វិទ្យាស្ថា ន អេសធី អលេម ិនស ៍St Clements Institute
គណិតវិភាគ Mathematical Analysis
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ប្លង់គោលនៃមុខវជិ្ជា ១. ចណំងគ ើងមខុវជិ្ជា គណិតវភិាគ (Mathematical Analysis) ២. គ្គឧូគទេស អុិត ផាៃី (ITH Phanny);
email: [email protected] ៣. គលខកដូមខុវជិ្ជា ៤. ចៃំៃួគគ្កឌតី ៣ (៤៥ គ ៉ោ ង កនងុ ១៥សបា្ត ហ)៍ 4. ខ្លមឹសារមខុ្វជិ្ជា : (COURSE DESCRIPTION)
មុខ្វជិ្ជា គណិតវទិ្យាវភិាគន េះ រតូវបា នរៀបចនំ ើង នោយមា នោលបំណង ន ើមបផី្តល់ ូវនេចក្តីណណនអំំពី គំ ិតនន ទាក្ទ់្យង ងឹគណិតវទិ្យា ិងមូលោា រគឹេះ េរមាបក់្មមវធិីឆ្ន េំក្ាមូលោា ន ើយវាក្ជ៏្ជ ឧបក្រណ៍ បនចេក្នទ្យេ នលើណផ្នក្ជ្ជនរចើ ន អាជីវក្មម ូចជ្ជ ៖ នេ ាក្ចិេ គណន យយ ងិជីវតិជ្ជក្ណ់េែង ិងវទិ្យាសាស្តេតេងគម ។ វាចាបន់ផ្តើម របធា បទ្យ ណ លម ិណម ជ្ជការគណន ូចជ្ជ គណិតវទិ្យាមូលោា , េមកីារ , អ ុគម ,៍ ពីជគណិត, មា៉ា រទ្យីេ , គណិតវទិ្យា ិរញ្ញ វតថុ ជ្ជន ើម ។ ភាពេមបរូណ៍ណបប ិង ភាពខុ្េោន ឹងមា ការអ ុវតតជ្ជនរចើ នៅក្នុងវគគេិក្ាន េះ។ ិេសតិ ងឹប តនមើលន ើញកា ណ់តចាេ់ពីរនបៀប ណ លពួក្នគ ក្ំពុងនរៀ គណិតវទិ្យា ថាវាអាចរតូវបា នគយក្នៅអ ុវតត ឹងបញ្ហា អាជីវក្មមជ្ជក្ណ់េតងយ៉ា ងណាខ្លេះ ។ ការអ ុវតត ឹងមា ចរមរេះ ូចជ្ជ ការអ ុវតតនលើ ពាណិជាក្មម, នេ ាក្ិចេ, េងគម ិរញ្ញ វតថុ ។ល។
5. នោលបណំងន មខុ្វជិ្ជា (LEARNING OBJECTIVES)
At the end of this course, students will be able to:
model situations described by linear or quadratic equations.
solve linear inequalities in one variable and to introduce interval notation.
model real-life situations in terms of inequalities.
solve equations and inequalities involving absolute values.
write sums in summation notation and evaluate such sums.
understand what functions and domains are.
introduce addition, subtraction, multiplication, division, and multiplication by a constant.
introduce inverse functions and properties.
graph equations and functions.
develop the notion of slope and different forms of equations of lines.
develop the notion of demand and supply curves and to introduce linear functions.
sketch parabolas arising from quadratic functions.
solve interest problems which require logarithms.
solve problems involving the time value of money.
solve problems with interest is compounded continuously.
introduce the notions of ordinary annuities and annuities due.
learn how to amortize a loan and set up an amortization schedule.
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6. រង្វវ យតនមលមខុ្វជិ្ជា លក្ខណៈវ ិចិឆយ័ ព ិទុជ្ជភាគរយ
វតតមា ិងការចូលរមួ ១០ ក្ិចេការេំនណរ បទ្យបង្វា ញ ិងលំហាតន់ធវើនៅផ្ទេះ ២០ រប ងពាក្ក់្ណាត លឆមាេ ២០ រប ងបញ្េបឆ់មាេ ៥០
េរបុ ១០០ 7. បលងន់មនរៀ េនងខប
WEEK LECTURE AND DESCRIPTION Time
1
1- Introduction to Mathematical Analysis
1.0 Why Study Mathematical Analysis?
1.1 Applications of Equations?
1.2 Linear Inequalities
3h Ch01
2
1.3 Applications of Inequalities
1.4 Absolute Value
1.5 Summation Notation
1.6 Sequences
3h Sample Quiz 1
3
2- Functions and Graphs 2.1 Functions
2.2 Special Functions
3h
Ch02
4
2.3 Combinations of Functions
2.4 Inverse Functions
2.5 Graphs in Rectangular Coordinates
3h
5
2.6 Symmetry
2.7 Translations and Reflections
2.8 Functions of Several Variables
3h Sample Quiz 2
6
3- Lines, Parabolas, and Systems
3.1 Lines
3.2 Applications and Linear Functions
3.3 Quadratic Functions
3h
Ch03
7
3.4 Systems of Linear Equations
3.5 Nonlinear Systems
3.6 Applications of Systems of Equations
3h
8
4- Exponential and Logarithmic Functions
4.1 Exponential Functions
4.2 Logarithmic Functions
3h Miterm Exam Ch01-03
9 4.3 Properties of Logarithms
4.4 Logarithmic and Exponential Equations 3h Ch04
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10
5- Mathematics of Finance
5.1 Compound Interest
5.2 Present Value
5.3 Interest Compounded Continuously
3h Sample Quiz 3
11 5.4 Annuities
5.5 Amortization of Loans 3h Ch05
12
6- Matrix Algebra
6.1 Matrices
6.2 Matrix Addition and Scalar Multiplication
6.3 Matrix Multiplication
3h Sample Quiz 4
13
6.4 Solving Systems by Reducing Matrices
6.5 Solving Systems by Reducing Matrices (continued)
6.6 Inverses
3h
Ch06-07
14
7- Linear Programming
7.1 Linear Inequalities in Two Variables
7.2 Linear Programming
3h
15 Comprehensive Review and Summary 3h Preparation for Final Exam
8. វធីិសាស្រ្តបង្រៀន មុខវជិ្ជា ងនេះ្រូវ្ិក្សាតាមវធីិសាស្រ្ត ដូចខារង្ោម៖
- កិ្សចចោរជ្ជបុគ្គល កិ្សចចោរ (Individual Work, Group Work, and Presentation)
- (In-Class Exercise Correction) - 5 Short Video Lectures
9. ឯក្សសារងោរ References 1) Introductory Mathematical Analysis for Business, Economics, and the Life and Social
Sciences, Ernest F Haeussler, Jr, Richard S Paul.
2) Mathematics for Economics and Finance Methods and Modeling, Martin Anthony and
Norman Biggs.
3) Essential mathematics for Economics and Business, Teresa Bradey and Paul Patton.
4) គណិតវទិ្យាវភិាគ សម្រាប់ឆ្ន ាំសិក្សាមូលដ្ឋា ន, លោក្ស លឹមយូលសង
10. MISCELLANEOUS NEEDED OR DESIRED In general, your participation, ideas, comments, suggestions, questions, grade challenges, etc.
are welcome. However, your care in these matters is expected. No part of your grade will be
based on anything other than your midterm and final exam, assignment, class participation and
attendance or attitude. You are encouraged to take advantage of this course or anything else
connected with the course, sound decision and to take one more useful step to your progress/
success and future job.
INTRODUCTORY MATHEMATICAL ANALYSISINTRODUCTORY MATHEMATICAL ANALYSISFor Business, Economics, and the Life and Social Sciences
Chapter 1 Chapter 1
Applications and More AlgebraApplications and More Algebra
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• To model situations described by linear or quadratic equations.
• To solve linear inequalities in one variable and to introduce interval notation.
Chapter 1: Applications and More Algebra
Chapter ObjectivesChapter Objectives
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• To model real-life situations in terms of inequalities.
• To solve equations and inequalities involving absolute values.
• To write sums in summation notation and evaluate such sums.
Chapter 1: Applications and More Algebra
Chapter OutlineChapter Outline
Applications of Equations
Linear Inequalities
Applications of Inequalities
Absolute Value
1.1)
1.2)
1.3)
1.4)
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Absolute Value
Summation Notation
1.6) Sequence
1.4)
1.5)
• Modeling: Translating relationships in the problems to mathematical symbols.
Chapter 1: Applications and More Algebra
1.1 Applications of Equations1.1 Applications of Equations
A chemist must prepare 350 ml of a chemical
Example 1 - Mixture
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A chemist must prepare 350 ml of a chemical
solution made up of two parts alcohol and three
parts acid. How much of each should be used?
Solution:Let n = number of milliliters in each part.
350
3505
35032
=
=+
n
nn
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 1 - Mixture
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Each part has 70 ml. Amount of alcohol = 2n = 2(70) = 140 mlAmount of acid = 3n = 3(70) = 210 ml
705
350 ==n
• Fixed cost is the sum of all costs that are independent of the level of production.
• Variable cost is the sum of all costs that are dependent on the level of output.
• Total cost = variable cost + fixed cost
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
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• Total cost = variable cost + fixed cost
• Total revenue = (price per unit) x (number of units sold)
• Profit = total revenue − total cost
The Anderson Company produces a product for
which the variable cost per unit is $6 and the fixed
cost is $80,000. Each unit has a selling price of
$10. Determine the number of units that must be
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 3 – Profit
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sold for the company to earn a profit of $60,000.
Solution:
Let q = number of sold units.
variable cost = 6q
total cost = 6q + 80,000
total revenue = 10q
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 3 – Profit
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total revenue = 10q
Since profit = total revenue − total cost
35,000 units must be sold to earn a profit of $60,000.
( )
q
q
=
=
+−=
000,35
4000,140
000,80610000,60
A total of $10,000 was invested in two business ventures, A and B. At the end of the first year, A and B yielded returns of 6%and 5.75 %, respectively, on the original investments. How was the original amount allocated if the total amount earned was $588.75?
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 5 – Investment
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$588.75?
Solution:
Let x = amount ($) invested at 6%.
( ) ( )( )
75.130025.0
75.5880575.057506.0
75.588000,100575.006.0
=
=−+
=−+
x
xx
xx
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 5 – Investment
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$5500 was invested at 6%
$10,000−$5500 = $4500 was invested at 5.75%.
5500
75.130025.0
=
=
x
x
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent
A real-estate firm owns the Parklane Garden Apartments, which consist of 96 apartments. At $550 per month, every apartment can be rented. However, for each $25 per month increase, there will be three vacancies with no possibility of filling them. The firm wants to receive $54,600 per month
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them. The firm wants to receive $54,600 per month from rent. What rent should be charged for each apartment?
Solution 1:
Let r = rent ($) to be charged per apartment.
Total rent = (rent per apartment) x (number of apartments rented)
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent
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Solution 1 (Con’t):( )
25
34050600,54
25
165032400600,54
25
550396600,54
−=
+−=
−−=
rr
rr
rr
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent
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Rent should be $650 or $700.
( )
( ) ( )( )( )
256756
500,224050
32
000,365,13440504050
0000,365,140503
34050000,365,1
25600,54
2
2
±=±
=
−−±=
=+−
−=
=
r
rr
rr
r
Solution 2:
Let n = number of $25 increases.
Total rent = (rent per apartment) x (number of apartments rented)
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent
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(number of apartments rented)
Solution 2 (Con’t):
( )( )
02410
0180075075
75750800,52600,54
39625550600,54
2
2
2
=+−
=+−
−+=
−+=
nn
nn
nn
nn
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent
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The rent charged should be either
550 + 25(6) = $700 or
550 + 25(4) = $650.
( )( )
4 or 6
046
02410
=
=−−
=+−
n
nn
nn
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities1.2 Linear Inequalities
• Supposing a and b are two points on the real-number line, the relative positions of two points are as follows:
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• We use dots to indicate points on a number line.
• Suppose that a < b and x is between a and b.
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
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• Inequality is a statement that one number is less than another number.
• Rules for Inequalities:
1. If a < b, then a + c < b + c and a − c < b − c.
2. If a < b and c > 0, then ac < bc and a/c < b/c.
3. If a < b and c < 0, then a(c) > b(c) and a/c > b/c.
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
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3. If a < b and c < 0, then a(c) > b(c) and a/c > b/c.
4. If a < b and a = c, then c < b.
5. If 0 < a < b or a < b < 0, then 1/a > 1/b .
6. If 0 < a < b and n > 0, then an < bn.
If 0 < a < b, then .nn ba <
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
• Linear inequality can be written in the form
ax + b < 0where a and b are constants and a ≠ 0
• To solve an inequality involving a variable is to find all values of the variable for which the
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find all values of the variable for which the inequality is true.
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
Example 1 – Solving a Linear Inequality
Solve 2(x − 3) < 4.
Solution:Replace inequality by equivalent inequalities.
( ) 432 <−x
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5
2
10
2
2
102
64662
462
<
<
<
+<+−
<−
x
x
x
x
x
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
Example 3 – Solving a Linear Inequality
Solve (3/2)(s − 2) + 1 > −2(s − 4).
( ) ( )42122
3 −−>+− ss
Solution:
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( ) ( )[ ]
( )[ ] ( )
7
20
207
16443
442232
422122
32
2
>
>
+−>−
−−>+−
−>
+−
s
s
ss
ss
s-s
The solution is ( 20/7 ,∞).
Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities1.3 Applications of Inequalities
Example 1 - Profit
• Solving word problems may involve inequalities.
For a company that manufactures aquarium heaters, the combined cost for labor and material is
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heaters, the combined cost for labor and material is $21 per heater. Fixed costs (costs incurred in a given period, regardless of output) are $70,000. If the selling price of a heater is $35, how many must be sold for the company to earn a profit?
Solution:
profit = total revenue − total cost
( )0 cost total revenue total >−
Let q = number of heaters sold.
Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities
Example 1 - Profit
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( )
5000
000,7014
0000,702135
>
>
>+−
q
q
After consulting with the comptroller, the president of the Ace Sports Equipment Company decides to take out a short-term loan to build up inventory. The company has current assets of $350,000 and current liabilities of $80,000. How much can the company borrow if the current ratio is to be no less
Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities
Example 3 – Current Ratio
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company borrow if the current ratio is to be no less than 2.5? (Note: The funds received are considered as current assets and the loan as a current liability.)
Solution:
Let x = amount the company can borrow.
Current ratio = Current assets / Current liabilities
We want,
x+000,350
Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities
Example 3 – Current Ratio
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The company may borrow up to $100,000.
( )
x
x
xx
x
x
≥
≥
+≥+
≥+
+
000,100
5.1000,150
000,805.2000,350
5.2000,80
000,350
• On real-number line, the distance of x from 0 is called the absolute value of x, denoted as |x|.
Chapter 1: Applications and More Algebra
1.4 Absolute Value1.4 Absolute Value
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DEFINITION
The absolute value of a real number x, written |x|, is defined by
<−
≥=
0 if ,
0 if ,
xx
xxx
Chapter 1: Applications and More Algebra
1.4 Absolute Value
Example 1 – Solving Absolute-Value Equations
a. Solve |x − 3| = 2
b. Solve |7 − 3x| = 5
c. Solve |x − 4| = −3
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Solution:a. x − 3 = 2 or x − 3 = −2
x = 5 x = 1
b. 7 − 3x = 5 or 7 − 3x = −5 x = 2/3 x = 4
Chapter 1: Applications and More Algebra
1.4 Absolute Value
Example 1 – Solving Absolute-Value Equations
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x = 2/3 x = 4
c. The absolute value of a number is never negative. The solution set is ∅.
Absolute-Value Inequalities
• Summary of the solutions to absolute-value inequalities is given.
Chapter 1: Applications and More Algebra
1.4 Absolute Value
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Chapter 1: Applications and More Algebra
1.4 Absolute Value
Example 3 – Solving Absolute-Value Equations
a. Solve |x + 5| ≥ 7
b. Solve |3x − 4| > 1
Solution:a. 75 or 75 −≥+−≤+ xx
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a.
We write it as , where ∪ is the union symbol.
b.
We can write it as .
2 12
75 or 75
≥−≤
−≥+−≤+
xx
xx
] [( )∞−∞− ,212, U
3
5 1
143 or 143
><
>−−<−
xx
xx
( )
∞∪∞− ,
3
51,
Properties of the Absolute Value
• 5 basic properties of the absolute value:
b
a
b
a
baab
=
⋅=
.2
.1
Chapter 1: Applications and More Algebra
1.4 Absolute Value
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• Property 5 is known as the triangle inequality.
baba
aaa
abba
bb
+≤+
≤≤−
−=−
.5
.4
.3
Chapter 1: Applications and More Algebra
1.4 Absolute Value
Example 5 – Properties of Absolute Value
( )
777777
77 c.
24224 b.
213737- a.
−−−−
−=−
=−=−
=⋅−=⋅
xx
Solution:
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( ) 323251132 g.
222 f.
5
3
5
3
5
3 e.
3
7
3
7
3
7 ;
3
7
3
7
3
7 d.
+−=+=≤==+−
≤≤
−=
−
−=
−
−
=−
−=
−
−=
−=
−
-
xxx
Chapter 1: Applications and More Algebra
1.5 Summation Notation1.5 Summation Notation
DEFINITION
The sum of the numbers ai, with i successively taking on the values m through n is denoted as
nmmm
n
i aaaaa ++++= ++∑ ...21
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nmmmmi
i aaaaa ++++= ++
=
∑ ...21
Evaluate the given sums.
a. b.
Solution:
a.
Chapter 1: Applications and More Algebra
1.5 Summation Notation
Example 1 – Evaluating Sums
( )∑=
−7
3
25n
n ( )∑=
+6
1
2 1j
j
( ) ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ]275265255245235257
−+−+−+−+−=−∑ n
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a.
b.
( ) ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ]
115
3328231813
275265255245235253
=
++++=
−+−+−+−+−=−∑=n
n
( ) ( ) ( ) ( ) ( ) ( ) ( )
97
3726171052
1615141312111 2222226
1
2
=
+++++=
+++++++++++=+∑=j
j
• To sum up consecutive numbers, we have
( )2
1
1
+=∑
=
nni
n
i
Chapter 1: Applications and More Algebra
1.5 Summation Notation
( )6
)12(1
1
2 ++=∑
=
nnni
n
i
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where n = the last number
61=i
Evaluate the given sums.
a. b. c.
Solution:
a.
Chapter 1: Applications and More Algebra
1.5 Summation Notation
Example 3 – Applying the Properties of Summation Notation
2847144471100
=⋅==∑∑
( )∑=
+100
1
35k
k ∑=
200
1
29k
k∑=
100
30
4j
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a.
b.
c.
( ) ( ) 550,2510032
1011005 3535
100
1
100
1
100
1
=+
⋅=+=+ ∑∑∑
=== kkk
kk
300,180,246
401201200999
200
1
2200
1
2 =
⋅⋅== ∑∑
== kk
kk
28471444130
=⋅==∑∑== ij
1.6 Sequence
• Arithmetic sequence
�An arithmetic sequence is a sequence (bk) defined recursively by
b1=a and, for each positive integer k, bk+1= d + bk
Example
1.5, 1.5+0.7 , 1.5+2*0.7, 1.5 +3*0.7, 1.5+ 4*0.7, 1.5+5*0.7
1.5, 2.2, 2.9, 3.6, 4.3, 5.0
• Geometric sequence
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• Geometric sequence
�A geometric sequence is a sequence (ck) defined recursively by
c1=a and, for each positive integer k, ck+1= ck*rExample
2 2*3 , 2*3*3, 2*3*3*3, 2*3*3*3*3
2, 6, 18, 48, 144
Sums of sequences
• Sum of an arithmetic sequence - first n term
First term – a, common difference – d
• Sum of an geometric sequence
First term – a, common ratio – r
)2)1((2
adnn
sn +−=
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First term – a, common ratio – r
- Sum to first n term
- Sum of an infinite geometric sequence
for
1,1
)1(≠
−
−= rfor
r
ras
n
n
∑∞
=
−
−=<
1
1
1,1
i
i
r
aarr
• Example 1
A rich woman would like to leave $100,000 a year, starting now, to be divided equally among all her direct descendants. She puts no time limit on his bequeathment and is able to invest for this long-term outlay of funds at 2% compounded
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outlay of funds at 2% compounded annually. How much must she invest now to meet such a long-term commitment?
• Solution:
Let us write R=100,000, set the clock to 0 now, and measure times is years from now. With these conventions we are to account for payments for R payments of R at times 0,1,2…..k,.. By making a single investment now. Then the investment must equal to the sum
....)02.1(......)02.1()02.1(21
+++++−−− k
RRRR
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....)02.1(......)02.1()02.1(21
+++++−−− k
RRRR
First term=a=R=100,000
Common ratio=r=(1.02)-1. Since |r|<1, we can evaluate the required investment as
000,100,5
02.1
11
000,100
1=
−
=− r
a
INTRODUCTORY MATHEMATICAL ANALYSISINTRODUCTORY MATHEMATICAL ANALYSISFor Business, Economics, and the Life and Social Sciences
Chapter 2 Chapter 2
Functions and GraphsFunctions and Graphs
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• To understand what functions and domains are.
• To introduce different types of functions.
• To introduce addition, subtraction, multiplication, division, and multiplication by a constant.
Chapter 2: Functions and Graphs
Chapter ObjectivesChapter Objectives
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constant.
• To introduce inverse functions and properties.
• To graph equations and functions.
• To study symmetry about the x- and y-axis.
• To be familiar with shapes of the graphs of six basic functions.
Functions
Special Functions
Combinations of Functions
Inverse Functions
Chapter 2: Functions and Graphs
Chapter OutlineChapter Outline
2.1)
2.2)
2.3)
2.4)
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Inverse Functions
Graphs in Rectangular Coordinates
Symmetry
Translations and Reflections
2.8) Functions of Several Variables
2.4)
2.5)
2.6)
2.7)
• A function assigns each input number to one output number.
• The set of all input numbers is the domain of the function.
• The set of all output numbers is the range.
Chapter 2: Functions and Graphs
2.1 Functions2.1 Functions
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• The set of all output numbers is the range.
Equality of Functions
• Two functions f and g are equal (f = g):
1.Domain of f = domain of g;
2. f(x) = g(x).
Chapter 2: Functions and Graphs2.1 Functions
Example 1 – Determining Equality of Functions
Determine which of the following functions are equal.
≠+
+=
−
−+=
2)( b.
)1(
)1)(2()( a.
xxg
x
xxxf
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=
≠+=
=
≠+=
1 if 3
1 if 2)( d.
1 if 0
1 if 2)( c.
x
xxxk
x
xxxh
Chapter 2: Functions and Graphs2.1 Functions
Example 1 – Determining Equality of Functions
Solution:When x = 1,
( ) ( )( ) ( )( ) ( ) 11
, 11
, 11
kf
hf
gf
≠
≠
≠
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By definition, g(x) = h(x) = k(x) for all x ≠ 1.Since g(1) = 3, h(1) = 0 and k(1) = 3, we conclude that
kh
hg
kg
≠
≠
=
,
,
Chapter 2: Functions and Graphs2.1 Functions
Example 3 – Finding Domain and Function Values
Let . Any real number can be used for x, so the domain of g is all real numbers.
a. Find g(z).
Solution:
2( ) 3 5g x x x= − +
2( ) 3 5g z z z= − +
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b. Find g(r2).
Solution:
c. Find g(x + h).Solution:
2 2 2 2 4 2( ) 3( ) 5 3 5
g r r r r r= − + = − +
2
2 2
( ) 3( ) ( ) 5
3 6 3 5
g x h x h x h
x hx h x h
+ = + − + +
= + + − − +
Chapter 2: Functions and Graphs2.1 Functions
Example 5 – Demand Function
Suppose that the equation p = 100/q describes the
relationship between the price per unit p of a certain
product and the number of units q of the product that
consumers will buy (that is, demand) per week at the
stated price. Write the demand function.
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Solution: pq
q =100
a
Chapter 2: Functions and Graphs
2.2 Special Functions2.2 Special Functions
Example 1 – Constant Function
• We begin with constant function.
Let h(x) = 2. The domain of h is all real numbers.
(10) 2 ( 387) 2 ( 3) 2h h h x= − = + =
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A function of the form h(x) = c, where c = constant, is a constant function.
(10) 2 ( 387) 2 ( 3) 2h h h x= − = + =
Chapter 2: Functions and Graphs
2.2 Special Functions
Example 3 – Rational Functions
Example 5 – Absolute-Value Function
a. is a rational function, since the
numerator and denominator are both polynomials.
b. is a rational function, since .
2 6( )
5
x xf x
x
−=
+
( ) 2 3g x x= +2 3
2 31
xx
++ =
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Example 5 – Absolute-Value Function
Absolute-value function is defined as , e.g. x
if 0
if 0
x xx
x x
≤ =
− <
Chapter 2: Functions and Graphs
2.2 Special Functions
Example 7 – Genetics
Two black pigs are bred and produce exactly five
offspring. It can be shown that the probability P that
exactly r of the offspring will be brown and the others
black is a function of r ,5
1 35!
4 4( ) 0,1,2,...,5
r r
P r r
−
= =
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On the right side, P represents the function rule. On
the left side, P represents the dependent variable.
The domain of P is all integers from 0 to 5, inclusive.
Find the probability that exactly three guinea pigs will
be brown.
( )4 4
( ) 0,1,2,...,5! 5 !
P r rr r
= =−
Chapter 2: Functions and Graphs
2.2 Special Functions
Example 7 – Genetic
Solution:
3 21 3 1 9
5! 120454 4 64 16
3!2! 6(2) 512(3)P
= ==
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Chapter 2: Functions and Graphs
2.3 Combinations of Functions2.3 Combinations of Functions
Example 1 – Combining Functions
• We define the operations of function as:
( )( ) ( ) ( )
( )( ) ( ) ( )
( )( ) ( ). ( )
( )( ) for ( ) 0
( )
f g x f x g x
f g x f x g x
fg x f x g x
f f xx g x
g g x
+ = +
− = −
=
= ≠
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Example 1 – Combining Functions
If f(x) = 3x − 1 and g(x) = x2 + 3x, find a. ( )( )
b. ( )( )
c. ( )( )
d. ( )g
1 e. ( )( )
2
f g x
f g x
fg x
fx
f x
+
−
Chapter 2: Functions and Graphs
2.3 Combinations of Functions
Example 1 – Combining Functions
Solution:2 2
2 2
2 3 2
2
a. ( )( ) ( ) ( ) (3 1) ( +3 ) 6 1
b. ( )( ) ( ) ( ) (3 1) ( +3 ) 1
c. ( )( ) ( ) ( ) (3 1)( 3 ) 3 8 3
( ) 3 1d. ( )
( ) 3
f g x f x g x x x x x x
f g x f x g x x x x x
fg x f x g x x x x x x x
f f x xx
g g x x x
+ = + = − + = + −
− = − = − − = − −
= = − + = + −
−= =
+
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2( ) 3
1 1 1 3 1e. ( )( ) ( ( )) (3 1)
2 2 2
g g x x x
xf x f x x
+
−= = − =
2
Composition
• Composite of f with g is defined by ( )( ) ( ( ))f g x f g x=o
Chapter 2: Functions and Graphs
2.3 Combinations of Functions
Example 3 – Composition
Solution:
2If ( ) 4 3, ( ) 2 1, and ( ) ,find
a. ( ( ))
b. ( ( ( )))
c. ( (1))
F p p p G p p H p p
F G p
F G H p
G F
= + − = + =
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Solution:2 2
2 2
2
a. ( ( )) (2 1) (2 1) 4(2 1) 3 4 12 2 ( )( )
b. ( ( ( ))) ( ( ))( ) (( ) )( ) ( )( ( ))
( )( ) 4 12 2 4 12 2
c. ( (1)) (1 4 1 3) (2) 2 2 1 5
F G p F p p p p p F G p
F G H p F G H p F G H p F G H p
F G p p p p p
G F G G
= + = + + + − = + + =
= = = =
= + + = + =
= + ⋅ − = = ⋅ + =
o
o o o o o
o
One-to-one function
• A function f that satisfies
For all a and b, if f(a)=f(b) then a=b
is called a one-to-one function
• Or
≠ ≠
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For all a and b, if a≠b then f(a)≠f(b)
Example
f(x)=x2, then f(-1)=f(1)=1 and -1≠1 show that the squaring function is not one-to-one.
Chapter 2: Functions and Graphs
2.4 Inverse Functions2.4 Inverse Functions
Example 1 – Inverses of Linear Functions
• An inverse function is defined as 1 1( ( )) ( ( ))f f x x f f x− −
= =
Show that a linear function is one-to-one. Find the
inverse of f(x) = ax + b and show that it is also linear.
Solution:
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Solution:
Assume that f(u) = f(v), thus .
We can prove the relationship,
au b av b+ = +
( )( )( ) ( ( ))
ax b b axg f x g f x x
a a
+ −= = = =o
( )( ) ( ( )) ( )x b
f g x f g x a b x b b xa
−= = + = − + =o
Chapter 2: Functions and Graphs
2.4 Inverse Functions
Example 3 – Inverses Used to Solve Equations
Many equations take the form f(x) = 0, where f is a
function. If f is a one-to-one function, then the
equation has x = f −1(0) as its unique solution.
Solution:
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Solution:
Applying f −1 to both sides gives .
Since , is a solution.
( )( ) ( )1 1 0f f x f− −
=1(0)f
−1( (0)) 0f f−
=
Chapter 2: Functions and Graphs
2.4 Inverse Functions
Example 5 – Finding the Inverse of a Function
To find the inverse of a one-to-one function f , solve
the equation y = f(x) for x in terms of y obtaining x =
g(y). Then f−1(x)=g(x). To illustrate, find f−1(x) if
f(x)=(x − 1)2, for x ≥ 1.
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Solution:
Let y = (x − 1)2, for x ≥ 1. Then x − 1 = √y and hence
x = √y + 1. It follows that f−1(x) = √x + 1.
Chapter 2: Functions and Graphs
2.5 Graphs in Rectangular Coordinates2.5 Graphs in Rectangular Coordinates
• The rectangular coordinate system provides a geometric way to graph equations in two variables.
• An x-intercept is a point where the graph intersects the x-axis. Y-intercept is vice versa.
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intersects the x-axis. Y-intercept is vice versa.
Chapter 2: Functions and Graphs
2.5 Graphs in Rectangular Coordinates
Example 1 – Intercepts and Graph
Find the x- and y-intercepts of the graph of y = 2x + 3, and sketch the graph.
Solution:
When y = 0, we have3
0 2 3 so that x x= + = −
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When y = 0, we have
When x = 0,
0 2 3 so that 2
x x= + = −
2(0) 3 3y = + =
Chapter 2: Functions and Graphs
2.5 Graphs in Rectangular Coordinates
Example 3 – Intercepts and Graph
Determine the intercepts of the graph of x = 3, and
sketch the graph.
Solution:There is no y-intercept, because x cannot be 0.
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There is no y-intercept, because x cannot be 0.
Chapter 2: Functions and Graphs
2.5 Graphs in Rectangular Coordinates
Example 7 – Graph of a Case-Defined Function
Graph the case-defined function
Solution:
if 0 < 3
( ) 1 if 3 5
4 if 5 < 7
x x
f x x x
x
≤
= − ≤ ≤ ≤
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Solution:
Use the preceding definition to show that the graph
of y = x2 is symmetric about the y-axis.
Chapter 2: Functions and Graphs
2.6 Symmetry2.6 Symmetry
Example 1 – y-Axis Symmetry
• A graph is symmetric about the y-axis when (-a, b) lies on the graph when (a, b) does.
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of y = x is symmetric about the y-axis.
Solution:
When (a, b) is any point on the graph, .
When (-a, b) is any point on the graph, .
The graph is symmetric about the y-axis.
2b a=
2 2( )a a b− = =
Chapter 2: Functions and Graphs
2.6 Symmetry
• Graph is symmetric about the x-axis when (x, -y) lies on the graph when (x, y) does.
• Graph is symmetric about the origin when (−x,−y) lies on the graph when (x, y) does.
• Summary:
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Chapter 2: Functions and Graphs
2.6 Symmetry
Example 3 – Graphing with Intercepts and Symmetry
Test y = f (x) = 1− x4 for symmetry about the x-axis,
the y-axis, and the origin. Then find the intercepts
and sketch the graph.
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Chapter 2: Functions and Graphs
2.6 Symmetry
Example 3 – Graphing with Intercepts and Symmetry
Solution:
Replace y with –y, not equivalent to equation.
Replace x with –x, equivalent to equation.
Replace x with –x and y with –y, not equivalent to equation.
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Thus, it is only symmetric about the y-axis.
Intercept at 41 0
1 or 1
x
x x
− =
= = −
Chapter 2: Functions and Graphs
2.6 Symmetry
Example 5 – Symmetry about the Line y = x
• A graph is symmetric about the y = x when (b, a) and (a, b).
Show that x2 + y2 = 1 is symmetric about the line
y = x.
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Solution:
Interchanging the roles of x and y produces
y2 + x2 = 1 (equivalent to x2 + y2 = 1).
It is symmetric about y = x.
Chapter 2: Functions and Graphs
2.7 Translations and Reflections2.7 Translations and Reflections
• 6 frequently used functions:
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Chapter 2: Functions and Graphs
2.7 Translations and Reflections
• Basic types of transformation:
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Chapter 2: Functions and Graphs
2.7 Translations and Reflections
Example 1 – Horizontal Translation
Sketch the graph of y = (x − 1)3.
Solution:
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2.8 Functions of Several Variables
• For any three sets X, Y and Z, the notation of a function f : X×Y�Z
• f is simply a rule which assign is assigns to each element (x,y) in X×Y at most one element of Z, denoted by f((x,y)).
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element of Z, denoted by f((x,y)).
• Example
f(x,y)=x+y is a function of two variables.
f(1,1)=1+1=2
f(2,3)=2+3=5
• Graphing a Plane
�In space, the graph of an equation of the form
Ax+By+Cz+D=0
where D is a constant and A, B, and C are constants that are not all zero, is a plane.
�Since three distinct points (not lying on the
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�Since three distinct points (not lying on the same line) determine a plane, a convenient way to sketch a plane is to first determine the points, if any, where the plane intercept the x-, y-, and x-axes. These points are called intercepts.
• Example 1
Sketch the plane 2x+3y+z=6.
The plane intersects the x-axis when y=0 and z=0. Thus 2x=6 which gives x=3.
Similarly, if x=z=0, then y=2;
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Similarly, if x=z=0, then y=2;
if x=y=0, then z=0.
Therefore, the intercepts are (3,0,0), (0,2,0) and (0,0,6). After these points are plotted, a plane is passed through them.
INTRODUCTORY MATHEMATICAL ANALYSISINTRODUCTORY MATHEMATICAL ANALYSISFor Business, Economics, and the Life and Social Sciences
Chapter 3 Chapter 3
Lines, Parabolas, and SystemsLines, Parabolas, and Systems
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• To develop the notion of slope and different forms of equations of lines.
• To develop the notion of demand and supply curves and to introduce linear functions.
• To sketch parabolas arising from quadratic functions.
Chapter 3: Lines, Parabolas and Systems
Chapter ObjectivesChapter Objectives
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• To sketch parabolas arising from quadratic functions.
• To solve systems of linear equations in both two and three variables by using the technique of elimination by addition or by substitution.
• To use substitution to solve nonlinear systems.
• To solve systems describing equilibrium and break-even points.
Lines
Applications and Linear Functions
Quadratic Functions
Systems of Linear Equations
3.1)
3.2)
3.3)
3.4)
Chapter 3: Lines, Parabolas and Systems
Chapter OutlineChapter Outline
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Systems of Linear Equations
Nonlinear Systems
Applications of Systems of Equations
3.4)
3.5)
3.6)
Slope of a Line
• The slope of the line is for two different points (x1, y1) and (x2, y2) is
Chapter 3: Lines, Parabolas and Systems
3.1 Lines3.1 Lines
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=
−
−=
change horizontal
change vertical
12
12
xx
yym
The line in the figure shows the relationship
between the price p of a widget (in dollars) and the
quantity q of widgets (in thousands) that consumers
will buy at that price. Find and interpret the slope.
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 1 – Price-Quantity Relationship
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Solution:
The slope is
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 1 – Price-Quantity Relationship
2
1
28
41
12
12 −=−
−=
−
−=
ppm
Equations of lines
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Equations of lines
• A point-slope form of an equation of the line through (x1, y1) with slope m is
( )1212
12
12
xxmyy
mxx
yy
−=−
=−
−
Find an equation of the line passing through (−3, 8) and (4, −2).
Solution:
The line has slope
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 3 – Determining a Line from Two Points
( ) 7
10
34
82−=
−−
−−=m
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Using a point-slope form with (−3, 8) gives
( ) 734 −−
( )[ ]
026710
3010567
37
108
=−+
−−=−
−−−=−
yx
xy
xy
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 5 – Find the Slope and y-intercept of a Line
• The slope-intercept form of an equation of the line with slope m and y-intercept b is .cmxy +=
Find the slope and y-intercept of the line with
equation y = 5(3 − 2x).
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Solution:
Rewrite the equation as
The slope is −10 and the y-intercept is 15.
( )
1510
1015
235
+−=
−=
−=
xy
xy
xy
a.Find a general linear form of the line whose
slope-intercept form is
Solution:
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 7 – Converting Forms of Equations of Lines
43
2+−= xy
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Solution:
By clearing the fractions, we have
01232
043
2
=−+
=−+
yx
yx
b. Find the slope-intercept form of the line having a
general linear form
Solution:
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 7 – Converting Forms of Equations of Lines
0243 =−+ yx
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Solution:
We solve the given equation for y,
2
1
4
3
234
0243
+−=
+−=
=−+
xy
xy
yx
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Parallel and Perpendicular Lines
• Parallel Lines are two lines that have the same slope.
• Perpendicular Lines are two lines with slopes m1 and m2 perpendicular to each other only if
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2
1
1
mm −=
Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 9 – Parallel and Perpendicular Lines
The figure shows two lines passing through (3, −2). One is parallel to the line y = 3x + 1, and the other is
perpendicular to it. Find the equations of these lines.
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Chapter 3: Lines, Parabolas and Systems
3.1 Lines
Example 9 – Parallel and Perpendicular Lines
Solution:
The line through (3, −2) that is parallel to y = 3x + 1 also has slope 3.
( ) ( )
932
332
−=+
−=−−
xy
xy
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For the line perpendicular to y = 3x + 1,
113 −= xy
( ) ( )
13
1
13
12
33
12
−−=
+−=+
−−=−−
xy
xy
xy
Chapter 3: Lines, Parabolas and Systems
3.2 Applications and Linear Functions3.2 Applications and Linear Functions
Example 1 – Production Levels
Suppose that a manufacturer uses 100 lb of material
to produce products A and B, which require 4 lb and
2 lb of material per unit, respectively.
Solution:
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Solution: If x and y denote the number of units produced of A and B, respectively,
Solving for y gives
0, where10024 ≥=+ yxyx
502 +−= xy
Chapter 3: Lines, Parabolas and Systems
3.2 Applications and Linear Functions
Demand and Supply Curves
• Demand and supply curves have the following trends:
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Chapter 3: Lines, Parabolas and Systems
3.2 Applications and Linear Functions
Example 3 – Graphing Linear Functions
Linear Functions
• A function f is a linear function which can be written as ( ) 0 where ≠+= abaxxf
Graph and . ( ) 12 −= xxf ( )215 t
tg−
=
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Graph and .
Solution:
( ) 12 −= xxf ( )3
215 ttg
−=
Chapter 3: Lines, Parabolas and Systems
3.2 Applications and Linear Functions
Example 5 – Determining a Linear Function
If y = f(x) is a linear function such that f(−2) = 6 and
f(1) = −3, find f(x).
Solution: The slope is .
( )3
21
6312 −=−−
−−=
−
−=
xx
yym
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The slope is .
Using a point-slope form:
( )3
2112
−=−−
=−
=xx
m
( )
( )[ ]
( ) xxf
xy
xy
xxmyy
3
3
236
11
−=
−=
−−−=−
−=−
Chapter 3: Lines, Parabolas and Systems
3.3 Quadratic Functions3.3 Quadratic Functions
Example 1 – Graphing a Quadratic Function
Graph the quadratic function .
• Quadratic function is written aswhere a, b and c are constants and
( ) 22++= bxaxxf
0≠a
( ) 1242+−−= xxxf
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Graph the quadratic function .
Solution: The vertex is .
( ) 124 +−−= xxxf
( )2
12
4
2−=
−
−−=−
a
b
( )( )2 and 6
260
1240 2
−=
−+=
+−−=
x
xx
xx
The points are
Chapter 3: Lines, Parabolas and Systems
3.3 Quadratic Functions
Example 3 – Graphing a Quadratic Function
Graph the quadratic function .
Solution:
( ) 762+−= xxxg
( )3
12
6
2=−=−
a
b
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23 ±=x
Chapter 3: Lines, Parabolas and Systems
3.3 Quadratic Functions
Example 5 – Finding and Graphing an Inverse
From determine the inverse
function for a = 2, b = 2, and c = 3.
Solution:
( ) cbxaxxfy ++==2
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Chapter 3: Lines, Parabolas and Systems
3.4 Systems of Linear Equations3.4 Systems of Linear Equations
Two-Variable Systems
• There are three different linear systems:
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• Two methods to solve simultaneous equations:
a) elimination by addition
b) elimination by substitution
Linear system(one solution)
Linear system(no solution)
Linear system(many solutions)
Chapter 3: Lines, Parabolas and Systems
3.4 Systems of Linear Equations
Example 1 – Elimination-by-Addition Method
Use elimination by addition to solve the system.
Solution: Make the y-component the same.
=+
=−
323
1343
xy
yx
=− 39129 yx
2011 Pearson Education, Inc.
Adding the two equations, we get . Use to find
Thus,
=+
=−
12128
39129
yx
yx
3=x
( )1
391239
−=
=−
y
y
−=
=
1
3
y
x
3=x
Chapter 3: Lines, Parabolas and Systems
3.4 Systems of Linear Equations
Example 3 – A Linear System with Infinitely Many Solutions
Solve
Solution: Make the x-component the same.
=+
=+
12
5
2
125
yx
yx
=+ 25 yx
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Adding the two equations, we get .
The complete solution is
−=−+−
=+
25
25
yx
yx
00 =
ry
rx
=
−= 52
Chapter 3: Lines, Parabolas and Systems
3.4 Systems of Linear Equations
Example 5 – Solving a Three-Variable Linear System
Solve
Solution: By substitution, we get
−=−−
=++−
=++
63
122
32
zyx
zyx
zyx
=+ 1573 zy
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Since y = -5 + z, we can find z = 3 and y = -2. Thus,
−+=
−=−
=+
63
5
1573
zyx
zy
zy
=
−=
=
1
2
3
x
y
z
Chapter 3: Lines, Parabolas and Systems
3.4 Systems of Linear Equations
Example 7 – Two-Parameter Family of Solutions
Solve the system
Solution:
Multiply the 2nd equation by 1/2 and add to the 1st
=++
=++
8242
42
zyx
zyx
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Multiply the 2nd equation by 1/2 and add to the 1st
equation,
Setting y = r and z = s, the solutions are
=
=++
00
42 zyx
sz
ry
srx
=
=
−−= 24
Chapter 3: Lines, Parabolas and Systems
3.5 Nonlinear Systems3.5 Nonlinear Systems
Example 1 – Solving a Nonlinear System
• A system of equations with at least one nonlinear equation is called a nonlinear system.
Solve (1)
(2)
=+−
=−+−
013
0722
yx
yxx
2011 Pearson Education, Inc.
(2)
Solution: Substitute Eq (2) into (1),
=+− 013 yx
( )
( )( )
7 or 8
2 or 3
023
06
071322
2
=−=
=−=
=−+
=−+
=−++−
yy
xx
xx
xx
xxx
Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations3.6 Applications of Systems of Equations
Equilibrium
• The point of equilibrium is where demand and supply curves intersect.
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Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Example 1 – Tax Effect on Equilibrium
Let be the supply equation for a
manufacturer’s product, and suppose the demand
equation is .
a. If a tax of $1.50 per unit is to be imposed on the
50100
8+= qp
65100
7+−= qp
2011 Pearson Education, Inc.
a. If a tax of $1.50 per unit is to be imposed on the
manufacturer, how will the original equilibrium price
be affected if the demand remains the same?
b. Determine the total revenue obtained by the
manufacturer at the equilibrium point both before and
after the tax.
Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Example 1 – Tax Effect on Equilibrium
Solution:
a. By substitution,
and100
50100
865
100
7
=
+=+−
q
qq ( ) 5850100100
8=+=p
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After new tax,
and( ) 70.5850.51100100
8=+=p
( )
90
65100
750.51100
100
8
=
+−=+
q
q
Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Example 1 – Tax Effect on Equilibrium
Solution:
b. Total revenue given by
After tax,( )( ) 580010058 === pqyTR
( )( ) 52839070.58 === pqy
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( )( ) 52839070.58 === pqyTR
Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Break-Even Points
• Profit (or loss) = total revenue(TR) – total cost(TC)
• Total cost = variable cost + fixed cost
• The break-even point is where TR = TC.
FCVCTC yyy +=
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• The break-even point is where TR = TC.
Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Example 3 – Break-Even Point, Profit, and Loss
A manufacturer sells a product at $8 per unit, selling
all that is produced. Fixed cost is $5000 and variable
cost per unit is 22/9 (dollars).
a. Find the total output and revenue at the break-even
point.
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b. Find the profit when 1800 units are produced.
c. Find the loss when 450 units are produced.
d. Find the output required to obtain a profit of
$10,000.
Chapter 3: Lines, Parabolas and Systems
3.6 Applications of Systems of Equations
Example 3 – Break-Even Point, Profit, and Loss
Solution:
a. We have
At break-even point,
50009
22
8
+=+=
=
qyyy
qy
FCVCTC
TR
22
= yy TCTR
2011 Pearson Education, Inc.
and
b.
The profit is $5000.
900
50009
228
=
+=
q
( ) 72009008 ==TRy
( ) ( ) 5000500018009
2218008 =
+−=− TCTR yy
INTRODUCTORY MATHEMATICAL ANALYSISINTRODUCTORY MATHEMATICAL ANALYSISFor Business, Economics, and the Life and Social Sciences
Chapter 4 Chapter 4
Exponential and Logarithmic Functions Exponential and Logarithmic Functions
2007 Pearson Education Asia
• To introduce exponential functions and their
applications.
• To introduce logarithmic functions and their
graphs.
Chapter 4: Exponential and Logarithmic Functions
Chapter ObjectivesChapter Objectives
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• To study the basic properties of logarithmic
functions.
• To develop techniques for solving logarithmic
and exponential equations.
Exponential Functions
Logarithmic Functions
Properties of Logarithms
Logarithmic and Exponential Equations
4.1)
4.2)
4.3)
4.4)
Chapter 4: Exponential and Logarithmic Functions
Chapter OutlineChapter Outline
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Logarithmic and Exponential Equations4.4)
• The function f defined by
where b > 0, b ≠ 1, and the exponent x is any real
number, is called an exponential function with
base b1.
Chapter 4: Exponential and Logarithmic Functions
4.1 Exponential Functions4.1 Exponential Functions
( ) xbxf =
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The number of bacteria present in a culture after t
minutes is given by .
a. How many bacteria are present initially?
b. Approximately how many bacteria are present
after 3 minutes?
Chapter 4: Exponential and Logarithmic Functions
4.1 Exponential Functions
Example 1 – Bacteria Growth
( )t
tN
=
3
4300
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Solution:
a. When t = 0,
b. When t = 3,
04
(0) 300 300(1) 3003
N
= = =
34 64 6400
(3) 300 300 7113 27 9
N
= = = ≈
Graph the exponential function f(x) = (1/2)x.
Solution:
Chapter 4: Exponential and Logarithmic Functions
4.1 Exponential Functions
Example 3 – Graphing Exponential Functions with 0 < b < 1
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Properties of Exponential Functions
Chapter 4: Exponential and Logarithmic Functions
4.1 Exponential Functions
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Solution:
Chapter 4: Exponential and Logarithmic Functions
4.1 Exponential Functions
Example 5 – Graph of a Function with a Constant Base2
Graph 3 .xy =
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Compound Interest
• The compound amount S of the principal P at the end of n
years at the rate of r compounded annually is given by
.(1 )n
S P r= +
Chapter 4: Exponential and Logarithmic Functions
4.1 Exponential Functions
Example 7 – Population Growth
The population of a town of 10,000 grows at the rate
of 2% per year. Find the population three years from
now.
Solution:
For t = 3, we have .3(3) 10,000(1.02) 10,612P = ≈
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For t = 3, we have .(3) 10,000(1.02) 10,612P = ≈
Chapter 4: Exponential and Logarithmic Functions
4.1 Exponential Functions
Example 9 – Population Growth
The projected population P of a city is given by
where t is the number of years after
1990. Predict the population for the year 2010.
Solution:
For t = 20,
0.05100,000 tP e=
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For t = 20,0.05(20) 1100,000 100,000 100,000 271,828P e e e= = = ≈
Chapter 4: Exponential and Logarithmic Functions
4.1 Exponential Functions
Example 11 – Radioactive Decay
A radioactive element decays such that after t days
the number of milligrams present is given by
.
a. How many milligrams are initially present?
0.062100 tN e
−=
( )0062.0==
−
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Solution: For t = 0, .
b. How many milligrams are present after 10 days?
Solution: For t = 10, .
( ) mg 100100 0062.0==
−eN
( ) mg 8.53100 10062.0==
−eN
Chapter 4: Exponential and Logarithmic Functions
4.2 Logarithmic Functions4.2 Logarithmic Functions
Example 1 – Converting from Exponential to Logarithmic Form
• y = logbx if and only if by=x.
• Fundamental equations are and logb xb x=log x
b b x=
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2
54
a. Since 5 25 it follows that log 25 2
b. Since 3 81 it follo
Exponential Form Logarithmic Form
= =
= 30
10
ws that log 81 4
c. Since 10 1 it follows that log 1 0
=
= =
Chapter 4: Exponential and Logarithmic Functions
4.2 Logarithmic Functions
Example 3 – Graph of a Logarithmic Function with b > 1
Sketch the graph of y = log2x.
Solution:
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Chapter 4: Exponential and Logarithmic Functions
4.2 Logarithmic Functions
Example 5 – Finding Logarithms
a. Find log 100.
b. Find ln 1.
c. Find log 0.1.
( ) 210log100log2
==
01ln =
110log1.0log 1−==
−
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d. Find ln e-1.
d. Find log366.
110log1.0log −==
1ln1ln 1 −=−=− ee
2
1
6log2
6log6log36 ==
Chapter 4: Exponential and Logarithmic Functions
4.2 Logarithmic Functions
Example 7 – Finding Half-Life
• If a radioactive element has decay constant λ, the
half-life of the element is given by
A 10-milligram sample of radioactive polonium 210
λ
2ln=T
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A 10-milligram sample of radioactive polonium 210
(which is denoted 210Po) decays according to the
equation. Determine the half-life of 210Po.
Solution:
daysλ
T 4.13800501.0
2ln2ln≈==
Chapter 4: Exponential and Logarithmic Functions
4.3 Properties of Logarithms4.3 Properties of Logarithms
• Properties of logarithms are:
nmmn bbb loglog)(log .1 +=
nmn
mbb logloglog .2 b −=
mrm br
b loglog 3. =b
mm
b
mm
a
ab
b
b
bb
log
loglog .7
1log .6
01log .5
log1
log 4.
=
=
=
−=
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Example 1 – Finding Logarithms
a.
b.
c.
d.
7482.18451.09031.07log8log)78log(56log =+≈+=⋅=
6532.03010.09542.02log9log2
9log =−≈−=
8062.1)9031.0(28log28log64log 2=≈==
3495.0)6990.0(2
15log
2
15log5log 2/1
=≈==
balog
Chapter 4: Exponential and Logarithmic Functions
4.3 Properties of Logarithms
Example 3 – Writing Logarithms in Terms of Simpler Logarithms
a.
wzx
wzx
zwxzw
x
lnlnln
)ln(lnln
)ln(lnln
−−=
+−=
−=
853/1
8585− −−
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b.
)]3ln()2ln(8ln5[3
1
)]3ln()2ln([ln3
1
)}3ln(])2({ln[3
1
3
)2(ln
3
1
3
)2(ln
3
)2(ln
85
85
853/1
85
3
85
−−−+=
−−−+=
−−−=
−
−=
−
−=
−
−
xxx
xxx
xxx
x
xx
x
xx
x
xx
Chapter 4: Exponential and Logarithmic Functions
4.3 Properties of Logarithms
Example 5 – Simplifying Logarithmic Expressions
a.
b.
c.
.3ln 3 xe x=
3
30
10log01000log1log 3
=
+=
+=+
89/89 8 7log7log ==
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c.
d.
e.
989/8
7
9 87 7log7log ==
1)3(log3
3log
81
27log 1
34
3
33 −==
=
−
0)1(1
10logln10
1logln 1
=−+=
+=+ −ee
Chapter 4: Exponential and Logarithmic Functions
4.3 Properties of Logarithms
Example 7 – Evaluating a Logarithm Base 5
Find log52.
Solution:
4307.02log
2log5log
2log5log
25
≈=
=
=
=
x
x
x
x
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4307.05log
2log≈=x
Chapter 4: Exponential and Logarithmic Functions
4.3 Properties of Logarithms
4.4 Logarithmic and Exponential Equations4.4 Logarithmic and Exponential Equations
• A logarithmic equation involves the logarithm of
an expression containing an unknown.
• An exponential equation has the unknown
appearing in an exponent.
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An experiment was conducted with a particular type
of small animal. The logarithm of the amount of
oxygen consumed per hour was determined for a
number of the animals and was plotted against the
logarithms of the weights of the animals. It was found
that
Chapter 4: Exponential and Logarithmic Functions
4.4 Logarithmic and Exponential Equations
Example 1 – Oxygen Composition
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that
where y is the number of microliters of oxygen
consumed per hour and x is the weight of the animal
(in grams). Solve for y.
xy log885.0934.5loglog +=
Solution:
Chapter 4: Exponential and Logarithmic Functions
4.4 Logarithmic and Exponential Equations
Example 1 – Oxygen Composition
)934.5log(log
log934.5log
log885.0934.5loglog
885.0
885.0
xy
x
xy
=
+=
+=
885.0934.5 xy =
2007 Pearson Education Asia
Chapter 4: Exponential and Logarithmic Functions
4.4 Logarithmic and Exponential Equations
Example 3 – Using Logarithms to Solve an Exponential Equation
Solution:
.124)3(5 1=+
−xSolve
ln4ln
4
124)3(5
71
371
1
=
=
=+
−
−
−
x
x
x
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61120.1
ln4ln371
≈
=−
x
x
In an article concerning predators and prey, Holling
refers to an equation of the form
where x is the prey density, y is the number of prey
attacked, and K and a are constants. Verify his claim
that
Chapter 4: Exponential and Logarithmic Functions
4.4 Logarithmic and Exponential Equations
Example 5 – Predator-Prey Relation
axyK
K=
−ln
)1( axeKy −−=
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Solution:
Find ax first, and thus
axyK
=−
ln
K
yKe
eK
y
eKy
ax
ax
ax
−=
−=
−=
−
−
−
1
)1(
axyK
K
axK
yK
axK
yK
=−
=−
−
−=−
ln
ln
ln
(Proved!)
INTRODUCTORY MATHEMATICAL ANALYSISINTRODUCTORY MATHEMATICAL ANALYSISFor Business, Economics, and the Life and Social Sciences
Chapter 5 Chapter 5
Mathematics of Finance Mathematics of Finance
2007 Pearson Education Asia
• To solve interest problems which require logarithms.
• To solve problems involving the time value of money.
Chapter 5: Mathematics of Finance
Chapter ObjectivesChapter Objectives
2007 Pearson Education Asia
• To solve problems with interest is compounded continuously.
• To introduce the notions of ordinary annuities and annuities due.
• To learn how to amortize a loan and set up an amortization schedule.
Compound Interest
Present Value
Interest Compounded Continuously
Annuities
5.1)
5.2)
5.3)
5.4)
Chapter 5: Mathematics of Finance
Chapter OutlineChapter Outline
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Annuities
Amortization of Loans
5.4)
5.5)
Chapter 5: Mathematics of Finance
5.1 Compound Interest5.1 Compound Interest
Example 1 – Compound Interest
• Compound amount S at the end of n interest periods at the periodic rate of r is as
( )nrPS += 1
Suppose that $500 amounted to $588.38 in a
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Suppose that $500 amounted to $588.38 in a savings account after three years. If interest was compounded semiannually, find the nominal rate of interest, compounded semiannually, that was earned by the money.
Solution:
There are 2 × 3 = 6 interest periods.
Chapter 5: Mathematics of Finance
5.1 Compound Interest
Example 1 – Compound Interest
( )
( )
38.588
500
38.5881
38.5881500
6
6
=+
=+
r
r
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The semiannual rate was 2.75%, so the nominal rate was 5.5 % compounded semiannually.
0275.01500
38.588
500
38.5881
6
6
≈−=
=+
r
r
How long will it take for $600 to amount to $900 at an annual rate of 6% compounded quarterly?
Solution:
The periodic rate is r = 0.06/4 = 0.015.
Chapter 5: Mathematics of Finance
5.1 Compound Interest
Example 3 – Compound Interest
( )015.1600900 =n
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It will take .
( )
( )
( )
233.27015.1ln
5.1ln
5.1ln015.1ln
5.1ln015.1ln
5.1015.1
015.1600900
≈=
=
=
=
=
n
n
n
n
27.233
4≈ 6.8083 = 6 years, 9
1
2 months
Chapter 5: Mathematics of Finance
5.1 Compound Interest
Example 5 – Effective Rate
Effective Rate
• The effective rate re for a year is given by
11 −
+=
n
en
rr
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To what amount will $12,000 accumulate in 15 years if it is invested at an effective rate of 5%?
Solution: ( ) 14.947,24$05.1000,1215
≈=S
Chapter 5: Mathematics of Finance
5.1 Compound Interest
Example 7 – Comparing Interest Rates
If an investor has a choice of investing money at 6% compounded daily or % compounded quarterly, which is the better choice?
Solution:
Respective effective rates of interest are
8
16
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Respective effective rates of interest are
The 2nd choice gives a higher effective rate.
%27.614
06125.01
and %18.61365
06.01
4
365
≈−
+=
≈−
+=
e
e
r
r
Chapter 5: Mathematics of Finance
5.2 Present Value5.2 Present Value
Example 1 – Present Value
• P that must be invested at r for n interest periods so that the present value, S is given by
Find the present value of $1000 due after three
( ) nrSP
−+= 1
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Find the present value of $1000 due after three years if the interest rate is 9% compounded monthly.
Solution:
For interest rate, .
Principle value is .( ) 15.764$0075.11000
36≈=
−P
0075.012/09.0 ==r
Chapter 5: Mathematics of Finance
5.2 Present Value
Example 3 – Equation of Value
A debt of $3000 due six years from now is instead to be paid off by three payments:
• $500 now,
• $1500 in three years, and
• a final payment at the end of five years.
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• a final payment at the end of five years.
What would this payment be if an interest rate of 6% compounded annually is assumed?
Chapter 5: Mathematics of Finance
5.2 Present Value
Solution:
The equation of value is
x + 500 1.06( )5
+1500 1.06( )2
= 3000 1.06( )−1
x ≈ $475.68
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Chapter 5: Mathematics of Finance
5.2 Present Value
Example 5 – Net Present Value
You can invest $20,000 in a business that guarantees you cash flows at the end of years 2, 3, and 5 as
Net Present Value
( ) investment Initial - values present of Sum NPV ValuePresent Net =
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you cash flows at the end of years 2, 3, and 5 as indicated in the table.
Assume an interest rate of 7% compounded annually and find the net present value of the cash flows.
Year Cash Flow
2 $10,000
3 8000
5 6000
Chapter 5: Mathematics of Finance
5.2 Present Value
Example 5 – Net Present Value
Solution:
( ) ( ) ( )
31.457$
000,2007.1600007.1800007.1000,10NPV532
−≈
−++=−−−
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Compound Amount under Continuous Interest
• The compound amount S is defined as
Chapter 5: Mathematics of Finance
5.3 Interest Compounded Continuously5.3 Interest Compounded Continuously
kt
k
rPS
+= 1
Effective Rate under Continuous Interest
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•Effective rate with annual r compounded continuously is .
Present Value under Continuous Interest
•Present value P at the end of t years at an annual r compounded continuously is .
1−= re er
rtSeP −=
Chapter 5: Mathematics of Finance
5.3 Interest Compounded Continuously
Example 1 – Compound Amount
If $100 is invested at an annual rate of 5% compounded continuously, find the compound amount at the end ofa. 1 year.
( )( ) 13.105$100 105.0≈== ePeS rt
2007 Pearson Education Asia
b. 5 years.
13.105$100 ≈== ePeS
( )( ) 40.128$100100 25.0505.0≈== eeS
Chapter 5: Mathematics of Finance
5.3 Interest Compounded Continuously
Example 3 – Trust Fund
A trust fund is being set up by a single payment so that at the end of 20 years there will be $25,000 in the fund. If interest compounded continuously at an annual rate of 7%, how much money should be paid into the fund initially?
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Solution:
We want the present value of $25,000 due in 20 years.
( )( )
6165$000,25
000,254.1
2007.0
≈=
==
−
−−
e
eSeP rt
Chapter 5: Mathematics of Finance
5.4 Annuities5.4 Annuities
Example 1 – Geometric Sequences
Sequences and Geometric Series
• A geometric sequence with first term a and common ratio r is defined as
0 where,...,,,, 132≠
− aarararara n
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Example 1 – Geometric Sequences
a. The geometric sequence with a = 3, common ratio 1/2 , and n = 5 is
432
2
13 ,
2
13 ,
2
13 ,
2
13 ,3
Chapter 5: Mathematics of Finance
5.4 Annuities
Example 1 – Geometric Sequences
b. Geometric sequence with a = 1, r = 0.1, and
n = 4.
c. Geometric sequence with a = Pe−kI , r = e−kI ,
n = d.
001.0 ,01.0 ,1.0 ,1
dkIkIkI PePePe −−− ,...,, 2
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dkIkIkI PePePe −−− ,...,, 2
Sum of Geometric Series
• The sum of a geometric series of n terms, with first term a, is given by
s = ari
i=0
n−1
∑ =a 1− r
n( )1− r
for r ≠1
Chapter 5: Mathematics of Finance
5.4 Annuities
Example 3 – Sum of Geometric Series
Find the sum of the geometric series:
Solution: For a = 1, r = 1/2, and n = 7
62
2
1...
2
1
2
11
++
++
s =a 1− r
n( )1− r
=
1 1−1
2
7
1−1
2
=127128
12
=127
64
2007 Pearson Education Asia
1−2
2
Chapter 5: Mathematics of Finance
5.4 Annuities
An annuity is a sequence of payments made at fixed periods of time over a given interval. The fixed period is called the payment period, and the given interval is the term of the annuity.
Example: the depositing of $100 in a savings account every three months for a year.
Present Value of an Annuity
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Present Value of an Annuity
• The present value of an annuity (A) (n
payments of R ($) each, interest rate per
period is r)is the sum of the present values of all the payments.
( ) ( ) ( ) nrRrRrRA
−−−++++++= 1...11
21
Chapter 5: Mathematics of Finance
5.4 Annuities
Example 5 – Present Value of Annuity
Find the present value of an annuity of $100 per month for years at an interest rate of 6% compounded monthly.
Solution: For R = 100, r = 0.06/12 = 0.005, n = ( )(12) = 42
2
13
13
2007 Pearson Education Asia
For R = 100, r = 0.06/12 = 0.005, n = ( )(12) = 42
From Appendix B, .
Hence,
A =100a42__
0.005
798300.37005.042
__ =a
( ) 83.3779$798300.37100 =≈A
2
13
Chapter 5: Mathematics of Finance
5.4 Annuities
Example 7 – Periodic Payment of Annuity
If $10,000 is used to purchase an annuity consisting of equal payments at the end of each year for the next four years and the interest rate is 6% compounded annually, find the amount of each payment.
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Solution: For A= $10,000, n = 4, r = 0.06,
91.2885$465106.3
000,10000,10
000,10
06.04
06.04
____
__
=≈==
=
aa
AR
Ra
rn
Chapter 5: Mathematics of Finance
5.4 Annuities
Example 9 – Amount of Annuity
Amount of an Annuity
• The amount S of ordinary annuity of R for n periods at r per period is
( )r
rRS
n11 −+
⋅=
2007 Pearson Education Asia
Find S consisting of payments of $50 at the end of every 3 months for 3 years at 6% compounded quarterly. Also, find the compound interest.
Solution: For R = 50, n = 4(3) = 12, r = 0.06/4 = 0.015,
( ) 06.652$041211.135050015.012
__ ≈≈=S
( ) 06.52$501206.652 Interest Compund =−=
Chapter 5: Mathematics of Finance
5.4 Annuities
Example 11 – Sinking Fund
A sinking fund is a fund into which periodic payments are made in order to satisfy a future obligation. A machine costing $7000 is replaced at the end of 8 years, at which time it will have a salvage value of $700. A sinking fund is set up. The amount in the fund at the end of 8 years is to be the difference
2007 Pearson Education Asia
the fund at the end of 8 years is to be the difference between the replacement cost and the salvage value. If equal payments are placed in the fund at the end of each quarter and the fund earns 8% compounded quarterly, what should each payment be?
Chapter 5: Mathematics of Finance
5.4 Annuities
Example 11 – Sinking Fund
Solution:
Amount needed after 8 years = 7000 − 700 = $6300.
For n = 4(8) = 32, r = 0.08/4 = 0.02, and S = 6300,the periodic payment R of an annuity is
6300 ____= Rs
2007 Pearson Education Asia
45.142$6300
6300
02.032
02.032
________
____
≈==
=
ss
SR
Rs
rn
Chapter 5: Mathematics of Finance
5.5 Amortization of Loans5.5 Amortization of Loans
Amortization Formulas
2007 Pearson Education Asia
Chapter 5: Mathematics of Finance
5.5 Amortization of Loans
Example 1 – Amortizing a Loan
A person amortizes a loan of $170,000 by obtaining a 20-year mortgage at 7.5% compounded monthly. Find
a.monthly payment,
b.total interest charges, and
2007 Pearson Education Asia
b.total interest charges, and
c.principal remaining after five years.
Chapter 5: Mathematics of Finance
5.5 Amortization of Loans
Example 1 – Amortizing a Loan
Solution:
a. Monthly payment:
b. Total interest charge:
( )51.1369$
00625.11
00625.0000,170
240≈
−=
−R
2007 Pearson Education Asia
c. Principal value:
( ) 40.682,158$000,17051.1369240 =−
( )74.733,147$
00625.0
00625.1151.1369
180
≈
−−
Chapter 6 Chapter 6
Matrix AlgebraMatrix Algebra
Matrices
Matrix Addition and Scalar Multiplication
Matrix Multiplication
Solving Systems by Reducing Matrices
6.1)
6.2)
6.3)
6.4)
Chapter 6: Matrix Algebra
Chapter OutlineChapter Outline
Solving Systems by Reducing Matrices
Solving Systems by Reducing Matrices (continued)
Inverses
Determinants and Cramer’s rule
6.4)
6.5)
6.6)
6.7)
Chapter 6: Matrix Algebra
6.1 Matrices6.1 Matrices
• A matrix consisting of m horizontal rows and n
vertical columns is called an m×n matrix or a matrix of size m×n.
n
n
aaa
aaa
......
...
...
21221
11211
• For the entry aij, we call i the row subscript and j the column subscript.
mnmm aaa ...
......
......
......
21
a. The matrix has size .
b. The matrix has size .
c. The matrix has size .
Chapter 6: Matrix Algebra
6.1 Matrices
Example 1 – Size of a Matrix
[ ]021 31×
−
49
15
61
23×
[ ]7 11×c. The matrix has size .
d. The matrix has size .
[ ]7 11×
−−
−
11126
865119
42731
53×
Chapter 6: Matrix Algebra
6.1 Matrices
Equality of Matrices
• Matrices A = [aij ] and B = [bij] are equal if they have the same size and aij = bij for each i and j.
Transpose of a Matrix
• A transpose matrix is denoted by AT.
Example 3 – Constructing Matrices
If , find .
Solution:
Observe that .
=
654
321A
=
63
52
41TA
( ) AATT
=
TA
Chapter 6: Matrix Algebra
6.2 Matrix Addition and Scalar Multiplication6.2 Matrix Addition and Scalar Multiplication
Matrix Addition
• Sum A + B is the m× n matrix obtained by adding corresponding entries of A and B.
Example 1 – Matrix Additiona.
b. is impossible as matrices are not of the same
size.
−=
++
+−
−+
=
−
−
+
68
83
08
0635
4463
2271
03
46
27
63
52
41
+
1
2
43
21
Chapter 6: Matrix Algebra
6.2 Matrix Addition and Scalar Multiplication
Example 3 – Demand Vectors for an Economy
Demand for the consumers is
For the industries is
What is the total demand for consumers and the
[ ] [ ] [ ]1264 1170 523 321 === DDD
[ ] [ ] [ ]0530 8020 410 === SEC DDD
What is the total demand for consumers and the
industries?
Solution:
Total:
[ ] [ ] [ ] [ ]182571264 1170523321 =++=++ DDD
[ ] [ ] [ ] [ ]1265005308020410 =++=++ SEC DDD
[ ] [ ] [ ]3031571265018257 =+
Chapter 6: Matrix Algebra
6.2 Matrix Addition and Scalar Multiplication
Scalar Multiplication
• Properties of Scalar Multiplication:
Subtraction of Matrices
• Property of subtraction is ( )AA 1−=−
Chapter 6: Matrix Algebra
6.2 Matrix Addition and Scalar Multiplication
Example 5 – Matrix Subtraction
a.
−
−
−
=
−+
−−−
+−
=
−
−
−
13
08
84
3203
1144
2662
30
14
26
23
14
62
b.
−−=
−−
−=−
52
80
42
66
10
262BAT
Chapter 6: Matrix Algebra
6.3 Matrix Multiplication6.3 Matrix Multiplication
Example 1 – Sizes of Matrices and Their Product
• AB is the m× p matrix C whose entry cij is given by
A = 3 × 5 matrix
njinji
n
k
jikjikij babababac +++==∑=
...22
1
11
Not commutative:In general:
A = 3 5 matrix
B = 5 × 3 matrix
AB = 3 × 3 matrix but BA = 5 × 5 matrix.
C = 3 × 5 matrixD = 7 × 3 matrixCD = undefined but DC = 7 × 5 matrix.
Not commutative:In general:
Chapter 6: Matrix Algebra
6.3 Matrix Multiplication
Example 3 – Matrix Products
a.
b.
[ ] [ ]32
6
5
4
321 =
[ ]
=
183
122
61
61
3
2
1
c.
d.
1833
−−
−
−
=
−
−−
−
−
1047
0110
11316
212
312
201
401
122
031
++
++=
2222122121221121
2212121121121111
2221
1211
2221
1211
babababa
babababa
bb
bb
aa
aa
Chapter 6: Matrix Algebra
6.3 Matrix Multiplication
Example 5 – Cost Vector
Given the price and the quantities, calculate the total
cost.
Solution:
[ ]432=P
C of units
B of units
Aof units
11
5
7
=Q
Solution:
The cost vector is
[ ] [ ]73
11
5
7
432 =
=PQ
Chapter 6: Matrix Algebra
6.3 Matrix Multiplication
Example 7 – Associative Property
If
compute ABC in two ways.
=
−=
−−
−=
11
20
01
211
103
43
21CBA
Solution 1: Solution 2:
Note that A(BC) = (AB)C.
( )
−−=
−
−−
−=
−−
−=
196
94
43
12
43
21
11
20
01
211
103
43
21BCA ( )
−−=
−
−−=
−−
−=
196
94
11
20
01
1145
521
11
20
01
211
103
43
21CAB
Chapter 6: Matrix Algebra
6.3 Matrix Multiplication
Example 9 – Raw Materials and Cost
Find QRC when
Solution:
[ ]975=Q
=
1358256
21912187
17716205
R
=
1500
150
800
1200
2500
C
2500
=
=
71650
81550
75850
1500
150
800
1200
1358256
21912187
17716205
RC
( ) [ ] [ ]900,809,1
71650
81550
75850
1275 =
== RCQQRC
Chapter 6: Matrix Algebra
6.3 Matrix Multiplication
Example 11 – Matrix Operations Involving I and O
If
compute each of the following.
Solution:
00
00
10
01
41
23
103
101
51
52
=
=
−
−=
= OIBA
31
22
41
23
10
01 a.
−−
−−=
−
=− AI
314110
−−
( )
=
−
=
−
=−
63
63
20
02
41
233
10
012
41
23323 b. IA
OAO =
=
00
00
41
23 c.
IAB =
=
−
−
=
10
01
41
23 d.
103
101
51
52
Chapter 6: Matrix Algebra
6.3 Matrix Multiplication
Example 13 – Matrix Form of a System Using Matrix Multiplication
Write the system
in matrix form by using matrix multiplication.
Solution:
=+
=+
738
452
21
21
xx
xx
Solution:
If
then the single matrix equation is
=
=
=
7
4
38
52
2
1B
x
xXA
=
=
7
4
38
52
2
1
x
x
BAX
Chapter 6: Matrix Algebra
6.4 Solving Systems by Reducing Matrices6.4 Solving Systems by Reducing Matrices
Elementary Row Operations
1. Interchanging two rows of a matrix
2. Multiplying a row of a matrix by a nonzero number
3. Adding a multiple of one row of a matrix to a different row of that matrix
Chapter 6: Matrix Algebra
6.4 Solving Systems by Reducing Matrices6.4 Solving Systems by Reducing Matrices
Properties of a Reduced Matrix
• All zero-rows at the bottom.
• For each nonzero-row, leading entry is 1 and all other entries in the column in which the leading entry appears are zeros.
• Leading entry in each row is to the right of the leading entry in any row above it.
*leading entry: the first nonzero entry in a nonzero-row
Chapter 6: Matrix Algebra
6.4 Solving Systems by Reducing Matrices
Example 1 – Reduced Matrices
For each of the following matrices, determine whether
it is reduced or not reduced.
2100
3010
f. 000
001
e. 000
d.
01
10c.
010
001b.
30
01 a.
Solution:
a. Not reduced b. Reduced
c. Not reduced d. Reduced
e. Not reduced f. Reduced
0000
2100f.
010
000e. 000
000d.
Chapter 6: Matrix Algebra
6.4 Solving Systems by Reducing Matrices
Example 3 – Solving a System by Reduction
By using matrix reduction, solve the system
Solution:Reducing the augmented coefficient matrix of the
=+
=+
−=+
1
52
132
yx
yx
yx
Reducing the augmented coefficient matrix of the system,
We have
−
1
5
1
11
12
32
−
0
3
4
00
10
01
−=
=
3
4
y
x
Chapter 6: Matrix Algebra
6.4 Solving Systems by Reducing Matrices
Example 5 – Parametric Form of a Solution
Using matrix reduction, solve
Solution:Reducing the matrix of the system,
=+−
=++
=+++
9633
22
06232
431
432
4321
xxx
xxx
xxxx
Reducing the matrix of the system,
We have and x4 takes on any real value.
− 9
2
10
6303
1210
6232
1
0
4
100
0010
001
21
25
−=
=
−=
421
3
2
425
1
1
0
4
xx
x
xx
Chapter 6: Matrix Algebra
6.5 Solving Systems by Reducing Matrices 6.5 Solving Systems by Reducing Matrices (continued)(continued)
Example 1 – Two-Parameter Family of Solutions
Using matrix reduction, solve
Solution:
=+−−
−=+++
−=+++
32
143
3552
4321
4321
4321
xxxx
xxxx
xxxx
Solution:
The matrix is reduced to
The solution is
−
0
2
1
0000
1210
3101
=
=
−−−=
−−=
sx
rx
srx
srx
4
3
2
1
22
31
Chapter 6: Matrix Algebra
6.5 Solving Systems by Reducing Matrices (Continue)
• The system
is called a homogeneous system if c1 = c2 = … = cm = 0.
=+++
=+++
mnmnmm
nn
cxaxaxa
cxaxaxa
...
.
.
.
.
...
2211
11212111
= cm = 0.
• The system is non-homogeneous if at least one of the c’s is not equal to 0.
Concept for number of solutions:
1. k < n � infinite solutions
2. k = n � unique solution
n: number of unkowns, k: number of nonzero-rows in a reduced matrix.
Chapter 6: Matrix Algebra
6.5 Solving Systems by Reducing Matrices (Continue)
Example 3 – Number of Solutions of a Homogeneous System
Determine whether the system has a unique solution
or infinitely many solutions.
Solution:
=−+
=−+
0422
02
zyx
zyx
Solution:
2 equations (k), homogeneous system, 3 unknowns (n).
The system has infinitely many solutions.
Chapter 6: Matrix Algebra
6.6 Inverses6.6 Inverses
Example 1 – Inverse of a Matrix
• When matrix CA = I, C is an inverse of A and A is invertible. I: identity matrix or unit matrix
Let and . Determine whether C is
an inverse of A.
=
73
21A
−
−=
13
27C
an inverse of A.
Solution:
Thus, matrix C is an inverse of A.
ICA =
=
−
−=
10
01
73
21
13
27
Chapter 6: Matrix Algebra
6.6 Inverses
Example 3 – Determining the Invertibility of a Matrix
Method to Find the Inverse of a Matrix
• When matrix is reduced, ,
- If R = I, A is invertible and A−1 = B.
- If R ≠ I, A is not invertible.
[ ] [ ]BRIA →→L
01Determine if is invertible.
Solution: We have
Matrix A is invertible where
=
22
01A
[ ]
=
10
01
22
01IA [ ]BI=
−211
01
10
01
−=
−
21
1
1
01A
− 12
01
20
01
Chapter 6: Matrix Algebra
6.6 Inverses
Example 5 – Using the Inverse to Solve a System
Solve the system by finding the inverse of the
coefficient matrix.
Solution:
−=−+
=+−
=−
1102
2 24
1 2
321
321
31
xxx
xxx
xx
Solution:
We have
For inverse,
The solution is given by X = A−1B:
−
−
−
=
1021
124
201
A
−
−
=−
115
4
229
29
2411A
−
−
−
=
−
−
−
−
=
4
17
7
1
2
1
115
4
229
29
241
3
2
1
x
x
x
6.7 Determinants and Cramer’s rule6.7 Determinants and Cramer’s rule
Matrix of minors
The determinant and inverse of matrices
greater than 2x2
Special cases
Determinant of a 2x2 matrix
Cramer’s Rule
Inverse of a Matrix
The Adjoint of a Matrix
To calculate the determinant
Cofactors
Matrix of minors
Determinant of a 2x2 matrix
eg15
23= 3x1 – 2x5 = -7
Take care with the negative signs
Special cases
(a) 1x1 matrix A = (a)
4
Then det A = a
eg A = (4) => = 4
(b) Singularity
123
41
−−−−−−−− = (-12 – (-12)) = 0
Three by three determinant
In order to find the determinant and
the inverse of a matrix when
n > 2 (3x3 or a 4x4 etc) we first need
nxnA
The determinant and inverse of
matrices greater than 2x2
n > 2 (3x3 or a 4x4 etc) we first need
to form the matrix of minors and
then the matrix of cofactors
These are needed to solve equations
in the form AX = B => X = A-1B
A =
141
325
142
(1) form the matrix of minors, M
Matrix of minors
232221
131211
MMM
MMM
MMM
M =
141
The minor =
141
325
142
11M = 14
32
= -10
333231
MMM
The minor =22M
11
12=> = 2x1 – 1x1 = 1
141
325
142
141
325
14233M = =
25
42= 2x2-4x5
= -16
−−−−
−−−−
16110
410
18210M =
(2) form matrix of cofactors
The cofactor of a given element is
the minor of that element
multiplied by either +1 or –1
according to its position in the
matrix
Cofactors
111
111
111
++++−−−−++++
−−−−++++−−−−
++++−−−−++++
Determinant of signs =
If the given element is in the same position
as +1 in the determinant, multiply by +1,
otherwise by -1otherwise by -1
−−−−−−−−
−−−−
−−−−−−−−
16110
410
18210
=> C =
−−−−
−−−−
16110
410
18210
+ - +
+ - +
- + -
This extends to any square matrix
Cij = Mij if (i + j) is even
Cij = - Mij if (i + j) is odd
x1
x(-1)
+ - + -1 2 3 4 j
ponm
lkji
hgfe
dcba+ - + -
- + - +
+ - + -
- + - +
1
2
3
4
1 2 3 4 j
i
−−−−
−−−−−−−−
410
18210
C =
142
To calculate the determinant
The determinant, D is the sum of the
products of each element in any row
with its cofactor.
−−−−−−−−
−−−−
16110
410C =A =
141
325
D = 2 x(-10) + 4 x (-2) + 1 x 18 = -10
= (a1,1x c1,1)+(a1,2x c1,2)+(a1,3x c1,3)D
A =
141
325
142
−−−−−−−−
−−−−
−−−−−−−−
16110
410
18210
C =
D = 5x0 + 2x1 + 3x(-4) = -10 or
D = 1x10 + 4x(-1) + 1 x(-16) = -10
Any row and its cofactors can be used to
calculate the determinant
or
This is not zero so this matrix has an inverse
The Adjoint of a Matrix
� Replacing each element by its cofactor
The Adj of matrix A is obtained by:
Adj A = CT The transpose of the
matrix of cofactors
� Then transposing
−−−−−−−−
−−−−
−−−−−−−−
16110
410
18210
C =
====
−−−−−−−−
−−−−−−−−
−−−−
16418
112
10010
Adj
Inverse of a Matrix
−−−−−−−−
−−−−
112
100101
A
AdjAA-1 =
A-1 =
−−−−−−−−
−−−−−−−−−−−−
16418
11210
1
−−−−
−−−−
−−−−
====
6.14.08.1
1.01.02.0
101
Alternative method for finding the
determinant
141
325
142A =
32 2535
Det = ∆∆∆∆ =
2x (1) + 4x (-1) + 1x (1)14
32
41
25
11
35
= 2(2-12)(1) + 4(5-3)(-1) + 1(20-2)(1)
= -20 – 8 + 18 = -10
Example 1The equilibrium for three related products
simplifies to the following simultaneous equations
2a + 5b - 3c = 10
-4a + 2b – c = 4
3a – 2b + 5c = 18(a) Express the system of equations in matrix (a) Express the system of equations in matrix
form(b) Using X = A-1 B, solve the equations
(a)
====
−−−−
−−−−−−−−
−−−−
18
4
10
c
b
a
523
124
352
A =Need determinant
and Adjoint of A
Det A = 12 −−−− 14 −−−−−−−− 24−−−−
2 + 5 (-1) + (-3)
−−−−
−−−−−−−−
−−−−
523
124
352
52
12
−−−−
−−−−
53 23 −−−−2 + 5 (-1) + (-3)
= 95
The determinant is not zero
therefore there is an inverse
24141
191919
2178
C
−−−−====
−−−−
−−−−−−−−
−−−−
523
124
352
A =
Check determinant using line 2
= (-4)(-19) + 2x19 – 1x19 = 95∆∆∆∆
CT =
24192
141917
1198
−−−−
A
AdjAA-1 =
A
AdjAA-1 = =
24192
141917
1198
95
1
−−−−
X = A-1 B =
−−−−
18
4
10
24192
141917
1198
95
1
182419295
====
528
498
22
95
1
c
b
a => a = 22/95 = 0.232
=> b = 498/95 = 5.24
=> c = 528/95 = 5.56
Divide by 95 after multiplying the matrices
Cramer’s Rule
a1x + b1y = d1
a2x + b2y = d2
Cramer’s rule is used to find the
solution to simultaneous equationsa1and a2 are coefficients of x
b1and b2 are coefficients of y
d1and d2 are constantsd1and d2 are constants
====∆∆∆∆ 22
11
ba
ba
22
11
bd
bd====∆∆∆∆ x
22
11
da
da====∆∆∆∆y
1 1 1
2 2 2
a b dx
a b dy
====
changed to matrix form
22
11
22
11
ba
ba
bd
bd
x = ====∆∆∆∆
∆∆∆∆
x
Replaces x coefficients
Replaces y coefficients
22
11
22
11
ba
ba
da
da
Learn
formula
y = ====∆∆∆∆
∆∆∆∆
y
3x + 4y = 15
5x – 2y = 18
∆∆∆∆ = = 3(-2) – (5x4)= -2625
43
−−−−
Can use either X = A-1 B
or Cramer’s rule to solve
simultaneous equations
Example 1
Solve
25 −−−−
x∆∆∆∆ =
218
415
−−−−
= 15(-2) – 18x4 = -102
y∆∆∆∆ = 185
153
= 3(18) - 5(15) = -21
x = 26
102x
−−−−
−−−−====
∆∆∆∆
∆∆∆∆
y = 26
21y
−−−−
−−−−====
∆∆∆∆
∆∆∆∆
= 3.923
= 0.808y = 26−−−−====
∆∆∆∆= 0.808
Solve x + 2y + 3z = 10x + y – z = 42x + z = 7
(a) Find all four determinants
∆∆∆∆ 111
321
−−−−==== ∆∆∆∆ = 1(1 – 0) + 2(1 -(-2))(-1)
Example 2
+0y
∆∆∆∆
102
111 −−−−====
x∆∆∆∆107
114
3210
−−−−====
Replace the x coefficients when
solving for x
∆∆∆∆ = 1(1 – 0) + 2(1 -(-2))(-1)
+ 3(0 – 2) = -11
x∆∆∆∆ = 10(1 - 0) + 2(4 - (-7)) (-1)
+ 3(0 - 7) = -33
172
141
3101
−−−−====y∆∆∆∆Replace the y
coefficients when
solving for y
y∆∆∆∆ = 1(4 – (-7)) + 10(1 – (-2))(-1) + 3(7 – 8) = -22
z∆∆∆∆ = 1(7 – 0) + 2(7 – 8)(-1) + 10(0 – 2)
= -11
Replace the z
coefficients when
solving for z702
411
1021
====z∆∆∆∆
3 11
33x x ====
−−−−
−−−−====
∆∆∆∆
∆∆∆∆==== 2
11
22y y ====
−−−−
−−−−====
∆∆∆∆
∆∆∆∆====
(b) Use Cramer’s Rule
1 11
11z z ====
−−−−
−−−−====
∆∆∆∆
∆∆∆∆====
=> x = 3, y = 2, z = 1
INTRODUCTORY MATHEMATICAL ANALYSISINTRODUCTORY MATHEMATICAL ANALYSISFor Business, Economics, and the Life and Social Sciences
Chapter 7 Chapter 7
Linear ProgrammingLinear Programming
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• To represent geometrically a linear inequality in two variables.
• To state a linear programming problem and solve it geometrically.
• To consider situations in which a linear programming problem exists.
Chapter 7: Linear Programming
Chapter ObjectivesChapter Objectives
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problem exists.
Linear Inequalities in Two Variables
Linear Programming
7.1)
7.2)
Chapter 7: Linear Programming
Chapter OutlineChapter Outline
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Chapter 7: Linear Programming
7.1 Linear Inequalities in 2 Variables7.1 Linear Inequalities in 2 Variables
• Linear inequality is written as
0
or 0
or 0
or 0
≥++
>++
≤++
<++
cbyax
cbyax
cbyax
cbyax
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where a, b, c = constants and
not both a and b are zero.
0≥++ cbyax
Chapter 7: Linear Programming
7.1 Linear Inequalities in 2 Variables
• A solid line is included in the solution and a
dashed line is not.
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Chapter 7: Linear Programming
7.1 Linear Inequalities in 2 Variables
Example 1 – Solving a Linear Inequality
Find the region defined by the inequality y ≤ 5.
Solution:
The region consists of the line y = 5 and with the half-plane below it.
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Example 3 – Solving a System of Linear Inequalities
and with the half-plane below it.
Solve the system
Solution:
The solution is the unshaded region.
−≥
+−≥
2
102
xy
xy
Chapter 7: Linear Programming
7.2 Linear Programming7.2 Linear Programming
• A linear function in x and y has the form
• The function to be maximized or minimized is called the objective function.
byaxZ +=
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Example 1 – Solving a Linear Programming Problem
Maximize the objective function Z = 3x + y subject
to the constraints
0
0
1232
82
≥
≥
≤+
≤+
y
x
yx
yx
Chapter 7: Linear Programming
7.2 Linear Programming
Example 1 – Solving a Linear Programming Problem
Solution:
The feasible region is nonempty and bounded.
Evaluating Z at these points, we obtain
( ) ( )( ) ( ) 12043
0003
=+=
=+=
BZ
AZ
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The maximum value of Z occurs when x = 4 and y = 0.
( ) ( )( ) ( )( ) ( ) 4403
11233
12043
=+=
=+=
=+=
DZ
CZ
BZ
Chapter 7: Linear Programming
7.2 Linear Programming
Example 3 – Unbounded Feasible Region
The information for a produce is summarized as
follows:
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If the grower wishes to minimize cost while still
maintaining the nutrients required, how many bags of
each brand should be bought?
Chapter 7: Linear Programming
7.2 Linear Programming
Example 3 – Unbounded Feasible Region
Solution:Let x = number of bags of Fast Grow bought
y = number of bags of Easy Grow bought
To minimize the cost functionSubject to the constraints
yxC 68 +=
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Subject to the constraints
There is no maximum value since the feasible region is unbounded .
0
0802
20025
16023
≥
≥
≥+
≥+
≥+
y
xyx
yx
yx