Z00_REND1011_11_SE_MOD6 PP2.QXD
2/21/11
12:48 PM
Page 1
MODULE
6
Calculus-Based
Optimization
LEARNING OBJECTIVES
After completing this module, students will be able to:
1. Find the slope of a curve at any point.
2. Find derivatives for several common types of
functions.
3. Find the maximum and minimum points on curves.
4. Use derivatives to maximize total revenue and other
functions.
MODULE OUTLINE
M6.1
M6.2
M6.3
M6.4
M6.5
M6.6
Introduction
Slope of a Straight Line
Slope of a Nonlinear Function
Some Common Derivatives
Maximum and Minimum
Applications
Summary • Glossary • Key Equations • Solved Problem • Self-Test • Discussion Questions and
Problems • Bibliography
M6-1
Z00_REND1011_11_SE_MOD6 PP2.QXD
M6-2
M6.1
2/21/11
12:48 PM
Page 2
MODULE 6 • CALCULUS-BASED OPTIMIZATION
Introduction
Although most of the quantitative techniques presented in this book are based on algebra, there
are many situations in which calculus and derivatives are helpful in finding the best solution to
a business problem. In this module, we begin by reviewing the slope of a line, as this is important in understanding the derivative of a function. Then we explain what a derivative is and how
it is used.
M6.2
Slope of a Straight Line
The equation of a straight line is
Y = a + bX.
Recall that the equation of a line can be written as
Y = a + bX
(M6-1)
where b is the slope of the line.
If we are given any two points, 1X1, Y12 and 1X2, Y22, that are on a line, the slope is
The slope of the line is the
change in Y over the change
in X.
Change in Y
Y2 - Y1
¢Y
=
=
Change in X
¢X
X2 - X1
b =
(M6-2)
where ¢ (delta) is used to represent “change in.”
For example, the slope of the line that passes through the points (2,3) and (4,7) is
b =
¢Y
7 - 3
4
=
= = 2
¢X
4 - 2
2
We can find the intercept, a, by using this slope and either point in the equation for a line. For
this example, if we select the point (2, 3), we have
Y = a + bX
3 = a + 2122
a = -1
The equation of this line is then
Y = -1 + 2X
This is illustrated in Figure M6.1. Notice that the slope of this line (and any other straight line)
is constant. We can pick any two points on the line and compute the slope using Equation M6-2.
For a nonlinear equation, the slope is not constant, as we explain in the next section.
FIGURE M6.1
Graph of Straight Line
14
Y = –1 + 2X
12
10
Y 8
(4,7)
⎧
⎪
⎨ ⌬Y = 7 – 3 = 4
⎪
⎩
6
4
–2
0
–2
–4
(2,3)
⎧
⎪
⎨
⎪
⎩
2
⌬X = 4 – 2 = 2
2
4
X
6
8
Z00_REND1011_11_SE_MOD6 PP2.QXD
2/21/11
12:48 PM
Page 3
M6.3
M6.3
SLOPE OF A NONLINEAR FUNCTION
M6-3
Slope of a Nonlinear Function
Figure M6.2 provides a graph of the function
Y = X2 - 4X + 6
If a curve is nonlinear, we can
find the slope at any point by
finding the slope of a tangent
line at that point.
Notice that this is not a straight line. To determine the slope of a curve at any point, we find the
slope of a line tangent to the curve at this point. For example, if we wish to find the slope of the
line when X = 3, we find a line that is tangent at that point. To find the slope of a line, we use
Equation M6-2, which requires us to have two points. We want the slope at X = 3, so we let
X1 = 3 and we find
Y1 = (3)2 - 4(3) + 6 = 3
Thus, one point will be (3,3). We wish to find another point close to this, so we will choose a
value of X2 close to X1 = 3. If we pick X2 = 5 for the other point (an arbitrary choice used simply to illustrate this process), we find
Y2 = 1522 - 4152 + 6 = 11
This other point is (5,11).
The slope of the line between these two points is
b =
¢Y
11 - 3
8
=
= = 4
¢X
5 - 3
2
To get a better estimate of the slope where X1 = 3, we choose a value of X2 even closer to
X1 = 3. Taking the point (3,3) with the point (4,6), we have
b =
¢Y
6 - 3
3
=
= = 3
¢X
4 - 3
1
Neither of these slopes would be exactly the same as the slope of the line tangent to this curve at
the point (3,3), as shown in Figure M6.3. However, we see that they each give us an estimate of
the slope of a tangent line. If we keep selecting points closer and closer to the point where
X = 3, we find slopes that are closer to the value we are trying to find.
To get a point very close to (3,3), we will use a value, ¢X, and add this to X = 3 to get the
second point. As we let ¢X get smaller, we find a point 1X + ¢X, Y22 closer to the original
point (3,3). From the original equation,
Y = X2 - 4X + 6
we have
Y1 = 32 - 4132 + 6 = 3
FIGURE M6.2
Graph of Quadratic
Function
12
Y = X 2 – 4X + 6
10
8
Y 6
4
2
–2
0
2
X
4
6
Z00_REND1011_11_SE_MOD6 PP2.QXD
M6-4
2/21/11
12:48 PM
Page 4
MODULE 6 • CALCULUS-BASED OPTIMIZATION
FIGURE M6.3
Graph of Tangent Line
and Other Lines
Connecting Points
20
Y = X 2 – 4X + 6
Line through
(3,3) and (5,11)
Line through
(3,3) and (4,6)
Tangent line at (3,3)
15
10
5
Y
–2
0
2
4
6
8
X
–5
–10
–15
and
Y2 = 13 + ¢X22 - 413 + ¢X2 + 6
= 19 + 6¢X + ¢X22 - 12 - 4¢X + 6 = ¢X2 + 2¢X + 3
Thus,
¢Y = Y2 - Y1 = 1¢X2 + 2¢X + 32 - 3 = ¢X2 + 2¢X
The slope is then
b =
The limit as ¢ X approaches 0
is used to find the slope of a
tangent line.
¢X1¢X + 22
¢Y
¢X2 + 2¢X
=
=
= ¢X + 2
¢X
¢X
¢X
As ¢X gets smaller, the value of ¢Y>¢X approaches the slope of the tangent line. The value of
the slope (b) in this example will approach 2. This is called the limit as ¢X approaches zero and
is written as
lim 1¢X + 22 = 2
¢X:0
It is obvious that at other points on the curve, the tangent line would have a different slope as Y
and ¢Y would not be the values given here.
To find a general expression for the slope of the tangent line at any point on a curve, we can
repeat this process for a general point X. Let X1 = X and X2 = X + ¢X, and let Y1 and Y2
represent the corresponding values for Y. For an equation of the form
Y = aX2 + bX + c
we let
Y1 = aX2 + bX + c
and
Y2 = a(X + ¢X)2 + b(X + ¢X) + c
Z00_REND1011_11_SE_MOD6 PP2.QXD
2/21/11
12:48 PM
Page 5
M6.4
SOME COMMON DERIVATIVES
M6-5
Expanding these expressions and simplifying, we find
¢Y = Y2 - Y1 = b(¢X) + 2aX(¢X) + c(¢X)2
Then
b(¢X) + 2aX(¢X) + c(¢X)2
¢X(b + 2aX + c¢X)
¢Y
=
=
= b + 2aX + c¢X
¢X
¢X
¢X
Taking the limit as ¢ X approaches zero, we have
lim (b + 2aX + c¢X) = b + 2aX
¢X:0
The derivative of a function is
used to find the slope of the
curve at a particular point.
This is the slope of the function at the point X, and it is called the derivative of Y. It is denoted
as Y¿ or dY>dX. The definition of a derivative is
Y¿ =
dY
¢Y
= lim e
f
¢X:0 ¢X
dX
(M6-3)
Fortunately, we will be using some common derivatives that are easy to remember, and it is not
necessary to go through this process of finding the limit every time we wish to find a derivative.
M6.4
Some Common Derivatives
Although derivatives exist for a number of different functional forms, we restrict our discussion
to the six most common ones. The references at the end of this module provide additional
information on derivatives. Table M6.1 gives a summary of the common derivatives. We provide
examples of each of these.
The derivative of a constant is 0.
If Y = c, then Y¿ = 0
c = constant
1.
(M6-4)
For example, if Y = 4, then Y¿ = 0. The graph of Y is a horizontal line, so the change in Y is
zero, regardless of the value of X.
If Y = Xn, then Y¿ = nXn - 1
2.
(M6-5)
For example,
if Y = X2, then Y¿ = 2X2 - 1 = 2X
if Y = X3, then Y¿ = 3X3 - 1 = 3X2
if Y = X9, then Y¿ = 9X9 - 1 = 9X8
If Y = cXn, then Y¿cnXn - 1
3.
TABLE M6.1
Some Common
Derivatives
FUNCTION
DERIVATIVE
Y = C
Y⬘ = 0
Y = Xn
Y⬘ = nXn⫺1
Y =
cXn
Y⬘ = cnXn⫺1
Y =
1
Xn
Y¿ =
-n
Xn + 1
Y = g(x) + h(x)
Y⬘ = g⬘(x) + h⬘(x)
Y = g(x) ⫺ h(x)
Y⬘ = g⬘(x) ⫺ h⬘(x)
(M6-6)
Z00_REND1011_11_SE_MOD6 PP2.QXD
M6-6
2/21/11
12:48 PM
Page 6
MODULE 6 • CALCULUS-BASED OPTIMIZATION
For example,
if Y = 4X3, then Y¿ = 4(3)X3 - 1 = 12X2
if Y = 2X4, then Y¿ = 2(4)X4 - 1 = 8X3
Note that Y =
If a function is the sum or
difference of two functions, the
derivative is the sum or
difference of the individual
derivatives of those two
functions.
1
- n
-n-1
= n+1
n , then Y¿ = - nX
x
X
If Y =
4.
(M6-7)
1
, is the same as Y = X - n. For example,
Xn
if Y =
1
-3
(or Y = X - 3), then Y¿ = -3X - 3 - 1 = -3X - 4 = 4
3
X
X
if Y =
2
-8
, then Y¿ = 2(-4)X - 4 - 1 = 5
4
X
X
If Y = g(x) + h(x), then Y¿ = g¿(x) + h¿(x)
5.
(M6-8)
For example,
if Y = 2X3 + X2 then Y¿ = 2(3)X3 - 1 + 2X2 - 1 = 6X2 + 2X
if Y = 5X4 + 3X2 then Y¿ = 5(4)X4 - 1 + 3(2)X2 - 1 = 20X3 + 6X
If Y = g(x) - h(x), then Y¿ = g¿(x) - h¿(x)
6.
(M6-9)
For example,
if Y = 5X3 - X2 then Y¿ = 5(3)X3 - 1 - 2X2 - 1 = 15X2 - 2X
if Y = 2X4 - 4X2 then Y¿ = 2(4)X4 - 1 - 4(2)X2 - 1 = 8X3 - 8X
The second derivative is the
derivative of the first derivative.
Second Derivatives
The second derivative of a function is the derivative of the first derivative. This is denoted as Y–
or d2Y>dX2. For example, if
Y = 6X4 + 4X3
then the first derivative is
dY
= 6(4)X4 - 1 + 4(3)X3 - 1
dX
= 24X3 + 12X2
Y¿ =
and
Y¿¿ =
d2Y
= 24(3)X3 - 1 + 12(2)X2 - 1
dX2
= 72X2 + 24X
The second derivative tells us about the slope of the first derivative, and it is used in finding the
maximum and minimum of a function. We use this in the next section.
M6.5
Maximum and Minimum
In using quantitative techniques in business, we often try to maximize profit or minimize cost. If
a profit function or cost function can be developed, taking a derivative may help us to find the
optimum solution. In dealing with nonlinear functions, we often look at local optimums, which
represent the maximums or minimums within a small range of X.
Z00_REND1011_11_SE_MOD6 PP2.QXD
2/21/11
12:48 PM
Page 7
M6.5
FIGURE M6.4
Graph of Curve with
Local Maximum and
Local Minimum
MAXIMUM AND MINIMUM
M6-7
Y
3
2
Y= 1
3 X – 4X + 12X + 3
A
B
0
0
A local maximum (or minimum)
is the highest (or lowest) point in
a neighborhood around that
point.
To find a local optimum, we find
the first derivative, set it equal to
0, and solve for X. This point is
called a critical point.
2
4
6
8
10
X
Figure M6.4 illustrates a curve where point A is a local maximum (it is higher than the
points around it), point B is a local minimum (it is lower than the points around it), but there is
no global maximum or minimum as the curve continues to increase without bound as X
increases, and it decreases without bound as X decreases. If we place limits on the maximum
and minimum values for X, then the endpoints can be checked to see if they are higher than any
local maximum or lower than any local minimum.
To find a local optimum, we take the derivative of the function and set it equal to zero.
Remember that the derivative gives the slope of the function. For a point to be a local maximum
or minimum, the tangent line must be a horizontal line, which has a slope of zero. Therefore,
when we set the derivative equal to zero and solve, we find a value of X that might be a local
maximum or minimum. Such a point is called a critical point.
The following function generated the graph in Figure M6.4:
Y =
1 3
X - 4X2 + 12X + 3
3
Point A gives a local maximum and point B gives a local minimum. To find the values of X
where these occur, we find the first derivative and set this equal to zero:
Y¿ = X2 - 8X + 12 = 0
Solving this for X, we factor this and have
(X - 2)(X - 6) = 0
so the critical points occur when X = 2 and X = 6.
The second derivative is
Y– = 2X - 8
At the first critical point, X = 2, so
Y– = 2(2) - 8 = -4
Because this is negative, this point is a local maximum. At the second critical point, where
X = 8,
Y– = 2(8) - 8 = 8
Because this is positive, this critical point is a local minimum.
Consider Figure M6.5, which is a graph of
Y = X3
Y¿ = 3X2
This derivative is equal to zero when X = 0. The second derivative is
Y– = 3(2)X2 - 1 = 6X
Z00_REND1011_11_SE_MOD6 PP2.QXD
M6-8
2/21/11
12:48 PM
Page 8
MODULE 6 • CALCULUS-BASED OPTIMIZATION
FIGURE M6.5
Graph of Function with
Point of Inflection at
Xⴝ0
Y
1.5
Y = X3
1
0.5
–1.5
–1 –0.5 0
0.5
1
1.5
X
–0.5
–1
–1.5
When X = 0, Y– = 6(0) = 0. Thus, this is neither a maximum nor a minimum but is a point of
inflection, as shown in Figure M6.5.
A critical point will be
1. a maximum if the second derivative is negative.
2. a minimum if the second derivative is positive.
3. a point of inflection if the second derivative is zero.
M6.6
Applications
There are many problems in which derivatives are used in business. We discuss a few of
these here.
Economic Order Quantity
The EOQ model is derived from
the total inventory cost function.
In Chapter 6, we show the formula for computing the economic order quantity (EOQ),
which minimizes cost when certain conditions are met. The total cost formula under these
conditions is
Total cost = (Total ordering cost) + (Total holding cost) + (Total purchase cost)
Q
D
TC =
Co + Ch + DC
Q
2
where
Q = order quantity
D = annual demand
Co = ordering cost per order
Ch = holding cost per unit per year
C = purchase (material) cost per unit
The variable is Q and all of the others are known constants. The derivative of the total cost with
respect to Q is
Ch
-DCo
dTC
+
=
2
dQ
2
Q
Z00_REND1011_11_SE_MOD6 PP2.QXD
2/21/11
12:48 PM
Page 9
M6.6
APPLICATIONS
M6-9
Setting this equal to zero and solving results in
Q = ;
2DCo
A Ch
We cannot have a negative quantity, so the positive value is the minimum cost. The second
derivative is
DCo
d2TC
=
2
dQ
Q3
If all the costs are positive, this derivative will be positive for any value of Q 7 0. Thus, this
point must be a minimum.
The formula for the EOQ model demonstrated here is the same formula used in Chapter 6.
This method of using derivatives to derive a minimum cost quantity can be used with many total
cost functions, even if the EOQ assumptions are not met.
Total Revenue
In analyzing inventory situations, it is often assumed that whatever quantity is produced can be
sold at a fixed price. However, we know from economics that demand is a function of the price.
When the price is raised, the demand declines. The function that relates the demand to the price
is called a demand function.
Suppose that historical sales data indicate that the demand function for a particular
product is
Q = 6,000 - 500P
where
Q = quantity demanded (or sold)
P = price in dollars
Total revenue equals total
demand times the selling price.
The total revenue function is
Total revenue = Price * Quantity
TR = PQ
Substituting for Q (using the preceding demand function) in this equation gives us
TR = P(6,000 - 500 P)
TR = 6,000 P - 500 P2
To maximize total revenue, find
the derivative of the total
revenue function, set it equal to
0, and solve.
A graph illustrating this total revenue function is provided in Figure M6.6. To find the price that
will maximize the total revenue, we find the derivative of total revenue:
TR¿ = 6,000 - 1,000P
Setting this equal to zero and solving, we have
P = 6
Thus, to maximize total revenue, we set the price at $6. The quantity that will be sold at this
price is
Q = 6,000 - 500 P = 6,000 - 500(6) = 3,000 units
The total revenue is
TR = PQ = 6(3,000) = $18,000
Z00_REND1011_11_SE_MOD6 PP2.QXD
M6-10
2/21/11
12:48 PM
Page 10
MODULE 6 • CALCULUS-BASED OPTIMIZATION
FIGURE M6.6
Total Revenue Function
20,000
Y = 6,000X – 500X 2
18,000
16,000
Revenue
14,000
12,000
10,000
8,000
6,000
4,000
2,000
0
0
5
10
15
Price
Summary
In this module we use the concept of limits to demonstrate that
the derivative is the slope of a curve. We use several common
derivatives to find the maximums and minimums of functions.
The local optimums are found by taking the derivative of a
function, setting it equal to zero, and solving. If the second derivative is negative at this point, the local optimum is a maximum. If the second derivative is positive at this point, the local
optimum is a minimum. If the second derivative is zero, the
point is called a point of inflection.
Derivatives can be used to find the EOQ, which minimizes
total inventory cost. A total revenue function is based on a demand curve. The derivative of this shows where revenues are
maximized.
Glossary
Critical Point A point on a curve where the first derivative
is equal to zero.
Derivative The slope of a function at a particular point. It is
the limit of the change in Y over the change in X, as the
change in X approaches 0.
Local Maximum The highest point in the neighborhood
around that point.
Local Minimum The lowest point in the neighborhood
around that point.
Point of Inflection A point on a curve where the first
derivative is equal to zero, but the point is neither a
maximum nor a minimum.
Second Derivative The derivative of the first derivative.
Slope The slope of a line is the change in the Y value divided
by the change in the X value.
Key Equations
(M6-1) Y = a + bX
Equation of a straight line.
Y2 - Y1
Change in Y
¢Y
=
=
Change in X
¢X
X2 - X1
Slope of a straight line found from two points.
(M6-2) b =
¢Y
dY
= lim e
f
¢X: 0 ¢X
dX
Definition of derivative.
(M6-3) Y¿ =
(M6-4) If Y = c where c is any constant, then Y¿ = 0
Derivative of a constant function.
(M6-5) If Y = Xn, then Y¿ = nXn - 1
Derivative of X raised to a power.
(M6-6) If Y = cXn, then Y¿cnXn - 1
Derivative of a constant times a function of X.
(M6-7) If Y = 1>Xn, then Y¿ = -nX - n - 1 = -n>Xn + 1
Derivative of 1>X to a power.
(M6-8) If Y = g(x) + h(x), then Y¿ = g¿(x) + h¿(x)
Derivative of the sum of two functions of X.
(M6-9) If Y = g(x) - h(x), then Y¿ = g¿(x) - h¿(x)
Derivative of the difference of two functions of X.
Z00_REND1011_11_SE_MOD6 PP2.QXD
2/21/11
12:48 PM
Page 11
SELF-TEST
M6-11
Solved Problem
Solved Problem M6-1
The following function relates the price for a product to the quantity sold:
P = 6 - 0.75Q
Find the price that will maximize total revenue.
Solution
Total revenue = Price * Quantity
TR = PQ
TR = (6 - 0.75Q)Q = 6Q - 0.75Q 2
So,
TR¿ = 6 - 0.75(2)Q
Setting this equal to zero and solving, we have
6 - 1.5Q = 0
or
Q = 4
Then,
P = 6 - 0.75Q = 6 - 0.75(4) = 6 - 3 = 3
Self-Test
䊉
䊉
䊉
Before taking the self-test, refer back to the learning objectives at the beginning of the module and the glossary at the end of
the module.
Use the key at the back of the book to correct your answers.
Restudy pages that correspond to any questions that you answered incorrectly or material you feel uncertain about.
1. The slope of a curve at a particular point can be found
a. by computing the first derivative of the function for
that curve.
b. by computing the second derivative of the function for
that curve.
c. by computing the change in X divided by the change
in Y.
d. by dividing the value of Y at that point by the value of X.
2. If the first derivative is set equal to zero and solved for X,
at that point (X) the curve could be
a. at a maximum.
b. at a minimum.
c. at an inflection point.
d. any of the above.
3. A critical point is a point where
a. the slope of the function is 0.
b. the Y value for the function is 0.
c. the X value for the function is 0.
d. the value of Y/X is 0.
4. A function is at a maximum at the point X = 5. At this
point,
a. the value of the second derivative is 0.
b. the value of the second derivative is negative.
c. the value of the second derivative is positive.
d. the value of the first derivative is negative.
5. A point of inflection occurs when
a. the first derivative is 0.
b. the second derivative is 0.
c. the first and the second derivatives are both 0.
d. the first and the second derivatives are both positive.
6. The EOQ model can be derived from the derivative of
a. the holding cost function.
b. the ordering cost function.
c. the purchase cost function.
d. the total inventory cost function.
7. To find the quantity that maximizes total revenue,
we should
a. set total revenue equal to total cost.
b. find the derivative of the total cost function.
c. find the derivative of the total profit function.
d. find the derivative of the total revenue function.
Z00_REND1011_11_SE_MOD6 PP2.QXD
M6-12
2/21/11
12:48 PM
Page 12
MODULE 6 • CALCULUS-BASED OPTIMIZATION
Discussion Questions and Problems
Discussion Questions
M6-1 Explain how to find the slope of a straight line.
M6-2 Explain how to find the slope of a nonlinear function
at a particular point, using limits.
M6-3 What is the procedure for finding the maximum or
minimum of a function? How is the second derivative used in this process?
M6-4 What are critical points?
Problems
M6-5 Find the derivative of the following functions.
(a) Y = 2X3 - 3X2 + 6
(b) Y = 4X5 + 2X3 - 12X + 1
(c) Y = 1>X2
(d) Y = 25>X4
M6-6 Find the second derivative for each of the functions
in Problem M6-5.
M6-7 Find the derivatives of the following functions.
(a) Y = X6 - 0.5X2 - 16
(b) Y = 5X4 + 12X2 + 10X + 21
(c) Y = 2>X3
(d) Y = 25>2X4
M6-8 Find the second derivative for each of the functions
in Problem M6-7.
M6-9 Find the critical point for the function Y = 6X2 5X + 4. Is this a maximum, minimum, or point of
inflection?
M6-10 Find the critical points for the function Y =
A 13 B X3 - 5X2 + 25X + 12. Identify each critical
point as a maximum or a minimum or a point of
inflection.
M6-11 Find the critical point for the function Y = X3 - 20.
Is this a maximum, minimum, or point of inflection?
M6-12 The total revenue function for a particular product is
TR = 1,200Q - 0.25Q 2. Find the quantity that will
maximize total revenue. What is the total revenue
for this quantity?
M6-13 The daily demand function for the AutoBright
car washing service has been estimated to be
Q = 75 - 2P, where Q is the quantity of cars and P
is the price. Determine what price would maximize
total revenue. How many cars would be washed at
this price?
M6-14 The total revenue function for AutoBright car washing service (see Problem M6-13) is believed to be
incorrect due to a changing demand pattern. Based
on new information, the demand function is now
estimated to be Q = 180 - 2P2. Find the price that
would maximize total revenue.
M6-15 The total cost function for the EOQ model is
TC =
Q
D
Co + Ch + DC
Q
2
In a particular inventory situation, annual demand
(D) is 2,000, ordering cost per order (Co) is $25,
holding cost per unit per year (Ch) is $10, and
purchase (material) cost per unit (C) is $40. Write
the total cost function with these values. Take the
derivative and find the quantity that minimizes cost.
M6-16 Find the second derivative of the total cost function
in Problem M6-15 and verify that the value at the
critical point is a minimum.
Bibliography
Barnett, Raymond, Michael Ziegler, and Karl Byleen. College Mathematics
for Business, Economics, Life Sciences, and Social Sciences, 9th ed.
Upper Saddle River, NJ: Prentice Hall, 2002.
Haeussler, Ernest, and Richard Paul. Introductory Mathematical Analysis for
Business, Economics and the Life and Social Sciences, 10th ed. Upper
Saddle River, NJ: Prentice Hall, 2002.