses are needed.
(i) Use Midpoint Sums.xlsm to plot f(x) = x2 + 1
over the interval [2, 2], and compute S200(f, [2, 2]). (ii) Repeat Part i for
the interval [ 0, 2]. (iii) Use the form of f(x) to explain the relationship
between your answers to Parts i and ii.
Midpoint Sums.xlsm
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Integration.
Integrals: page 5
Integration, Integrals
4 .7

Example 2. Evaluate
2
x e
x
dx , rounded to 4 decimal places.
1 .3
We first use
Graphing.xlsm to plot f over [1.3,
4.7]. From the plot, it appears that
the area of the region between the
x-axis and the graph of f, over the
given interval might be roughly
1.4.
0.6
0.5
0.4
f(x) 0.3
0.2
Graphing.xlsm
0.1
0
0
1
2
3
4
5
x
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Integration.
Integrals: page 6
Integration, Integrals
Next, we create an Excel file Integration Example.xlsx. Read the
following description and then go to that file.
On the first sheet, n = 50, we have entered values for a, b, and n. Cell
E6 is defined by the formula =(C6-B6)/D6. The numbers 0, 1, …, 50 are
entered in Column B. =$B$6+B9*$E$6 is entered in Cell C9, and then
dragged down to Cell C59.
Integration Example.xlsx
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Integration, Integrals
Integration.
Integrals: page 7
We entered =(C9+C10)/2 in Cell D10 and then dragged this down to
Cell D59. =D10^2*EXP(-D10) was entered in Cell E10 and then dragged
down to Cell E59. =E10*$E$6 was entered in Cell F10 and then dragged
down to Cell F59. Finally, =SUM(F10:F59) was entered in Cell F6.
S50(f, [1.3, 4.7]) = 1.4097, when rounded to 4 decimal places.
Integration Example.xlsx
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Integration.
Integrals: page 8
Integration, Integrals
Sheets n =100 and n = 200 in Integration Example.xlsx are made in
a similar way. As they show, S100(f, [1.3, 4.7]) = 1.4096 and S200(f, [1.3, 4.7]) =
1.4096, when rounded to 4 decimal places. Since the values of the midpoint
sums do not appear to be changing once n is over 100, we can assume that
4 .7

2
x e
x
dx  1 . 4096 ,
1 .3
rounded to 4 decimal places.
(i) Use Graphing.xlsm to plot f(x) = x2 + 6x +22
over the interval [2.5, 3.1]. (ii) Based on this graph, what is your estimate
3 .1
for

 x
2
 6  x  22 dx ?
 2 .5
Integration Example.xlsx
Excel interprets
x2 as (x)2. You must enter the
expression as (x2).
Graphing.xlsm
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Integration.
Integrals:
page 9
Integration,
Integrals
Use Excel and the method of Example 2 to
3 .1

evaluate
 x
2
 6  x  22 dx ,
rounded to 2 decimal places.
 2 .5
(i) Use Graphing.xlsm to plot f(x) = x2ex over the
interval [0, 6.5]. (ii) Based on this graph, what is your estimate for
6 .5

x
2
e
x
dx ?
0
Use Excel and the method of Example 2 to
6 .5
evaluate

x
2
e
x
dx , rounded to 4 decimal places.
0
Graphing.xlsm
Excel
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Integration.
page 10
Integration,Integrals:
Integrals
(i) Modify the method of Example 2 to create an
Excel file that allows you to enter a positive value for b, and then evaluate
b
 2  x  1 dx ,
Graphing.xlsm
0
rounded to 4 decimal places. (ii) Use your work from Part i to evaluate
the integral for b = 1, 1.5, 2, 2.5, and 3. (iii) Based upon your work in
Part ii, can you guess a formula for the exact value of
b
 2  x  1 dx ,
0
in terms of b?
Excel
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Integration. Integrals:
page 11
Integration, Integrals
Click here for
information on running the
macro in Integrating.xlsm.
Creating Excel files to evaluate
sums provides a good understanding of
integration, but it is rather tedious and time
consuming. Carrying out this much work, while in the middle of a business
application, could easily distract us from the basic business problem.
What we are doing is called numerical integration. Many software
packages, but not Excel, have built-in functions that perform numerical
integration. To meet this need, we have created an Excel file,
Integrating.xlsm, that serves as a numerical integration utility.
To use Integrating.xlsm, open the file and follow the instructions in
the blue box. Once you have entered the formula for f(x), you can see the
graph of f over any interval and can get a numerical value for the integral of f
over any other interval.
From now on, unless otherwise indicated, we will assume that all
integrations are performed with Integrating.xlsm.
Integrating.xlsm
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Integration.Integration,
Integrals:
Integrals page 12
Excel interprets x2 as (x)2.
You must enter the expression as (x2).
Use Integrating.xlsm to evaluate the following.
3 .1

 x
2
 6  x  22 dx
 2 .5
Use Integrating.xlsm to evaluate the following.
10

 x  3 3  e  x dx
0
Use Integrating.xlsm to evaluate the following.
4

x
2
x
 e dx
2
Integrating.xlsm
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Integration.
Integrals: page 13
Integration, Integrals
Use Integrating.xlsm to evaluate the following.
5
e
 0 . 5 x
2
dx
5
Use Integrating.xlsm to evaluate the following.
2

2  x  1 dx
0
Use Integrating.xlsm to evaluate the following.
20

0 .5  e
 0 . 5 x
dx
0
Integrating.xlsm
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Integration, Integrals
Integration.
Integrals: page 14
When entering a function in Integrating.xlsm, you can use the letters
s, t, u, v, or w as constants. If this is done, you must enter values for any letters
that you use.
For example, suppose that you want to plot and integrate f(x) = x s,
where s is a number that you enter in Cell O23. You would enter the following
function in Cell C20.
= x^s
Accept the function and then click on the Integrate button at the right
of the plot. Since Excel is set for automatic calculation, entering a new number
in Cell O23 will automatically change the function, redraw the plot, and
recompute the integral.
If values are entered in Cells O23:O27; then constants s, t, u, v, or w
may also be used to describe the range, [A, B], of the plot; and the limits, a and
b, of integration. If these constant names are used, they must be entered in
Cells H20:I20 and Cells K20:L20, preceded with "=" signs. For example, =s.
Integrating.xlsm
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Integration.
Integrals: page 15
Integration, Integrals
1
Use Integrating.xlsm to evaluate

s
for s = 1,
x dx
0
2, 5, 10, 20, and 100. Leave the parameter s as a constant. Changing s and
recomputing the sheet should change the plot and integral correspondingly.
What happens to the value of the integral as s increases?
u
(i) Use Integrating.xlsm to evaluate  
0
2
u
2
x
2
dx
u
for u = 0.1, 0.5, 1, 2, 5, and 10. Leave the parameter u as a constant.
Changing u and recomputing the sheet should change the plot
correspondingly. (ii) What happens to the value of the integral as u
increases? (iii) Use the plots of f(x) = (2/u2)x + 2/u and the values of f(0)
to explain your answer to Part ii.
Integrating.xlsm
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Integration. Integrals:
page 16
Integration, Integrals
200
(i) Use Integrating.xlsm to evaluate

x

e
 x /
dx
0
for  = 0.5, 1, 1.5, 2, 3, 5, 8, and 10. Leave the parameter  as a constant.
Changing  and recomputing the sheet should change the plot correspondingly. (ii) What happens to the value of the integral as  changes? Hint: use
the built-in constant, s, as a replacement for .
(i) Leaving b, , and  as positive constants that
b
you can set later, use Integrating.xlsm to evaluate
   x
 1
e
  x

dx .
0
(ii) Display the graph and integral approximation for b = 2,  = 1, and  = 2.
(iii) Display the graph and integral approximation for b = 3,  = 2, and  =
3.5. (iv) Given any values for  > 1 and  > 1, what happens to the value of
the integral as b increases? Hint: use the built-in constants to replace b, ,
and . The integrand in the above integral is the probability density function
for a class of distributions named after the Swedish physicist Waloddi
Weibull.
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Integrating.xlsm
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Integration.
Integrals: page 17
Integration, Integrals
It is time for disclaimers and warnings. The
mathematical theory of integration is much more sophisticated than the
simple plan that we have presented. In many cases it is possible to evaluate
integrals exactly, rather than using only numerical approximation.
When approximation is used, there are much
better routines than our plan of midpoint sums. One
of these, called Simpson’s Rule, is used in Integrating.xlsm.
Our goal is to have integration as a tool that
can be used to solve business problems. For this
purpose, the numerical understanding that we have
developed and our integrating utility will serve quite
well.
Integrating.xlsm
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If there is a way
to evaluate integrals
exactly, why don’t we
use that, and not bother
with approximation?
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Integration, Applications
Integration.
Applications
3. APPLICATIONS
All of our work in defining and evaluating integrals can now pay off
by providing information about business problems. We will start by revisiting
Example 1 from the section Area Under A Curve.
Example 3. Recall that the demand function for buffalo steak dinners
is given by D(q) = 0.0000018q2  0.0002953q + 30.19. We can use the
computation cells in the sheet Income of Dinners.xlsm to find that the graph
of D reaches the q-axis at approximately q = 4,014.
If every person who would buy a buffalo steak dinner in a given
week were to pay the maximum price that he or she would be willing to pay,
then the restaurants’ total possible revenue would be given by the area of the
region that lies between the graph of D(q) and the q-axis. Using the animation
in Area Under A Curve, we saw that the total possible revenue was $79,999.
We can now see how the computations were done in the animation.
Dinners.xlsm
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Integration, Applications
$32
D(q) = 0.0000018q2
 0.0002953q + 30.19
$24
Demand Function
$16
Total Possible
Revenue
$8
$0
0
1000
2000
4 , 014
4 , 014


D ( q ) dq 
0
3000
4000
5000
Integrating.
xlsm allows us to
confirm that the total
possible revenue is
2
 0 . 0000018  q  0 . 0002953  q  30 . 19 dq  $ 79 , 999 .
0
Recall that, if dinners are sold at a single price, then the restaurants’
expected revenue is represented by the area of a rectangular region under the
graph of the demand function.
Integrating.xlsm
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Integration.
Applications:
page 3
Integration,
Applications
For example, to sell 2,300 dinners they would have to be priced at
D(2,300) = $19.99 each. In this case the expected revenue would be
($19.99)(2,300) = $45,977.
$32
Consumer
$24
Surplus
$19.99
D(q) = 0.0000018q2
 0.0002953q + 30.19
Demand Function
$16
q = 2,300
$8
D(2,300) = $19.99
Revenue
$45,977
Not
Sold
$0
0
1000
2000
3000
4000
5000
2,300
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Applications:
page 4
Integration,
Applications
Consumer surplus is represented by the area of the region above the
revenue rectangle and under the graph of the demand function. This lost
revenue is the total extra amount of money that people who bought the dinners
would have been willing to pay. Note that the amount of consumer surplus
depends upon the number of dinners that are sold.
Looking at the plot on the preceding page, and using
Integrating.xlsm, we can compute the consumer surplus for the sale of 2,300
buffalo steak dinners.
2 , 300
Consumer
Surplus

 D ( q ) dq  R ( q )  $ 61 ,356  $ 45 ,977  $ 15 ,379 .
0
For completeness we will also compute the amount of revenue that is
lost from dinners which were not sold, because people were not willing to pay
the fixed price of $19.99.
4 . 014
Not

Sold
 D ( q ) dq  $ 18 ,643
2 . 300
Integrating.xlsm
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page 5
Integration,
Applications
As a check on our work, note that the sum of revenue, consumer
surplus, and money from dinners not sold is equal to the total possible
revenue.
$45,977 + $15,379 + $18,643 = $79,999
Our revenue animation is no longer “magic.” We now know how to
do all of its calculations.
Assuming that 1,800 buffalo steak dinners are
sold, use the restaurants’ demand function to compute (i) the total possible
revenue, (ii) the revenue, (iii) the consumer surplus, and (iv) the lost
revenue from dinners that were not sold.
We have seen that the maximum profit from the
sale of buffalo steak dinners would result from selling 2,025 dinners. At this
sales level, use the restaurants’ demand function to compute (i) the total
possible revenue, (ii) the revenue, (iii) the consumer surplus, and (iv) the
lost revenue from dinners that were not sold.
Integrating.xlsm
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Integration, Applications
Integration.
Applications: page 6
Revenue computations for an arbitrary demand function work in the
same way as those for the buffalo steak dinners.
Let D(q) give the price per unit for a good, that would result in the
sale of q units, and let qmax be the maximum number of units that could be sold
at any price. That is, D(qmax) = 0. The total possible revenue is given by
Total Possible

Revenue
qmax
 D(q ) dq.
0
If qsold units are sold, then the revenue will be qsoldD(qsold). The
following formulas give consumer surplus and lost revenue from units not
sold.
q sold
Consumer

Surplus

D( q ) dq  q sold  D( q sold )
qmax
Not

Sold
0
 D(q ) dq
q sold
It is clear that
revenue + consumer surplus + not sold = total possible revenue.
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Integration.
Applications: page 7
Suppose that the demand function for a good is
given by D(q) = 0.1q + 150. (i) Use Graphing.xlsm to plot D(q) and
estimate the total possible revenue. Explain how you arrived at your
estimate. (ii) Use Integrating.xlsm to compute the total possible revenue.
Refer to the information in Exercise 31 and
suppose that qsold = 400 units are sold. Compute (i) the revenue, (ii) the
consumer surplus, and (iii) the lost revenue from units not sold.
Suppose that D(q) = 0.00006q2 + 250 is the
demand function for a certain model of audio speaker. (i) Use
Graphing.xlsm to plot D(q) and estimate the total possible revenue. Explain
how you arrived at your estimate. (ii) Use Integrating.xlsm to compute the
total possible revenue.
Refer to the information in Exercise 33 and
suppose that qsold = 950 units are sold. Compute (i) the revenue, (ii) the
consumer surplus, and (iii) the lost revenue from units not sold.
Graphing.xlsm
Integrating.xlsm
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Integration, Applications
Integration.
Applications: page 8
DEMAND FUNCTION
500
400
D (q ) in $
The
demand function for a new type
of car alarm system is shown in
the adjacent plot. Note that the
quantities sold are given in
thousands. (i) Estimate the
total possible revenue, in
dollars. (ii) Estimate the
consumer surplus, in dollars,
that would result from selling
300,000 alarms.
300
200
100
0
0
100
200
300 400
q in K's
500
600
(i) Use your team’s Marketing Project data to
compute the total possible revenue. (ii) Compute the consumer surplus for
the quantity of goods that maximizes profit. (This is Question 5 in your
team’s project.) Use the same units as in the work on the Class Project in
the sheet Functions of Marketing Focus.xlsm.
Excel
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Integration,
Applications
Integration.
Applications:
page 9
In addition to the computation of consumer surplus and total possible
revenue, there are a great many applications of integration to business
problems. We will conclude this section with one such application. The
important uses of integration in probability will be considered in our work on
Project 2.
Example 4. The Plastic-Is-Us Toy Company considers its incoming
revenue as an income stream, rather than as a huge collection of discrete
payments. At a time t years from the start of its fiscal year on July 1, the
company expects to receive revenue at the rate of A(t) million dollars per year.
Records from past years indicate that Plastic-Is-Us can model its revenue rate
with A(t) = 110t5 + 330t4  330t3 + 110t2 +3.174 million dollars per year.
Using Graphing.xlsm to plot A(t), we see that the company expects
its largest rate of revenue in late November, at the height of the Christmas
shopping season. The slowest rate of revenue flow appears to be during the
summer months.
Graphing.xlsm
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Integration, Applications
8
6
A(t) 4
2July 1
0
0
Oct. 1
Jan. 1
April 1
July 1
0.25
0.5
0.75
1
t
The chief financial officer would like to compute the total amount of
revenue that Plastic-Is-Us will receive in one year. The income stream, A(t), is
a rate of change in money, given in dollars per year. Since the units along the
t-axis are years, the area of a region under the graph of A(t) is given in
(millions of dollars/year)(years) = millions of dollars.
T
Since
 A ( t ) dt
gives the area between the t-axis and the graph of
0
A(t), over the interval [0, T], it can be shown that the integral gives the total
amount of money, in millions of dollars, that will be received from the income
stream in the first T years.
Graphing.xlsm
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Integration, Applications
Looking at the graph of the preceding page, we note that the average
value of the income stream A(t) appears to be roughly 5 million dollars per
year. Hence in one year, the company can expect a total income of
approximately 5 million dollars.
More precisely, we can use Integrating.xlsm to compute the total
income received by Plastic-Is-Us during the period from 0 to 1 year.
(Remember that we must use x, not t, as the variable of integration in
Integrating.xlsm.)
1

5
4
3
2
 110  x  330  x  330  x  110  x  3 . 174 dx  5 . 007 million dollars
0
This is very close to our estimate of 5 million dollars for the
company’s total yearly income.
The same type of reasoning allows us to compute the total revenue
from any income stream.
Graphing.xlsm
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Integration, Applications
The total revenue, in dollars, received from an income stream of
A(t) dollars per year, starting now and continuing for the next T years is
T
given by
 A(t ) dt .
0
In addition to the total revenue, a company would often like to know
the present value of its income stream during the next T years (0  t  T),
assuming that money earns interest at some annual rate r, compounded
continuously.
We will develop a structure for this computation. Recall that A
dollars, invested for t years at an annual rate r, compounded continuously, has
a present value of Ae-rt dollars. Consider an income stream of A(t) dollars per
year and let t  be a fixed time, in years.
If we get A(t ) dollars per year for a very short period of time, say t
years, we will get A(t )t dollars at approximately time t . The present value
of this amount is A(t )t e-rt  = A(t )e-rt t dollars.
Graphing.xlsm
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Integration, Applications
Suppose that we receive the income stream for the next T years.
Integration can be used to “sum” the amounts A(t )e-rt t over the interval 0
 t  T.
Suppose that money earns at an annual rate, r, compounded
continuously. The present dollar value of an income stream of A(t)
dollars per year, starting now and continuing for the next T years is
T
given by

A( t )  e
 r t
dt .
0
Example 5. We return to the Plastic-Is-Us Toy Company that we
considered in Example 4. Recall that they have an income stream of A(t) =
110t5 + 330t4  330t3 + 110t2 +3.174 million dollars per year. The
management of Plastic-Is-Us would like to know the present value of its
income stream during the next year (0  t  1), assuming that money earns
interest at an annual rate of 5.5%, compounded continuously.
Graphing.xlsm
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Integration, Applications
Applying the integral formula for present value to Plastic-Is-Us, we
use Integrating.xlsm to find that the present value of their income stream for
one year, starting on July 1, is
1
5
4
3
2
 0 . 055 t

 110  t  330  t  330  t  110  t  3 . 174   e
dt  4 . 879

0
million dollars.
How can we tell if this is a sensible value? We have seen that the
total revenue from the income stream during a single year is 5.007 million
dollars. Since this money will be received in the future, its present value is
somewhat less than 5.007 million dollars. Our computed present value of
4.879 million dollars seems to be quite reasonable.
Integrating.xlsm
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page 15
Integration, Applications
(i) What total revenue will the Plastic-Is-Us Toy
Company receive from its income stream during the period from July 1 to
January 1? (ii) Assuming that money earns 5.5% compounded continuously,
what is the present value, on July 1, of that income stream?
(i) What total revenue will the Plastic-Is-Us Toy
Company receive from its income stream during the period from July 1 to
October 1? (ii) Assuming that money earns 5.5% compounded continuously,
what is the present value, on July 1, of that income stream?
(i) What is the present value of a constant income
stream of A(t) = $950,000 per year for the next three years at 6%,
compounded continuously? (ii) How much total revenue will be received
during the three year period. (You should be able to answer Part ii without
using integration!)
Integrating.xlsm
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Integration. Applications:
Integration, Applicationspage 16
What is the
present value of a constant income
stream of A(t) = $5,450,000 per year for
the next five and one-half years at 5.7%,
compounded continuously? (ii) How
much total revenue will be received
during the five and one-half year period.
(You should be able to answer Part ii
without using integration!)
(i) What is the present value of an income stream
of A(t) = 348,000,000e0.07t dollars per year for the next two years at 5%,
compounded continuously? (ii) How much total revenue will be received
during the two year period.
Integrating.xlsm
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Integration, Calculus
Integration.
Calculus
4. CALCULUS*
The topics of differentiation and integration are the main parts of the
mathematical subject called calculus. Income streams allow us to explore a
very useful and important relationship between integration and differentiation.
Example 6. Suppose that we invest $250,000 at an annual rate of
6.25%, compounded continuously. What is the instantaneous rate of change
in the value of our investment per year after three and one-half years? That is,
at what rate, in dollars per year, will we earn interest after 3.5 years?
Recall from Part 1 of Mathematics for Business Decisions that the
future value, F, of P dollars, invested for t years at a continuous annual rate r
is given by F = Pert. Thus, F(t) = $250,000e0.0625t represents the value of our
investment after t years.
* This section, and the corresponding section Calculus of Normal Distributions in
Project 2, are not needed for the study of other material in Mathematics for Business Decisions.
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Integration, Calculus
Integration.
Calculus: page 2
Since the derivative of a function gives its instantaneous rate of
change, F(3.5) gives the rate, in dollars per year, at which we will earn interest
after 3.5 years. Differentiating.xlsm shows that F(3.5)  $311,130 and F(3.5)
 $19,445.63.
After three and one-half years, the value of our investment will have
grown to $311,130, and we will be earning interest at a rate of $19,445.63 per
year.
Instead of considering the change in value at just one time, we can
use Differentiating.xlsm to plot F(t) over a period of ten years.
$40,000
Dollars Per Year
$30,000
F '(t ) $20,000
$10,000
$0
0
Differentiating.xlsm
2
4
6
t
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Integration.
Calculus: page 3
Since F(t) is in dollars per year, we can think of this derivative as an
income stream, which we will call A(t) = F(t).
As we saw in the section Applications, the total income from this
stream, during the first T years, is given by the integral of A(t) from 0 to T.
T
Thus,
 A ( t ) dt
gives the total amount of interest that the investment will earn
0
in the first T years.
At the end of T years the full value of the investment will be the
original principal of $250,000 plus the total amount of interest that has
T
accumulated, $ 250 , 000 
 A ( t ) dt .
0
This same full value of the investment after T years, can also be
computed with the formula, F(t), for future value. Thus,
T
F ( T )  $ 250 , 000 
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 A ( t ) dt .
0
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Integration, Calculus
Integration.
Calculus: page 4
We now subtract $250,000 from both sides of the equation.
T
F ( T )  $ 250 , 000 
 A ( t ) dt
0
T
F ( T )  $ 250 , 000 
 A ( t ) dt
0
Both sides of the last equation are functions of T. Since, the
derivative of a difference is the difference of the derivatives, the derivative of
the left side of the equation, with respect to T, is F(T), plus the derivative of
$250,000. But F(T) = A(T) and the derivative of a constant is equal to 0, so
the derivative of F(T) + $250,000, with respect to T, is equal to A(T) + 0 =
A(T).
T
T
Since F ( T )  $ 250 , 000 
 A ( t ) dt , both F(T)  $250,000 and  A ( t ) dt
0
0
must have the same derivatives.
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Integration, Calculus
Integration. Calculus:
page 5
We have just found that, if A(T) is the income stream resulting from
interest on an investment with future value F(T) after T years, then the
T
derivative of
 A(t ) dt , with respect to T, is equal to A(T).
0
This exhibits an inverse relationship between integration and
differentiation. If we first integrate A(t), and then differentiate the integral, we
return to the original function, A(t). It is an amazing fact, discovered
independently by Isaac Newton (1642-1727) and Gottfried Wilhelm Leibniz
(1646-1716), that this relationship holds for a large class of functions.
There is nothing special about using the letter A as the name for a
function, and it would make no difference if we replaced the variable of
integration, t, with any other letter. The important part of our result is that we
are differentiating with respect to the letter that is the upper limit of
x
integration. Thus, for many functions, f, the derivative of  f ( u ) du , with
0
respect to x, is equal to f(x).
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Integration, Calculus
Integration. Calculus:
page 6
In fact, it can be shown that the same result holds if 0 is replaced by
any other number, a. When stated formally, the inverse connection between
integration and differentiation is called the Fundamental Theorem of
Calculus.
Fundamental Theorem of Calculus. For many of the functions, f,
x
which occur in business applications, the derivative of
 f ( u ) du ,with
a
respect to x, is f(x). This holds for any number a and any x, such that the
closed interval between a and x is in the domain of f.
We will return to this theorem when working with probability in
Project 2. For now, we will consider two examples. The first of these uses a
simple function, but requires a good understanding of integrals as areas.
Example 7. Let f(u) = 2 for all values of u. If x  1, then integral of f
from 1 to x is the area of the region over the interval [1, x], between the u-axis
and the graph of f.
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Integration, Calculus
Integration.
Calculus: page 7
3
(1, 2)
2
(x, 2)
The region whose area
is represented by the integral is
rectangular, with height 2 and
width x  1. Hence, its area is
2(x  1) = 2x  2, and
x
f (u )
1
 f (u) du
2
1
x1
0
x
0
1
2
3
x 4
u
5
 f (u) du  2  x  2.
1
In the section Properties and Applications of Differentiation, we saw
that the derivative of f(x) = mx + b is equal to m, for all values of x. Thus, the
x
derivative of  f (u ) du, with respect to x, is equal to 2. As predicted by the
1
Fundamental Theorem of Calculus, this is also the value of f(x).
The next example uses the definition of a derivative as the limit of
difference quotients.
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Integration, Calculus
Integration. Calculus:
page 8
Example 8. Recall the income stream of A(t) = 110t5 + 330t4 
330t3 + 110t2 +3.174 million dollars per year that was expected by the
Plastic-Is-Us toy company in Example 4 of Applications. Let G(T) be the total
income that is expected during the first T years, for 0  T  1. Picking a time T
= 0.5 years, we will check that the instantaneous rate of change of G(T), with
respect to T, is the same as A(T).
T
Note that G (T )   A(t ) dt. We now wish to compute G(0.5). Recall
0
that G(T) is approximated by the difference quotient
G (T  h )  G (T  h )
2h
,
for small values of h. We will let h = 0.0001, and use Integrating.xlsm to
evaluate G(0.5 + 0.0001) and G(0.5  0.0001). Integrating.xlsm rounds the
numerical values of integrals to four decimal places. For the present
calculation, we gain extra precision by copying the values from Cell N20 and
keeping all of their decimal places.
G(0.5 + 0.0001) = G(0.5001)  2.79078611562868
G(0.5  0.0001) = G(0.4999)  2.78946381564699
Integrating.xlsm
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Integration,
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Integration.
Calculus:
page 9
These give a value of 6.6115 for the difference quotient
G ( 0 . 5001 )  G ( 0 . 4999 )
0 . 0002
,
rounded to four decimal places. This is the
instantaneous rate of change in total income after 0.5 years. Integrating.xlsm
T
shows the same value for A(0.5). Noting that G (T )   A(t ) dt , we have
0
verified the Fundamental Theorem of Calculus. At T = 0.5, the derivative of
T
 A(t ) dt, with respect to T, is equal to A(T).
0
Let f(u) = 1/2 for all values of u. (i) If x  2,
x
use the method of Example 7 to find a formula for

f (u ) du.
(ii) Differ-
2
entiate your result from Part i, with respect to x, and verify that the
derivative is equal to f(x) for all values of x.
Graphing.xlsm
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Integration, Calculus:
Calculus
Integration.
page 10
Consider the income stream A(t) for the PlasticIs-Us toy company that was last discussed in Example 8, and let
T
G (T ) 
 A(t ) dt.
0
(i) Use the method of Example 8 to evaluate the derivative of G(T), with
respect to T, for T = 0.75 years. (ii) Show that your result from Part i is
equal to A(0.75).
A company’s expected income stream is
modeled by A(t) = 742,000,000e0.06t dollars per year, for the next three
years. Let G(T) be the total income that they will receive during the first T
years of this period.. (i) Use the method of Example 8 to evaluate the
derivative of G(T), with respect to T, for T = 0.75 years. (ii) Show that your
result form Part i is equal to A(0.75).
Integrating.xlsm
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Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
Drives
on the project
What can integration tell
us about marketing 12-GB drives?
The graph of D(q) shows that demand
is positive for values of q between 0 and
approximately 2,500 thousand drives. Using
Solver in Rows 64 and 65 in the sheet
Functions of Marketing Focus.xlsm, we find
that D(2,480.767) = 0. Integration can be used
to compute the total possible revenue that
could result from sales of 12-GB drives.
Integration, Focus
Marketing
Focus.xlsm
Class Project
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Integrating.xlsm shows that
2 , 480 . 767
2 , 480 . 767
D ( q ) dq 
 0.00005349


on the project
0
0
Marketing
Calculus, Mathematics,
Tests, Homework, Computers
2
Computer
 q  0.03440302  q  414.534444 91 dq
Drives
 650,288.49 72
Since D(q) is in dollars per drive and q is in thousands of drives,
the value of the integral is in thousands of dollars. We are considering
revenue in millions of dollars, so we divide by 1,000. The total possible
revenue is (650,288.4972 thousand dollars)/1,000  650.288 million dollars.
Question 5 asks for the consumer surplus, at the production level
that maximizes profit. We have seen that this involves selling 1,262.274
thousand drives, priced at $285.88 each.
We first use Integrating.xlsm to evaluate the integral of D(q) over
the interval from 0 to 1,262.274.
Integration,Focus.xlsm
Focus
Marketing
Class Project
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1, 262 . 274
1, 262 . 274


D ( q ) dq 
 0.00005349  q
2
 0.03440302  q  414.534444 91 dq
Marketing
Calculus, Mathematics,
Tests, Homework, Computers
0
0
Computer
 459,988.11 02
on the
project
Drives
As with the total possible revenue, the value of the integral is in
thousands of dollars. To express the possible revenue from selling
1,262.274 thousand drives in millions of dollars, we must divide by 1,000.
(459,988.1102 thousand dollars)/1,000  459.9881102 million dollars.
Our expected revenue from selling 1,262.274 thousand drives,
priced at $285.88 each, is 1,262.274285.88  360,858.8911 thousand
dollars. Thus, the predicted revenue is 360,858.8911/1,000  360.8588911
million dollars.
The consumer surplus when 1,262.274 thousand drives are sold is
the possible revenue from sales minus the revenue that we expect, at a price
of $285.88 per drive. This gives a consumer surplus of (459.9881102
million dollars)  (360.8588911 million dollars)  99.129 million dollars.
Integration,Focus.xlsm
Focus
Marketing
Class Project
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Marketing
Calculus, Mathematics,
Tests, Homework, Computers
Computer
on the projectWHAT SHOULD YOU DO? Drives
Basic Computation. Each team should now answer Question 5 by
computing the consumer surplus for its team project. Use the same units as
we have used in the Class Project.
Further Exploration. What more useful information can you
extract from your team’s marketing data? What graphical and numerical
displays can you produce to help make business decisions? Teams might
want to explore the sensitivity of their conclusions to small changes in the
marketing data and assumptions. Other areas of investigation could include
the relationship of maximum profit to changes in demand and costs.
Prepare Reports. When all work is completed, prepare files for
your team’s written and oral reports.
Integration,
Focus
Marketing
Focus.xlsm
Class Project
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Marketing Example,
Double Pricing Double Pricing
Marketing
Example.
1. DOUBLE PRICING
In the buffalo steak dinner example that we have been following,
Solver can be used in the computation cells of the sheet Solver in
Dinners.xlsm. We find that D(4,014) = 0. Hence, the total possible revenue is
4 , 014
4 , 014


0
D ( q ) dq 
2
 0 . 0000018  q  0 . 0002953  q  30 . 19 dq  $ 79 , 999 .
0
We have seen that the restaurant chain would obtain a maximum
weekly profit of $14,052 by selling 2,025 dinners, priced at $22.21 each.
Under this plan, the mangers note that 2,025 people are buying the dinners for
less than they would have been willing to pay and that 4,014  2,025 = 1,989
people who would have been willing to buy dinners at a lower price are not
paying money to the restaurants.
Looking at the situation from another point of view, $79,999 
2,025$22.21  $35,024 of potential revenue is being lost. How might the
restaurants collect some of this money and increase their profit?
T C I
(material continues)

Marketing Example.Marketing,
Double
Pricing: page 2
Double
After consultation with the managers and chefs at the various
restaurants, the management team decided to up-grade the buffalo dinners in
1/4 of the establishments in the chain. The fancier entree will be promoted as
more desirable and will be sold at a higher price than the regular buffalo steak
dinner.
It is believed that the improved image of the up-scale dinners will
make people who can afford the higher prices choose to pay more and eat at
places which sell the up-graded dinners. We will assume that the chain has
enough restaurants to allow a choice of dining level for all of its potential
customers and that both fixed and variable costs are now distributed
uniformly among all of the establishments.
To spread out its clientele evenly, the chain decides to price the
dinners in such a fashion that up-graded meals will account for 1/4 of the total
number of buffalo steak dinners which will be purchased. In making their
plans, the managers assume that the demand function for a total of q buffalo
steak dinners will remain unchanged under the new plan. Hence, we still have
2
D ( q )   0 . 0000018  q  0 . 0002953  q  30 . 19 .
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Marketing
Example.
Double Pricing: page 3
Marketing,
Double
Some increase in costs will be necessary if the up-scale entree is to
have an image that will support higher prices. Managers and chefs estimate
that up-grading an entree will raise its share of the original fixed costs by
50% and its share of the original total variable costs by 10%.
Since 25% of the dinners will be up-graded, fixed costs will now be
(0.75)$9,000 for the remaining regular meals and (1.5)(0.25)($9,000) for the
up-scale meals. Recall that the variable cost for preparing a total of q dinners
is 177q0.633. Regular meals will still account for 3/4 of this cost,
0.75177q0.633. Up-scale meals accounted for the other 1/4 of the original
variable costs. Since production costs are 10% higher for up-graded entrees, it
will cost (1.1)(0.25)177q0.633 to prepare their share of the q dinners. This
information can be combined to form a new cost function.
C ( q )  0 . 75  9 , 000  1 . 5  0 . 25  9 , 000  0 . 75  177  q
 10 ,125  181 . 425  q
0 . 633
 1 . 1  0 . 25  177  q
0 . 633
0 . 633
Before we can analyze profits, we must construct the new revenue
function.
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Marketing
Example. Double Pricing: page 4
Marketing, Double
Since it is planned to sell a total of q dinners, up-scale dinners will be
priced so that q/4 dinners will be sold at these places. This means a price of
D(q/4) dollars per meal, generating (q/4)D(q/4) dollars of revenue. The
remaining (3/4)q regular dinners will have to be priced at a level that will
allow the total sales of all q dinners. These will generate an additional
(3/4)qD(q) dollars of revenue. These two components of revenue are shown
under a plot of the demand
Demand Function
function at q = 2,000.
$32
$24
D(q)
For example, if
2,000 dinners are produced,
500 will be up-scale entrees,
priced at D(500) = $29.59,
each. The remaining 1,500
regular dinners will be priced
at D(2,000) = $22.40, each.
$16
Regular
revenue
$8
$0
0
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1,000
Up-scale
revenue
2,000
q
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3,000
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Marketing
Example. Double Pricing: page 5
Marketing, Double
The double pricing plan is analyzed in the Up-Scale sheet of the
Excel file Dinners.xlsm. To see the regions of regular and up-scale revenue
for q dinners, enter q in Cell H24 of that sheet. Experiment with Up-Scale to
confirm the following fact. Since no more that 4,014 dinners can be sold, a
value of q which is greater than or equal to 4,014 yields the same revenue as
selling q/4 dinners.
The total revenue, R(q), that the chain will receive from producing q
dinners is the sum of the up-scale and regular income.
R ( q )  0 . 25  q  D ( 0 . 25  q )  0 . 75  q  D ( q )
A plot of the revenue and cost functions for the double pricing plan is
shown on the next page.
As expected, revenue recovers some after q = 4,014. However, such
production levels are absurd, since no more than 4,014 dinners can be sold.
Obviously, the cost of these unsold meals continues to increase.
Dinners.xlsm
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Marketing
Example.
Marketing,
Double
Double Pricing: page 6
Revenue
Cost
Revenue & Cost Functions
R(q) & C(q)
$100,000
$80,000
$60,000
$40,000
$20,000
$0
0
2,000
4,000
6,000
8,000
q
10,000 12,000 14,000 16,000
It is clear that a positive profit will occur only when q is between
approximately 700 and 3,600.
Using Solver in the Computation box of the sheet Up-Scale, we find
that a maximum weekly profit of $16,923 can be expected to result from
selling 582 up-scale dinners at $29.41 each and 2,328  582 = 1746 regular
dinners at $19.75 each.
Dinners.xlsm
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Marketing Example.
Marketing, Double
Double Pricing: page 7
Profit Function
$20,000
P(q)
$16,000
$12,000
$8,000
$4,000
$0
0
1,000
2,000
q
3,000
4,000
The plan for up-grading the entree in some of the restaurants in the
chain and then selling the buffalo steak dinners at two different prices gives
an improvement over the single price plan. The maximum weekly profit of
$16,923 is up $2,871 from the maximum weekly profit of $14,052 under the
single price plan.
Dinners.xlsm
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Marketing
Double Pricing: page 8
Marketing,Example.
Double
(i) Suppose that the restaurant chain wants to sell a
total of 1,000 dinners. Use Solver in the sheets Solver and Up-Scale of
Dinners.xlsm to decide whether the single or double pricing plan would
yield a greater weekly profit. (ii) Repeat Part i for 3,000 dinners.
(i) Use Solver in the sheet Up-Scale of
Dinners.xlsm to find the total number of buffalo dinners that the restaurants
should sell in order to price the up-scale dinners at $29.95. (ii) What price
should be put on the regular dinners to match the up-scale price of $29.95?
(i) Use Solver in the sheet Up-Scale of
Dinners.xlsm to find the total number of dinners that should be sold in order
to maximize the weekly revenue. (ii) How many of these dinners should be
of each type, how should they be priced, and what maximum revenue could
be expected?
Dinners.xlsm
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Marketing,
Optimizing Conversion Optimizing Conversion
Marketing
Example.
2. OPTIMIZING CONVERSION
Having seen that up-grading 1/4 of the buffalo dinners, and instituting
a two-price system could increase profits, the managers wondered if 1/4 is the
optimum fraction of establishments to convert to up-scale. Suppose that a
fraction, r, of the dinners are up-graded, and that dinners are priced so that rq
up-scale dinners out of a total of q dinners are sold. The remaining (1  r)q
dinners will be regular meals.
The same type of reasoning that we used with r = 0.25 can be used to
derive the following formulas for the new revenue and cost functions.
R( q )  r  q  D( r  q )  (1  r )  q  D( q )
C ( q )  1  0.5  r   9,000  1  0.1  r   177  q
0.633
The sheet Up-Scale of Dinners.xlsm is set up to compute demand,
revenue, cost, and profit for any value of r that is entered in Cell H23.
Dinners.xlsm
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Marketing
Example. Optimizing Conversion: page 2
Marketing, Optimizing
In Double Pricing, we only used the sheet Up-Scale with r = 0.25.
As a check on our more general analysis of the situation, we can enter r = 0 in
Cell H23. This means that none of the entrees would be up-graded. Since no
meals will be sold at higher prices, and there has been no increase in costs, we
should obtain the same results that we found under the single pricing plan that
was discussed in earlier work.
Setting r = 0 and using Solver to find the maximum profit for an
integer number of dinners, yields the following.
C om p u tation
q
D (q )
R (q )
C (q )
P (q )
2,025
$22.21
$44,977.06
$30,925.04
$14,052.02
rq
D (r  q )
0
$30.19
These are exactly the same results that we found for single pricing.
At the other extreme, we can see what would happen if we converted
all of the meals to up-scale entrees. To do this, we let r = 1 in Up-Scale.
Dinners.xlsm
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Marketing Example.
Optimizing Conversion: page 3
Marketing, Optimizing
After maximizing profit with Solver, we see that the weekly profit
from the up-graded meals would be only $7,371. This is roughly half of the
profit that was expected from the original single pricing. The ability to sell
buffalo steak dinners at a slightly higher price does not off-set the increased
costs of preparing the dinners.
C om p u tation
q
D (q )
R (q )
C (q )
P (q )
1,992
$22.46
$44,738.82
$37,368.00
$7,370.82
rq
D (r  q )
1992
$22.46
Of the plans investigated so far, up-grading 25% of the dinners has
predicted the largest maximum weekly profit for the chain. Could they expect
to do any better? Suppose that 50% of the meals were up-graded. We can
enter 0.5 for r in Up-Scale, and then use Solver to find the maximum profit.
The resulting $19,010 is the highest value that we have yet found.
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Marketing
Example.
Marketing,
Optimizing Optimizing Conversion: page 4
C om p u tation
q
D (q )
R (q )
C (q )
P (q )
2,580
$17.45
$57,095.77
$38,086.00
$19,009.77
rq
D (r  q )
1290
$26.81
It is now clear that, for each fraction, r, of meals that will be upgraded, there are two new functions. Let Pmax(r) be the maximum weekly
profit, and let qmax(r) be the total number of dinners that should be prepared
and sold in order to realize that profit.
The sheet Fraction in the Excel file Dinners.xlsm shows values of
qmax(r) and Pmax(r) for values of r ranging from 0.00 to 1.00, in hundredths.
The graphs and significant data have been copied from that sheet onto the
following two pages.
Dinners.xlsm
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MarketingMarketing,
Example.
Optimizing
Optimizing Conversion: page 5
2,548
2,555
2,562
2,569
2,574
2,580
2,584
2,588
2,591
2,594
2,596
2,597
2,598
2,597
2,596
2,594
Dinners.xlsm
$20,000
$16,000
P max (r )
$18,868
$18,913
$18,950
$18,978
$18,998
$19,010
$19,012
$19,004
$18,987
$18,959
$18,921
$18,872
$18,812
$18,741
$18,658
$18,565
P max (r)
0.45
0.46
0.47
0.48
0.49
0.50
0.51
0.52
0.53
0.54
0.55
0.56
0.57
0.58
0.59
0.60
q max (r )
$12,000
$8,000
$4,000
$0
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
r
Dinners For Maximum Profit
3,000
2,500
q max (r)
r
Maximum Profit
2,000
1,500
1,000
500
0
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
r
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Marketing Example.
We see that
the highest
Marketing,
Optimizing weekly profit
from buffalo steak dinners, $19,012, will result Optimizing Conversion: page 6
from converting 51% of the restaurant chain’s
establishments to up-scale units. To find the pricing structure for this plan, we
return to the sheet Up-Scale, enter r = 0.51, and use Solver to find the
maximum profit.
C om p u tation
q
D (q )
R (q )
C (q )
P (q )
2,584
$17.41
$57,194.71
$38,182.92
$19,011.80
rq
D (r  q )
1318
$26.67
$32
$24
D(q)
The chain should plan on
selling 1,318 up-scale dinners,
priced at $26.67, and it can expect
to sell 2,584  1,318 = 1,266
regular dinners, priced at $17.41.
Demand Function
$16
$8
$0
0
Dinners.xlsm
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1,000
2,000
q
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C
4,000
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Marketing
Example.
Optimizing Conversion: page 7
Marketing,
Optimizing
(i) Use Solver in the sheet Up-Scale of
Dinners.xlsm to find the total number of dinners that should be sold in order
to maximize weekly profit, if 60% of the chain’s establishments are upgraded. (ii) How many of these dinners should be of each type, how should
they be priced, and what maximum profit could be expected?
What reduction in maximum profit would the
restaurant chain experience in they up-graded 40% of the dinners, rather than
the optimal 51%?
(i) What fraction of the buffalo dinners should be
up-graded, in order to obtain the maximum weekly revenue from the entree?
(ii) How many dinners of each type should be sold, how should they be
priced, and what maximum revenue could be expected?
Dinners.xlsm
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Business Mathematics Students:
Backup Plan
You never can tell.
It might really help.
Back up your studying!
omphaloskepsis: “The act of contemplating
one’s navel, as an exercise for mystics.”
(Greek, omphalos, the navel + Greek,
skepsis, a viewing.)
Webster’s New World Dictionary
Second College Edition, 1980
Back to Differentiation
For further mathematical study of differentiation, it is common to define the derivative in terms of
one-sided difference quotients,
f ' ( x )  lim
h0
The Derivative
The Derivative
f ( x  h)  f ( x )
h
where h is allowed to assume either positive or negative values.
For the purposes of numerical approximation, our symmetric
definition of the derivative is preferable to the above theoretical definition.
Although it will not be of concern in our work, it should be noted that
the theoretical difference quotient for a function, f, at a given value of x,
may not approach any fixed number as h takes on smaller and smaller
values. This happens if the graph of f has a corner at the point (x, f(x)). In this
case f (x) does not exist and we say that f is not differentiable at x. For
example, the absolute value function, f(x) = |x|, is not differentiable at x = 0.
(material ends)
Back to Differentiation
Symbolic Differentiation
Symbolic Differentiation
The definition of a derivative may be used to establish the following
general formulas for differentiation.
 If r  0 is a real number and f(x) = x r, then f (x) = rx r1.
 If a > 0 is a real number, and f(x) = ax, then f (x) = axln(a). In
particular, the derivative of ex is the same function, ex.
 If b  1 is a positive real number, and f(x) = logbx, then f (x) =
(1/x)logbe. In particular, the derivative of ln(x) is 1/x.
These formulas may be used in conjunction with the rules for
differentiation which we discussed earlier.
Example 1. Let f(x) = 5.7x3 + 1.2x2 + 5.3x + 8.1. Now
f (x) = 5.7(3x31) + 1.2(2x21) + 5.3(1x11) + 0
= 17.1x2 + 2.4x + 5.3.
Thus, for example, f (6) = 17.162 + 2.46 + 5.3 = 635.3.
(material continues)
Back to Differentiation

Symbolic
Differentiation: page 2
Symbolic Differentiation
Example 2. Return to the buffalo
steak dinner example that was considered earlier. The demand and cost
functions for the dinners were given by D(q) = 0.0000018q2  0.0002953q +
30.19 and C(q) = 9,000 + 177q0.663, respectively. This yields the following
profit function.
P(q) = R(q)  C(q) = qD(q)  C(q)
= 0.0000018q3  0.0002953q2 + 30.19q  9,000  177q0.633
The marginal profit, P, can be computed with our rules and formulas
for differentiation.
P(q) = 0.0000018(3q31)  0.0002953(2q21) + 30.19
 0  177(0.633q0.6331)
= 0.0000054q2  0.0005906q + 30.19  112.041q0.367
Unfortunately, after all of this work, there is no closed form way to
solve the equation P(q) = 0, and find an exact value for q. The only way that
we could use the formula for marginal profit would be to use some type of
numerical approximation to solve P(q) = 0. This type of dilemma is one of
the reasons numerical differentiation is used in Mathematics for Business
Decisions.
 Back to Differentiation
(material ends)
E-L M-P Q-T U-Z
Index
Index
County administrator example
2-123, 2-171, 2-179
Course files,
needed 28
running 13
Cumulative distribution
function 2-2,
A-B
Bicycle shorts example 2-138
Binomial random variable 2-3,
2-13, 2-71
Buffalo steak dinner example
1-35, 1-41, 1-64, 1-120
1-139, 1-173
Bug in random number
generator 2-176
D
C
Central limit theorem 2-117
Confidence interval 2-128
Consumer surplus 1-138
Control buttons 15
Cost function 1-41
marginal 1-64, 1-68
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Data Analysis, running 81
Demand function 1-35
marginal 1-70
Derivative 1-80, 1-219, 1-220
properties 1-89
Difference quotient 1-68
Differentiation 1-80, 1-219,
1-220
properties 1-89

A-D M-P Q-T U-Z
Index: Index
page 2
E
Expected value,
continuous random variable
2-38
finite random variable 2-12
Exponential random variable
2-27, 2-42, 2-77
F-G-H

Files needed 28
Final reports,
Oral 35
Written 40
Focus pages,
Demand, Revenue, Cost,
and Profit 1-52
Differentiation 1-106
Focus pages (continued),
Distributions 2-58
Integration 1-199
Normal Distributions 2-159
Simulating Normal Random
Variables 2-184
Using Solver 1-130
Variance 2-94
Fundamental Theorem of
Calculus 1-194, 2-156
I-J-K
Income stream 1-181, 1-189
Integral 1-157
L
Loading Data Analysis 81
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
A-D E-L Q-T U-Z
Index:Index
page 3
Normal random variable 2-107,
2-119, 2-143
M
Macros 83
Marginal,
cost 1-64, 1-68
demand 1-70
profit 1-70
revenue 1-70
Mean,
continuous random variable
2-38
finite random variable 2-12
sample 2-46
Midpoint sums,
defined 1-149
animation 1-152
O
Omphaloskepsis 1-218
P
Parameter 2-47, 2-81
Preliminary reports 33
Present value of income stream
1-185
Probability density function,
definition 2-16
Probability mass function,
definition 2-2
T
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
A-D E-L M-P U-Z
Index: page
Index 4
P (continued)
Profit function 1-44
marginal 1-70
Project 1,
class project 54
team projects 58
Project 2,
class project 62
team projects 68
S-T
Sample mean 2-46
Standard deviation,
continuous 2-75
finite random variable 2-70
sample 2-86
sample mean 2-92
Standard normal random
variable 2-107, 2-119
Q
Quotation 2-227
R
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Random variable,
binomial 2-3, 2-13, 2-71
exponential 2-27, 2-42, 2-77
Running animations 23
Reports,
final oral 35
final written 40
preliminary 33
Revenue function 1-37
marginal 1-70
T
C
(S-T continues)
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A-D E-L M-P Q-T
Index: page
Index 5
S-T (continued)
Standardized random variable
2-82
Symbol font 77
U
Uniform random variable 2-27
V-W-X-Y-Z
Variance,
continuous random variable
2-75
finite random variable 2-69
sample 2-85
sample mean 2-92
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T
C
Marketing Computer Drives.
Executive Level Thinking
Executive Level Thinking
The mathematical predictions for marketing models are dependent on
the assumptions that we put into them. In the real world, our estimates for
things such as the size of the potential national market and the future costs of
production are almost never exactly correct. The goal is to build mathematical
models that are robust. This means that their predications do not fluctuate
widely from small errors in our data estimates.
Executive level thinking involves understanding the role which our
assumptions play in the predictions of our mathematical models. One of the
more subtle assumptions in the project on Marketing Computer Drives is our
choice of a 2nd degree polynomial as a model for the demand function.
In order for us to study revenue and profit, we must have a formula
for our demand function. Unfortunately, business settings never provide
formulas. We must construct these mathematically, using sound business
judgment. Excel does this by fitting a trend line to our test market data points.
There is no way to mathematically determine the “best fitting” trend line.
(material continues)
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Marketing Computer
Drives. Executive Level Thinking: page 2
Thinking
We must select a type of function from the list of those that are
available in Excel. Once the computer is given a specific type of function, it
can find the “best fitting” trend line of that type. If we select a type of
function that does not reflect the underlying demand, then Excel’s answer will
give us false information about the marketing problem.
As is the case with most test market data, our six points do not
strongly suggest any type of function. We used a 2nd degree polynomial for
our trend line, since it appears to fit the data quite well and its graph is a
common shape for demand functions. How crucial is this choice in the
outcome of the model? High level managers in marketing departments
consider such questions very carefully.
The best way to see the effect of our trend line choice is to investigate
the predictions that would be made by a different type of function. One simple
and useful such choice would be a linear model. For this we ask Excel to find
the best fitting trend line of the form D(q) = ax + b. The results are shown in
the worksheet Linear Trend Line in Marketing Focus.xlsm. The graph on the
following slide is copied from that worksheet.
(material continues)
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Marketing Computer
Drives. Executive Level Thinking: page 3
Thinking
Demand Functions
$500
Linear
D(q)
$400
Second Degree
$300
$200
$100
$0
0
400
800
1,200
1,600
2,000
2,400
2,800
3,200
q (K's)
Looking at the two types of demand functions, we see that both
models have very similar values in the middle range where they are fitted to
the test market data. They differ for both quite small and quite large sales
levels. The business interpretation of this is that the linear model predicts
higher prices than the quadratic model at extreme sales levels. Which, if
either, is correct? This is a business decision that a company
must make.
 Close
(material continues)
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Marketing Computer
Drives Executive Level Thinking: page 4
Thinking
To see how robust a marketing model we have created, we will check
the differences in the predictions of the two types of demand functions. A plot
of the two revenue functions is helpful.
Revenue Functions
$500
Linear
R(q) (M's)
$400
Second Degree
$300
$200
$100
$0
0
400
800
1,200
1,600
2,000
2,400
2,800
3,200
q (K's)
(material continues)
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Marketing Computer
Drives. Executive Level Thinking: page 5
Thinking
Using Solver in the worksheet Linear Trend Line, we find a
maximum revenue of $369,705,000 from selling 1,503,905 drives at a price of
$245.83 per drive. This is up less than 1% from the maximum revenue of
$366,155,000 that was predicted by the quadratic trend line for demand.
The wider spread of the linear model’s revenue graph on the right side
reflects the model’s predictions of more sales at very low prices. This is
probably irrelevant for any practical business decisions.
In general, our greatest interest is in profit. Graphs of the profit
functions determined by the two forms of trend lines for demand are shown in
a single plot on the next slide. In the critical region near their maximum
values they both have very similar “two humped” shapes. Using Solver and
the computational cells in the worksheet Linear Trend Line, we can compute
the price and quantity to sell that would maximize profit. These are compared
with the corresponding values for the quadratic trend line model for demand.
(material continues)
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Marketing Computer
Drives. Executive Level Thinking: page 6
Thinking
P(q) (M's)
Profit Functions
$70
$60
$50
$40
$30
$20
$10
$0
-$10 0
-$20
Linear
Second Degree
400
800
1,200
1,600
2,000
q (K's)
M aximu m P rofit
Price P er D rive Q u an tity to S ell
P rofit
Qu ad ratic Dem an d
$285.88
1,262,274
$42,176,000
L in ear Dem an d
$281.83
1,283,669
$41,552,000
-1.4%
1.7%
-1.5%
C h an ge
(material continues)
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Marketing Computer
Drives. Executive Level Thinking: page 7
Thinking
This is good news. Given the uncertain nature of our estimates for
production costs and the randomness of the test marketing process, a variation
of less than 2% in the predicted outcomes is of no practical business
consequence. We have developed a robust model, whose results do not
fluctuate widely depending upon our choice of a trend line form.
(material ends)
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