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- 1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences ©2007 Pearson Education Asia Chapter 1Chapter 1 Applications and More AlgebraApplications and More Algebra
- 2. ©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
- 3. ©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
- 4. ©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
- 5. ©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)
- 6. ©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
- 7. ©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 ml Amount of acid = 3n = 3(70) = 210 ml 70 5 350 3505 35032 == = =+ n n nn Chapter 1: Applications and More Algebra 1.1 Applications of Equations Example 1 - Mixture
- 8. ©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
- 9. ©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
- 10. ©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 qq = = +−= 000,35 4000,140 000,80610000,60 Chapter 1: Applications and More Algebra 1.1 Applications of Equations Example 3 – Profit
- 11. ©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 $588.75? Chapter 1: Applications and More Algebra 1.1 Applications of Equations Example 5 – Investment
- 12. ©2007 Pearson Education Asia Solution: Let x = amount ($) 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
- 13. ©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 $54,600 per month from rent. What rent should be charged for each apartment?
- 14. ©2007 Pearson Education Asia Solution 1: Let r = rent ($) 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
- 15. ©2007 Pearson Education Asia Solution 1 (Con’t): Rent should be $650 or $700. ( ) ( ) ( ) ( )( ) ( ) 25675 6 500,224050 32 000,365,13440504050 0000,365,140503 34050000,365,1 25 34050 600,54 25 165032400 600,54 25 5503 96600,54 2 2 ±= ± = −−± = =+− −= − = +− = − −= r rr rr r r r r r r Chapter 1: Applications and More Algebra 1.1 Applications of Equations Example 7 – Apartment Rent
- 16. ©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
- 17. ©2007 Pearson Education Asia Solution 2 (Con’t): The rent charged should be either 550 + 25(6) = $700 or 550 + 25(4) = $650. ( )( ) ( )( ) 4or6 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
- 18. ©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:
- 19. ©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
- 20. ©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 <
- 21. ©2007 Pearson Education Asia Chapter 1: Applications and More Algebra 1.2 Linear Inequalities • Linear inequality can be written in the form ax + b < 0 where 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.
- 22. ©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
- 23. ©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 42212 2 3 2 4212 2 3 > > +−>− −−>+− −> +− −−>+− s s ss ss s-s ss The solution is ( 20/7 ,∞). Solution:
- 24. ©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?
- 25. ©2007 Pearson Education Asia Solution: profit = total revenue − total cost ( ) 5000 000,7014 0000,702135 0costtotalrevenuetotal > > >+− >− q q qq Let q = number of heaters sold. Chapter 1: Applications and More Algebra 1.3 Applications of Inequalities Example 1 - Profit
- 26. ©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
- 27. ©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 5.2 000,80 000,350 Chapter 1: Applications and More Algebra 1.3 Applications of Inequalities Example 3 – Current Ratio
- 28. ©2007 Pearson Education Asia • On real-number line, the distance of x from 0 is called the absolute value of x, denoted as |x|. DEFINITION The absolute value of a real number x, written |x|, is defined by <− ≥ = 0if, 0if, xx xx x Chapter 1: Applications and More Algebra 1.4 Absolute Value1.4 Absolute Value
- 29. ©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
- 30. ©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
- 31. ©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
- 32. ©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 . 212 75or75 ≥−≤ −≥+−≤+ xx xx ] [( )∞−∞− ,212, 3 5 1 143or143 >< >−−<− xx xx ( ) ∞∪∞− , 3 5 1,
- 33. ©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
- 34. ©2007 Pearson Education Asia Chapter 1: Applications and More Algebra 1.4 Absolute Value Example 5 – Properties of Absolute Value ( ) ( ) 323251132g. 222f. 5 3 5 3 5 3 e. 3 7 3 7 3 7 ; 3 7 3 7 3 7 d. 77c. 24224b. 213737-a. +−=+=≤==+− ≤≤ − = − − = − − = − − = − − = − = − −=− =−=− =⋅−=⋅ - xxx xx Solution:
- 35. ©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 mi i aaaaa ++++= ++ = ∑ ...21
- 36. ©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 25 n n ( )∑= + 6 1 2 1 j j ( ) ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ] 115 3328231813 27526525524523525 7 3 = ++++= −+−+−+−+−=−∑=n n ( ) ( ) ( ) ( ) ( ) ( ) ( ) 97 3726171052 1615141312111 222222 6 1 2 = +++++= +++++++++++=+∑=j j
- 37. ©2007 Pearson Education Asia • To sum up consecutive numbers, we have where n = the last number. ( ) 2 1 1 + =∑= nn i n i Chapter 1: Applications and More Algebra 1.5 Summation Notation
- 38. ©2007 Pearson Education Asia Evaluate the given sums. a. b. c. Solution: a. b. c. ( ) ( ) 550,251003 2 101100 53535 100 1 100 1 100 1 =+ ⋅ =+=+ ∑∑∑ === kkk kk 300,180,24 6 401201200 999 200 1 2 200 1 2 = ⋅⋅ == ∑∑ == kk kk Chapter 1: Applications and More Algebra 1.5 Summation Notation Example 3 – Applying the Properties of Summation Notation 28471444 71 1 100 30 =⋅== ∑∑ == ij ( )∑= + 100 1 35 k k ∑= 200 1 2 9 k k∑= 100 30 4 j

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