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INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
2007 Pearson Education Asia
Chapter 1 Chapter 1 Applications and More AlgebraApplications and More Algebra
2007 Pearson Education Asia
INTRODUCTORY MATHEMATICAL ANALYSIS
0. Review of Algebra
1. Applications and More Algebra
2. Functions and Graphs
3. Lines, Parabolas, and Systems
4. Exponential and Logarithmic Functions
5. Mathematics of Finance
6. Matrix Algebra
7. Linear Programming
8. Introduction to Probability and Statistics
2007 Pearson Education Asia
9. Additional Topics in Probability
10. Limits and Continuity
11. Differentiation
12. Additional Differentiation Topics
13. Curve Sketching
14. Integration
15. Methods and Applications of Integration
16. Continuous Random Variables
17. Multivariable Calculus
INTRODUCTORY MATHEMATICAL ANALYSIS
2007 Pearson Education Asia
• 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
2007 Pearson Education Asia
Chapter 1: Applications and More Algebra
Chapter OutlineChapter Outline
Applications of Equations
Linear Inequalities
Applications of Inequalities
Absolute Value
Summation Notation
1.1)
1.2)
1.3)
1.4)
1.5)
2007 Pearson Education Asia
• 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
2007 Pearson Education Asia
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
2007 Pearson Education Asia
• 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
2007 Pearson Education Asia
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
2007 Pearson Education Asia
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
2007 Pearson Education Asia
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 $588.75?
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 5 – Investment
2007 Pearson Education Asia
Solution:
Let x = amount ($) 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
2007 Pearson Education Asia
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 $54,600 per month from rent. What rent should be charged for each apartment?
2007 Pearson Education Asia
Solution 1:
Let r = rent ($) 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
2007 Pearson Education Asia
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
2007 Pearson Education Asia
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
2007 Pearson Education Asia
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
2007 Pearson Education Asia
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:
2007 Pearson Education Asia
• 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
2007 Pearson Education Asia
• 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
2007 Pearson Education Asia
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.
2007 Pearson Education Asia
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
2007 Pearson Education Asia
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:
2007 Pearson Education Asia
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?
2007 Pearson Education Asia
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
2007 Pearson Education Asia
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
2007 Pearson Education Asia
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
2007 Pearson Education Asia
• 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
2007 Pearson Education Asia
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
2007 Pearson Education Asia
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
2007 Pearson Education Asia
Absolute-Value Inequalities
• Summary of the solutions to absolute-value inequalities is given.
Chapter 1: Applications and More Algebra
1.4 Absolute Value
2007 Pearson Education Asia
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,
2007 Pearson Education Asia
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
2007 Pearson Education Asia
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:
2007 Pearson Education Asia
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
2007 Pearson Education Asia
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
2007 Pearson Education Asia
• 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
2007 Pearson Education Asia
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
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