St Clements Institute - pith-edupith-edu.weebly.com/.../math_analysis-powerpoint.pdf · 1) Introductory Mathematical Analysis for Business, Economics, and the Life and Social Sciences,

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.

mailto:[email protected]

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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

Page 3 of 3

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

2011 Pearson Education, Inc.

• 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


• 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 OutlineChapter Outline

Applications of Equations

Linear Inequalities

Applications of Inequalities

Absolute Value

1.1)

1.2)

1.3)

1.4)


Absolute Value

Summation Notation

1.6) Sequence

1.4)

1.5)

• Modeling: Translating relationships in the problems to mathematical symbols.


1.1 Applications of Equations1.1 Applications of Equations

A chemist must prepare 350 ml of a chemical

Example 1 - Mixture


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


1.1 Applications of Equations

Example 1 - Mixture


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





• 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



Example 3 – Profit


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



Example 3 – Profit


total revenue = 10q

Since profit = total revenue − total cost

35,000 units must be sold to earn a profit of $60,000.

( )

q

q

qq

=

=

+−=

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?



Example 5 – Investment


$588.75?

Solution:

Let x = amount ($) invested at 6%.

( ) ( )( )

75.130025.0

75.5880575.057506.0

75.588000,100575.006.0

=

=−+

=−+

x

xx

xx



Example 5 – Investment


$5500 was invested at 6%

$10,000−$5500 = $4500 was invested at 5.75%.

5500

75.130025.0

=

=

x

x



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


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)





Solution 1 (Con’t):( )

25

34050600,54

25

165032400600,54

25

550396600,54

−=

+−=

−−=

rr

rr

rr





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)





(number of apartments rented)

Solution 2 (Con’t):

( )( )

02410

0180075075

75750800,52600,54

39625550600,54

2

2

2

=+−

=+−

−+=

−+=

nn

nn

nn

nn





The rent charged should be either

550 + 25(6) = $700 or

550 + 25(4) = $650.

( )( )

4 or 6

046

02410

=

=−−

=+−

n

nn

nn


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:


• We use dots to indicate points on a number line.

• Suppose that a 0, then ac < bc and a/c < b/c.

3. If a b(c) and a/c > b/c.




3. If a b(c) and a/c > b/c.

4. If a 1/b .

6. If 0 < a 0, then an < bn.

If 0 < a < b, then .nn ba <



• 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


find all values of the variable for which the inequality is true.



Example 1 – Solving a Linear Inequality

Solve 2(x − 3) < 4.

Solution:Replace inequality by equivalent inequalities.

( ) 432 <−x


5

2

10

2

2

102

64662

462

<

<

<

+<+−

<−

x

x

x

x

x




Solve (3/2)(s − 2) + 1 > −2(s − 4).

( ) ( )42122

3 −−>+− ss

Solution:


( ) ( )[ ]

( )[ ] ( )

7

20

207

16443

442232

422122

32

2

>

>

+−>−

−−>+−

−>

+−

s

s

ss

ss

s-s

The solution is ( 20/7 ,∞).


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


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.



Example 1 - Profit


( )

5000

000,7014

0000,702135

>

>

>+−

q

q

qq

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



Example 3 – Current Ratio


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



Example 3 – Current Ratio


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|.


1.4 Absolute Value1.4 Absolute Value


DEFINITION

The absolute value of a real number x, written |x|, is defined by

<−

≥=

0 if ,

0 if ,

xx

xxx


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


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


1.4 Absolute Value



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.


1.4 Absolute Value



1.4 Absolute Value


a. Solve |x + 5| ≥ 7

b. Solve |3x − 4| > 1

Solution:a. 75 or 75 −≥+−≤+ xx


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


1.4 Absolute Value


• Property 5 is known as the triangle inequality.

baba

aaa

abba

bb

+≤+

≤≤−

−=−

.5

.4

.3


1.4 Absolute Value

Example 5 – Properties of Absolute Value

( )

777777

77 c.

24224 b.

213737- a.

−−−−

−=−

=−=−

=⋅−=⋅

xx

Solution:


( ) 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


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


nmmmmi

i aaaaa ++++= ++

=

∑ ...21

Evaluate the given sums.

a. b.

Solution:

a.



Example 1 – Evaluating Sums

( )∑=

−7

3

25n

n ( )∑=

+6

1

2 1j

j

( ) ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ]275265255245235257

−+−+−+−+−=−∑ n


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



( )6

)12(1

1

2 ++=∑

=

nnni

n

i


where n = the last number

61=i

Evaluate the given sums.

a. b. c.

Solution:

a.



Example 3 – Applying the Properties of Summation Notation

2847144471100

=⋅==∑∑

( )∑=

+100

1

35k

k ∑=

200

1

29k

k∑=

100

30

4j


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


• 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 +−=


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


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


....)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


Chapter 2 Chapter 2

Functions and GraphsFunctions and Graphs


• 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



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



2.1)

2.2)

2.3)

2.4)


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.


2.1 Functions2.1 Functions


• 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


=

≠+=

=

≠+=

1 if 3

1 if 2)( d.

1 if 0

1 if 2)( c.

x

xxxk

x

xxxh


Example 1 – Determining Equality of Functions

Solution:When x = 1,

( ) ( )( ) ( )( ) ( ) 11

, 11

, 11

kf

hf

gf

≠

≠

≠


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

≠

≠

=

,

,


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= − +


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

+ = + − + +

= + + − − +


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.


Solution: pq

q =100

a


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= − = + =


A function of the form h(x) = c, where c = constant, is a constant function.

(10) 2 ( 387) 2 ( 3) 2h h h x= − = + =



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

++ =


Example 5 – Absolute-Value Function

Absolute-value function is defined as , e.g. x

if 0

if 0

x xx

x x

≤ =

− <



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

−

= =


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

= =−



Example 7 – Genetic

Solution:

3 21 3 1 9

5! 120454 4 64 16

3!2! 6(2) 512(3)P

= ==



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

+ = +

− = −

=

= ≠



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

+

−




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

+ = + = − + = + −

− = − = − − = − −

= = − + = + −

−= =

+


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



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

= + − = + =


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

≠ ≠


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.


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:


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



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:


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−

=



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.


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.


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.


intersects the x-axis. Y-intercept is vice versa.



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= + = −


When y = 0, we have

When x = 0,

0 2 3 so that 2

x x= + = −

2(0) 3 3y = + =



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.


There is no y-intercept, because x cannot be 0.



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

≤

= − ≤ ≤ ≤


Solution:

Use the preceding definition to show that the graph

of y = x2 is symmetric about the y-axis.


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.


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− = =


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:



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.



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.


Thus, it is only symmetric about the y-axis.

Intercept at 41 0

1 or 1

x

x x

− =

= = −


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.


Solution:

Interchanging the roles of x and y produces

y2 + x2 = 1 (equivalent to x2 + y2 = 1).

It is symmetric about y = x.


2.7 Translations and Reflections2.7 Translations and Reflections

• 6 frequently used functions:




• Basic types of transformation:




Example 1 – Horizontal Translation

Sketch the graph of y = (x − 1)3.

Solution:


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)).


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


�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;


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.


Chapter 3 Chapter 3

Lines, Parabolas, and SystemsLines, Parabolas, and Systems


• 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



• 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)




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


3.1 Lines3.1 Lines


=

−

−=

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.


3.1 Lines

Example 1 – Price-Quantity Relationship


Solution:

The slope is


3.1 Lines

Example 1 – Price-Quantity Relationship

2

1

28

41

12

12 −=−

−=

−

−=

qq

ppm

Equations of lines


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


3.1 Lines

Example 3 – Determining a Line from Two Points

( ) 7

10

34

82−=

−−

−−=m


Using a point-slope form with (−3, 8) gives

( ) 734 −−

( )[ ]

026710

3010567

37

108

=−+

−−=−

−−−=−

yx

xy

xy


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).


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:


3.1 Lines

Example 7 – Converting Forms of Equations of Lines

43

2+−= xy


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:


3.1 Lines

Example 7 – Converting Forms of Equations of Lines

0243 =−+ yx


Solution:

We solve the given equation for y,

2

1

4

3

234

0243

+−=

+−=

=−+

xy

xy

yx


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


2

1

1

mm −=


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.



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


For the line perpendicular to y = 3x + 1,

113 −= xy

( ) ( )

13

1

13

12

33

12

−−=

+−=+

−−=−−

xy

xy

xy


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:


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



Demand and Supply Curves

• Demand and supply curves have the following trends:




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−

=


Graph and .

Solution:

( ) 12 −= xxf ( )3

215 ttg

−=



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


The slope is .

Using a point-slope form:

( )3

2112

−=−−

=−

=xx

m

( )

( )[ ]

( ) xxf

xy

xy

xxmyy

3

3

236

11

−=

−=

−−−=−

−=−


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



Solution: The vertex is .

( ) 124 +−−= xxxf

( )2

12

4

2−=

−

−−=−

a

b

( )( )2 and 6

260

1240 2

−=

−+=

+−−=

x

xx

xx

The points are



Example 3 – Graphing a Quadratic Function


Solution:

( ) 762+−= xxxg

( )3

12

6

2=−=−

a

b


23 ±=x



Example 5 – Finding and Graphing an Inverse

From determine the inverse

function for a = 2, b = 2, and c = 3.

Solution:

( ) cbxaxxfy ++==2



3.4 Systems of Linear Equations3.4 Systems of Linear Equations

Two-Variable Systems

• There are three different linear systems:


• 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)



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


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



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


Adding the two equations, we get .

The complete solution is

−=−+−

=+

25

25

yx

yx

00 =

ry

rx

=

−= 52



Example 5 – Solving a Three-Variable Linear System

Solve

Solution: By substitution, we get

−=−−

=++−

=++

63

122

32

zyx

zyx

zyx

=+ 1573 zy


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



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


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


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


(2)

Solution: Substitute Eq (2) into (1),

=+− 013 yx

( )

( )( )

7 or 8

2 or 3

023

06

071322

2

=−=

=−=

=−+

=−+

=−++−

yy

xx

xx

xx

xxx


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.




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


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.




Solution:

a. By substitution,

and100

50100

865

100

7

=

+=+−

q

qq ( ) 5850100100

8=+=p


After new tax,

and( ) 70.5850.51100100

8=+=p

( )

90

65100

750.51100

100

8

=

+−=+

q

q




Solution:

b. Total revenue given by

After tax,( )( ) 580010058 === pqyTR

( )( ) 52839070.58 === pqy


( )( ) 52839070.58 === pqyTR



Break-Even Points

• Profit (or loss) = total revenue(TR) – total cost(TC)


• The break-even point is where TR = TC.

FCVCTC yyy +=


• The break-even point is where TR = TC.



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.


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.



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


and

b.

The profit is $5000.

900

50009

228

=

+=

q

qq

( ) 72009008 ==TRy

( ) ( ) 5000500018009

2218008 =

+−=− TCTR yy


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



• 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)




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.


4.1 Exponential Functions4.1 Exponential Functions

( ) xbxf =


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?



Example 1 – Bacteria Growth

( )t

tN

=

3

4300


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:



Example 3 – Graphing Exponential Functions with 0 < b < 1


Properties of Exponential Functions




Solution:



Example 5 – Graph of a Function with a Constant Base2

Graph 3 .xy =


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= +



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 = ≈


For t = 3, we have .(3) 10,000(1.02) 10,612P = ≈



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=


For t = 20,0.05(20) 1100,000 100,000 100,000 271,828P e e e= = = ≈



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==

−


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


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=


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

=

= =



Example 3 – Graph of a Logarithmic Function with b > 1

Sketch the graph of y = log2x.

Solution:




Example 5 – Finding Logarithms

a. Find log 100.

b. Find ln 1.

c. Find log 0.1.

( ) 210log100log2

==

01ln =

110log1.0log 1−==

−


d. Find ln e-1.

d. Find log366.

110log1.0log −==

1ln1ln 1 −=−=− ee

2

1

6log2

6log6log36 ==



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


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≈==


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.

=

=

=

−=


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


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− −−


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



Example 5 – Simplifying Logarithmic Expressions

a.

b.

c.

.3ln 3 xe x=

3

30

10log01000log1log 3

=

+=

+=+

89/89 8 7log7log ==


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



Example 7 – Evaluating a Logarithm Base 5

Find log52.

Solution:

4307.02log

2log5log

2log5log

25

≈=

=

=

=

x

x

x

x


4307.05log

2log≈=x



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.


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


4.4 Logarithmic and Exponential Equations

Example 1 – Oxygen Composition


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:



Example 1 – Oxygen Composition

)934.5log(log

log934.5log

log885.0934.5loglog

885.0

885.0

xy

x

xy

=

+=

+=

885.0934.5 xy =




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


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



Example 5 – Predator-Prey Relation

axyK

K=

−ln

)1( axeKy −−=


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!)


Chapter 5 Chapter 5

Mathematics of Finance Mathematics of Finance


• To solve interest problems which require logarithms.

• To solve problems involving the time value of money.

Chapter 5: Mathematics of Finance



• 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)




Annuities

Amortization of Loans

5.4)

5.5)


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


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.




( )

( )

38.588

500

38.5881

38.5881500

6

6

=+

=+

r

r


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.




( )015.1600900 =n


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



Example 5 – Effective Rate

Effective Rate

• The effective rate re for a year is given by

11 −

+=

n

en

rr


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



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


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


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


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


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.


• a final payment at the end of five years.

What would this payment be if an interest rate of 6% compounded annually is assumed?


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



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 =


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


5.2 Present Value

Example 5 – Net Present Value

Solution:

( ) ( ) ( )

31.457$

000,2007.1600007.1800007.1000,10NPV532

−≈

−++=−−−


Compound Amount under Continuous Interest

• The compound amount S is defined as


5.3 Interest Compounded Continuously5.3 Interest Compounded Continuously

kt

k

rPS

+= 1

Effective Rate under Continuous Interest


•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 −=



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


b. 5 years.

13.105$100 ≈== ePeS

( )( ) 40.128$100100 25.0505.0≈== eeS



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?


Solution:

We want the present value of $25,000 due in 20 years.

( )( )

6165$000,25

000,254.1

2007.0

≈=

==

−

−−

e

eSeP rt


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



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


5.4 Annuities


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


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


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


1−2

2


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


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


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


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


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.


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


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 −+

⋅=


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 =−=


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


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?


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


45.142$6300

6300

02.032

02.032

________

____

≈==

=

ss

SR

Rs

rn


5.5 Amortization of Loans5.5 Amortization of Loans

Amortization Formulas



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


b.total interest charges, and

c.principal remaining after five years.


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


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


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)


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 .


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×


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


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



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 =+



Scalar Multiplication

• Properties of Scalar Multiplication:

Subtraction of Matrices

• Property of subtraction is ( )AA 1−=−



Example 5 – Matrix Subtraction

a.

−

−

−

=

−+

−−−

+−

=

−

−

−

13

08

84

3203

1144

2662

30

14

26

23

14

62

b.

−−=

−−

−=−

52

80

42

66

10

262BAT


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:



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



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



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



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



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



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


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


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



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.



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



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


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


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.


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.


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


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


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


Chapter 7 Chapter 7

Linear ProgrammingLinear Programming


• 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



problem exists.

Linear Inequalities in Two Variables

Linear Programming

7.1)

7.2)





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


where a, b, c = constants and

not both a and b are zero.

0≥++ cbyax


7.1 Linear Inequalities in 2 Variables

• A solid line is included in the solution and a

dashed line is not.



7.1 Linear Inequalities in 2 Variables


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.


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


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 +=


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



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


The maximum value of Z occurs when x = 4 and y = 0.

( ) ( )( ) ( )( ) ( ) 4403

11233

12043

=+=

=+=

=+=

DZ

CZ

BZ



Example 3 – Unbounded Feasible Region

The information for a produce is summarized as

follows:


If the grower wishes to minimize cost while still

maintaining the nutrients required, how many bags of

each brand should be bought?



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 +=


Subject to the constraints

There is no maximum value since the feasible region is unbounded .

0

0802

20025

16023

≥

≥

≥+

≥+

≥+

y

xyx

yx

yx

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