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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
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.
• 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
2011 Pearson Education, Inc.
Chapter 1: Applications and More Algebra
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)
2011 Pearson Education, Inc.
• 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
solution made up of two parts alcohol and three
parts acid. How much of each should be used?
Example 1 - Mixture
2011 Pearson Education, Inc.
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
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 1 - Mixture
2011 Pearson Education, Inc.
• 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
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
2011 Pearson Education, Inc.
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.
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 3 – Profit
2011 Pearson Education, Inc.
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
000,35
4000,140
000,80610000,60
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 3 – Profit
2011 Pearson Education, Inc.
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 $588.75?
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 5 – Investment
2011 Pearson Education, Inc.
Solution:
Let x = amount ($) 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
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 5 – Investment
2011 Pearson Education, Inc.
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 $54,600 per month from rent. What rent should be charged for each apartment?
2011 Pearson Education, Inc.
Solution 1:
Let r = rent ($) 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
2011 Pearson Education, Inc.
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
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent
2011 Pearson Education, Inc.
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
2011 Pearson Education, Inc.
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
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent
2011 Pearson Education, Inc.
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:
2011 Pearson Education, Inc.
• 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.
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
2011 Pearson Education, Inc.
• 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 .
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
nn ba
2011 Pearson Education, Inc.
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 inequality is true.
2011 Pearson Education, Inc.
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.
5 2
10
2
2
102
64662
462
432
x
x
x
x
x
x
2011 Pearson Education, Inc.
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).
7
20
207
16443
442232
422122
32
42122
3
s
s
ss
ss
s-s
ss
The solution is ( 20/7 ,∞).
Solution:
2011 Pearson Education, Inc.
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 $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?
2011 Pearson Education, Inc.
Solution:
profit = total revenue − total cost
5000
000,7014
0000,702135
0 cost total revenue total
q
q
Let q = number of heaters sold.
Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities
Example 1 - Profit
2011 Pearson Education, Inc.
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.)
Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities
Example 3 – Current Ratio
2011 Pearson Education, Inc.
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
Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities
Example 3 – Current Ratio
2011 Pearson Education, Inc.
• 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
Chapter 1: Applications and More Algebra
1.4 Absolute Value1.4 Absolute Value
2011 Pearson Education, Inc.
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
2011 Pearson Education, Inc.
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 .
Chapter 1: Applications and More Algebra
1.4 Absolute Value
Example 1 – Solving Absolute-Value Equations
2011 Pearson Education, Inc.
Absolute-Value Inequalities
• Summary of the solutions to absolute-value inequalities is given.
Chapter 1: Applications and More Algebra
1.4 Absolute Value
2011 Pearson Education, Inc.
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.
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,
2011 Pearson Education, Inc.
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
Chapter 1: Applications and More Algebra
1.4 Absolute Value
2011 Pearson Education, Inc.
Chapter 1: Applications and More Algebra
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:
2011 Pearson Education, Inc.
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
mii aaaaa
...21
2011 Pearson Education, Inc.
Evaluate the given sums.
a. b.
Solution:
a.
b.
Chapter 1: Applications and More Algebra
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
2011 Pearson Education, Inc.
• To sum up consecutive numbers, we have
where n = the last number
2
1
1
nni
n
i
Chapter 1: Applications and More Algebra
1.5 Summation Notation
6
)12(1
1
2
nnni
n
i
2011 Pearson Education, Inc.
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
Chapter 1: Applications and More Algebra
1.5 Summation Notation
Example 3 – Applying the Properties of Summation Notation
2847144471
1
100
30
ij
100
1
35k
k
200
1
29k
k
100
30
4j
2011 Pearson Education, Inc.
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
2011 Pearson Education, Inc.
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
2011 Pearson Education, Inc.
• 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?
2011 Pearson Education, Inc.
• 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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