INTRODUCTORY MATHEMATICAL ANALYSIS For Business, Economics, and the Life and Social Sciences 2011 Pearson Education, Inc. Chapter 1 Applications and More

INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences

2011 Pearson Education, Inc.

Chapter 1 Chapter 1 Applications and More AlgebraApplications and More Algebra


• To model situations described by linear or quadratic equations.

• To solve linear inequalities in one variable and to introduce interval notation.

• 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 ObjectivesChapter Objectives



Chapter OutlineChapter Outline

Applications of Equations

Linear Inequalities

Applications of Inequalities

Absolute Value

Summation Notation

1.6) Sequence

1.1)

1.2)

1.3)

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

solution made up of two parts alcohol and three

parts acid. How much of each should be used?

Example 1 - Mixture


Solution:Let n = number of milliliters in each part.

Each part has 70 ml. Amount of alcohol = 2n = 2(70) = 140 mlAmount of acid = 3n = 3(70) = 210 ml

705

350

3505

35032

n

n

nn


1.1 Applications of Equations

Example 1 - Mixture


• 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 sold for the company to earn a profit of $60,000.



Example 3 – Profit


Solution:

Let q = number of sold units.

variable cost = 6q

total cost = 6q + 80,000

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



Example 3 – Profit


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


Solution:

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

$5500 was invested at 6%

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

5500

75.130025.0

75.5880575.057506.0

75.588000,100575.006.0

x

x

xx

xx



Example 5 – Investment




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 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):

Rent should be $650 or $700.

256756

500,224050

32

000,365,13440504050

0000,365,140503

34050000,365,1

25

34050600,54

25

165032400600,54

25

550396600,54

2

2

r

rr

rr

rr

rr

rr





Solution 2:

Let n = number of $25 increases.

Total rent = (rent per apartment) x (number of apartments rented)





Solution 2 (Con’t):

The rent charged should be either

550 + 25(6) = $700 or

550 + 25(4) = $650.

4 or 6

046

02410

0180075075

75750800,52600,54

39625550600,54

2

2

2

n

nn

nn

nn

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 < b and x is between a and b.

• Inequality is a statement that one number is less than another number.


1.2 Linear Inequalities


• 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.

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




• 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 inequality is true.




Example 1 – Solving a Linear Inequality

Solve 2(x − 3) < 4.

Solution:Replace inequality by equivalent inequalities.

5 2

10

2

2

102

64662

462

432

x

x

x

x

x

x




Example 3 – Solving a Linear Inequality

Solve (3/2)(s − 2) + 1 > −2(s − 4).

7

20

207

16443

442232

422122

32

42122

3

s

s

ss

ss

s-s

ss

The solution is ( 20/7 ,∞).

Solution:



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 $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

5000

000,7014

0000,702135

0 cost total revenue total

q

q

qq

Let q = number of heaters sold.


1.3 Applications of Inequalities

Example 1 - Profit


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 than 2.5? (Note: The funds received are considered as current assets and the loan as a current liability.)



Example 3 – Current Ratio


Solution:

Let x = amount the company can borrow.

Current ratio = Current assets / Current liabilities

We want,

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



Example 3 – Current Ratio


• On real-number line, the distance of x from 0 is called the absolute value of x, denoted as |x|.

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

0 if ,

0 if ,

xx

xxx


1.4 Absolute Value1.4 Absolute Value



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

c. The absolute value of a number is never negative. The solution set is .


1.4 Absolute Value



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.

We write it as , where is the union symbol.

b.

We can write it as .

2 12

75 or 75

xx

xx

,212,

3

5 1

143 or 143

xx

xx

,3

51,


Properties of the Absolute Value

• 5 basic properties of the absolute value:

• Property 5 is known as the triangle inequality.

baba

aaa

abba

b

a

b

a

baab

.5

.4

.3

.2

.1


1.4 Absolute Value



1.4 Absolute Value

Example 5 – Properties of Absolute Value

323251132 g.

222 f.

5

3

5

3

5

3 e.

3

7

3

7

3

7 ;

3

7

3

7

3

7 d.

77 c.

24224 b.

213737- a.

-

xxx

xx

Solution:



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

mii aaaaa

...21


Evaluate the given sums.

a. b.

Solution:

a.

b.


1.5 Summation Notation

Example 1 – Evaluating Sums

7

3

25n

n

6

1

2 1j

j

115

3328231813

275265255245235257

3

n

n

97

3726171052

1615141312111 2222226

1

2

j

j


• To sum up consecutive numbers, we have

where n = the last number

2

1

1

nni

n

i



6

)12(1

1

2

nnni

n

i


Evaluate the given sums.

a. b. c.

Solution:

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



Example 3 – Applying the Properties of Summation Notation

2847144471

1

100

30

ij

100

1

35k

k

200

1

29k

k

100

30

4j


1.6 Sequence

• Arithmetic sequenceAn 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 sequenceA geometric sequence is a sequence (ck) defined

recursively by

c1=a and, for each positive integer k, ck+1= ck*r

Example

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

- Sum to first n term

- Sum of an infinite geometric sequence

for

)2)1((2

adnn

sn

1,1

)1(

rforr

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 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 kRRRR

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

02.11

1

000,100

1

ra

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INTRODUCTORY MATHEMATICAL ANALYSIS For Business, Economics, and the Life and Social Sciences 2011 Pearson Education, Inc. Chapter 1 Applications and More