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4. Inequalities. 4.1 Solving Linear Inequalities Problem Basic fee: $20 Basic fee: $20 Per minute: 5¢ Per minute: 5¢ Budget: $40 Budget: $40 How many.

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Presentation on theme: "4. Inequalities. 4.1 Solving Linear Inequalities Problem Basic fee: $20 Basic fee: $20 Per minute: 5¢ Per minute: 5¢ Budget: $40 Budget: $40 How many."— Presentation transcript:

1 4. Inequalities

2 4.1 Solving Linear Inequalities Problem Basic fee: $20 Basic fee: $20 Per minute: 5¢ Per minute: 5¢ Budget: $40 Budget: $40 How many minutes?: x How many minutes?: x x ≤ 40 x ≤ 400

3 Notations Closed interval [a, b] = {x | a ≤ x ≤ b} [a, b] = {x | a ≤ x ≤ b} Open interval (a, b) = {x | a < x < b} (a, b) = {x | a < x < b} ab ab

4 Notations Infinite interval [a, ∞ ) = {x | a ≤ x < ∞ } [a, ∞ ) = {x | a ≤ x < ∞ } Infinite interval (- ∞, b] = {x | - ∞ < x ≤ b} (- ∞, b] = {x | - ∞ < x ≤ b} a b

5 Your Turn Express in set-builder notation. [a, b) [a, b) (- ∞, b) (- ∞, b)

6 Solving Inequalities in One Variable 0.05x + 20 ≤ x ≤ 20 x ≤ 20/0.05 x ≤ 400 [0, 400] (Interval notation) {x | x ≤ 400} (set-builder notation)

7 Properties of Inequalities Addition a < b → a + c < b + c a < b → a - c < b – c a < b → a + c < b + c a < b → a - c < b – c Positive Multiplication (c > 0) a < b → ac < bc a < b → a /c < b/c a < b → ac < bc a < b → a /c < b/c Negative Multiplication (c < 0) a < b → ac ≥ bc a < b → a /c ≥ b/c a < b → ac ≥ bc a < b → a /c ≥ b/c

8 Example 1 -2x – 4 > x x > 9 (-1/3)(-3x) 9 (-1/3)(-3x) < (-1/3)9 x < -3 (- ∞,-3)

9 Example 2 (x + 3) (x – 2) ≥ (x + 3) 4(x – 2) ≥ x + 9 ≥ 4x – x ≥ -14 x ≤ 14 (-∞, 14]

10 Special Cases x > x + 1 {x | x > x + 1} {x | x > x + 1} What kind of set is this? What kind of set is this? x < x + 1 {x } x < x + 1} {x } x < x + 1} What kind of set is this? What kind of set is this?

11 4.2 Compound Inequalities Intersection of Sets Given set A and B, intersection A and B, A ∩ B = {x | x ε A AND x ε B} Given set A and B, intersection A and B, A ∩ B = {x | x ε A AND x ε B} A ∩ B A ∩ B B

12 Compound Inequalities Union of Sets Given set A and B, union of A and B, A U B = {x | x ε A OR x ε B} Given set A and B, union of A and B, A U B = {x | x ε A OR x ε B} A UB A U B A

13 Intersection of Sets Given: A = {1, 2, 3,5, 9} B = {3, 5, 9, 10, 12} A ∩ B = {5, 9} Given: A = {x | x ≥ 3} B = {x | x ≤ 10} A ∩ B = {x | x ≥ 3 AND x ≤ 10 } [ ] 310

14 Union and Intersection of Sets Given: A = set of all male students at CUH A = set of all male students at CUH B = set of all female students at CUH B = set of all female students at CUH C = set of all freshman students at CUH C = set of all freshman students at CUH Draw a diagram of: A ∩ B A ∩ B A U B A U B A ∩ C A ∩ C A U C A U C

15 Solving Compound Inequality Given: 2x – 7 > 3 AND 5x – 4 3 AND 5x – 4 < 6 What does it mean: Solve the compound inequality? It means: Find the set of x so that both inequalities are true Solution Set: {x | 2x – 7 > 3 AND 5x – 4 3 AND 5x – 4 < 6}

16 Solving an AND Compound Inequality 2x – 7 ≥ 3 AND 5x – 4 ≤ 6 2x ≥ x ≤ x ≥ 10 5x ≤ 10 x ≥ 10/2 x ≤ 10/5 x ≥ 5 x ≤ 2 Solution Set: { Φ } 25

17 Solving an AND Compound Inequality -3 < 2x + 1 ≤ 3 This means: (-3 < 2x + 1 AND 2x + 1 ≤ 3) -3 – 1 < 2x + 1 – 1 ≤ < 2x ≤ 2 -1 < x ≤ 1 Solution Set: { x | -1 < x ≤ 1 } 1

18 Solving an OR Compound Inequality Given: 2x – 3 < 7 OR 35 – 4x ≤ 3 2x < 4 OR -4x ≤ -32 x < 4 OR x ≥ 8 Take the union of solution sets {x | x < 4 U x ≥ 8} = {x | x < 4 or x ≥ 8} 48

19 Solving an OR Compound Inequality Given: 3x – 5 ≤ 13 OR 5x + 2 > -3 3x ≤ 18 OR 5x > -5 x ≤ 6 OR x > -1 Take the union of solution sets {x | x ≤ 6 U x > -1} = {x | x ≤ 6 or x > -1} = R 6

20 Your Turn Find the following sets 1){a, b, c, d, e} ∩ {b, c, 2, 3, x, y} 2){a, b, c, d, e} U {b, c, 2, 3, x, y} Solve the following 1)3 ≤ 4x – 3 < 19 2)3x 10

21 4.3 Equations & Inequalities Involving Absolute Values Absolute value of A -- |A| -- where A is any algebraic expression: |A| = c  A = c or A = -c, where c > 0 |A| = c  A = c or A = -c, where c > 0 |2x – 3| = 11 2x – 3 = 11 or 2x – 3 = cc AA |A|

22 Solving Equation Involving Absolute Value Solve for x: |2x – 3| = 11 2x – 3 = 11 or 2x – 3 = -11 2x = 14 2x = -8 x = 7 x = - 4 {4, 7} Solve for x: 5|1 - 4x| -15 = 0 |1 – 4x| = 15/5 = 3 1 – 4x = 3 or 1 – 4x = -3 -4x = 2 -4x = -4 x = -1/2 x = 1 {-1/2, 1}

23 Equation With 2 Absolute Values Solve for x: |2x – 7| = |x + 3| 2x – 7 = (x + 3) or 2x – 7 = -(x + 3) x = 10 2x – 7 = -x – 3 3x = 4 x = 4/3 {4/3, 10}

24 Solving Absolute Value Inequality (Using Boundary Points) Solve and graph: |2x + 3| ≥ 5 1.Solve the equation 2x + 3 = 5 or 2x + 3 = -5 x = 1 x = -4 2.Locate the boundary points 3.Choose a test value in each interval and substitute in the inequality -4 1

25 Solve and graph: |2x + 3| ≥ 5 2.Locate the boundary points 3.Choose a test value in each interval and substitute in inequality Interval Test value CheckConclusion (-∞, -4) |2 ∙(-5) + 3| ≥ 5 |-7| ≥ 5 true (-∞, -4) in solution set (-4, 1) 0 |2 ∙ 0 + 3| ≥ 5 |3| ≥ 5 false (-4, 1) not in solution set (1, ∞) 2 |2 ∙ 2 + 3| ≥ 5 |7| ≥ 5 true (1, ∞) in solution set

26 Solve and graph: |2x + 3| ≥ Write the solution set. Check for boundaries. Preliminary Solution: (-∞, -4) U (1, ∞) Because |2x + 3| = 5, we need to include the solution set of this equation (i.e., boundaries): x = -4, 1. (This was found in step 1.) Final Solution: (-∞, -4] U [1, ∞) - 4 1

27 Using Boundary Points Solve and graph: |2x -5| ≥ 3 1.Solve the equation 2x – 5 = 3 or 2x – 5 = -3 x = 4 x = 1 2.Locate the boundary points 3.Choose a test value in each interval and substitute in inequality 1 4

28 Solve and graph: |2x - 5| ≥ 3 2.Locate the boundary points 3.Choose a test value in each interval and substitute in inequality 1 4 Interval Test value CheckConclusion (-∞, 1) 0 |2 ∙ 0 – 5| ≥ 3 |-5| ≥ 3 true (-∞, 1) in solution set (1, 4) 2 |2 ∙ 2 - 5| ≥ 3 |-1| ≥ 3 false (1, 4) not in solution set (4, ∞) 5 |2 ∙ 5 - 5| ≥ 3 |5| ≥ 5 true (4, ∞) in solution set

29 Solve and graph: |2x – 5| ≥ Write the solution set. Check for boundaries. Preliminary Solution: (-∞, 1) U (4, ∞) Because |2x - 5| = 3, we need to include the solution set of this equation (i.e., boundaries): x = 1, 4 (This was found in step 1.) Final Solution: (-∞, 1] U [4, ∞) 1 4

30 Solving Absolute Value Inequality (Using Compound Inequalities) Note: Solution set of |x| < 2 is (-2, 2) (-2, 2)  -2 < x < 2 Solution set of |x| < 2 is (-2, 2) (-2, 2)  -2 < x < 2 Solution set of |x| > 2 is (-∞, -2) U (2, ∞) Solution set of |x| > 2 is (-∞, -2) U (2, ∞) (-∞, -2) U (2, ∞)  x 2

31 Solving Absolute Value Inequality (Using Compound Inequalities) Solve: |x – 4| < 3 -3 < x – 4 < 3 1 < x < 7

32 Solving Absolute Value Inequality (Using Compound Inequalities) Solve: |2x + 3| ≥ 5 2x + 3 ≥ 5 or 2x + 3 ≤ -5 2x ≥ 2 or 2x ≤ -8 x ≥ 1 or x ≤

33 Your Turn Solve inequalities using equivalent compound inequalities 1.|x – 2| < 5 2.|2x – 5| ≥ 3

34 4.4 Linear Inequalities in 2 Variables Solve: 2x – 3y ≥ 6 1.Graph: 2x – 3y = 6 To find y-intercept To find x-intercept y = 0 x = 0 2x = 6 -3y = 6 x = 3 y = -2 (0, -2) (3, 0) 2x – 3y > 6 2x – 3y = 6 2x – 3y < 6

35 2.Choose a test point in one half-plane and check with original inequality. 2x – 3y ≥ 6 Choose A (0, 0) as a test point 0 – 0 ≥ 6 0 ≥ 6 false—A is outside the solution set

36 3.If A(0, 0) is not in solution set, the other half- plane is the solution set of 2x – 3y ≥ 6 Because of ≥, include the boundary line in the graph of the solution set. 2x – 3y < 6 2x – 3y = 6 2x – 3y > 6 A(0, 0) Graph of : {x | 2x – 3y ≥ 6}

37 Your Turn Graph the following inequality: 1.4x – 2y ≥ 8 2.x/4 + y/2 < 1

38 Graphing System of Linear Inequalities Graph solution set of: x – y < 1 2x + 3y ≥ 12 Graph equations x – y = 1 2x + 3y = 12 x-intercept: x-intercept: y = 0 y = 0 x – 0 = 1 2x + 0 = 12 x = 1 x = 6 (1, 0) (6, 0) Graph equations x – y = 1 2x + 3y = 12 x-intercept: x-intercept: y = 0 y = 0 x – 0 = 1 2x + 0 = 12 x = 1 x = 6 (1, 0) (6, 0)

39 Graphing System of Linear Inequalities Graph equalities x – y = 1 2x + 3y = 12 x-intercept: x-intercept: (1, 0) (6, 0) y-intercept: y-intercept: x = 0 x = 0 0 – y = y = 12 -y = 1 3y = 12 y = -1 y = 4 (0, -1) ( 0, 4) Points for: x – y = 1 Points for: 2x + 3y = 12 (1, 0) (0, -1) (6, 0) (0, 4) Graph equalities x – y = 1 2x + 3y = 12 x-intercept: x-intercept: (1, 0) (6, 0) y-intercept: y-intercept: x = 0 x = 0 0 – y = y = 12 -y = 1 3y = 12 y = -1 y = 4 (0, -1) ( 0, 4) Points for: x – y = 1 Points for: 2x + 3y = 12 (1, 0) (0, -1) (6, 0) (0, 4)

40 Graphing System of Linear Inequalities 2x + 3y = 12x – y = 1 (0, 0) (1, 0) (0, 4) (6, 0)

41 Graphing System of Linear Inequalities Choose a point and check with original inequalities. Pick (0, 0) Pick (0, 0) Part of x – y < 1? Part of 2x + 3y ≥ 12? Check: Check: 0 – 0 < 1? ≥ 12? 0 < 1? 0 ≥ 12? true false this half-plane other half-plane Choose a point and check with original inequalities. Pick (0, 0) Pick (0, 0) Part of x – y < 1? Part of 2x + 3y ≥ 12? Check: Check: 0 – 0 < 1? ≥ 12? 0 < 1? 0 ≥ 12? true false this half-plane other half-plane

42 Graphing System of Linear Inequalities 2x + 3y = 12 x – y = 1 (0, 0)

43 Your Turn Graph the solution set of the system: x – 3y < 6 2x + 3y ≥ -6

44 4.5 Linear Programming Problem: A division of a furniture company specializes in manufacturing bookcases and computer desks. A division of a furniture company specializes in manufacturing bookcases and computer desks. The division makes $25 per bookcase and $55 per desk. The division makes $25 per bookcase and $55 per desk. To maintain quality, the division can make a maximum of 80 bookcases and desks (total) per day To maintain quality, the division can make a maximum of 80 bookcases and desks (total) per day

45 4.5 Linear Programming Problem (cont.) Because of customer demands, between 30 and 80 bookcases must be made daily. Because of customer demands, between 30 and 80 bookcases must be made daily. Furthermore, at least 10 and not more than 30 desks must be made per day Furthermore, at least 10 and not more than 30 desks must be made per day How many bookcases and desks must be made each day to maximize profit? How many bookcases and desks must be made each day to maximize profit?

46 4.5 Linear Programming Solution 1. Use variables to represent quantities x = number of bookcases per month y = desks per month z = profit for month 2. Form objective function z = 25x + 55y 3. Write constraints as inequalities x + y ≤ ≤ x ≤ ≤ y ≤ Graph the inequalities

47 4.5 Linear Programming 4. Graph the inequalities 1. x + y ≤ 80 x + y = 80 line passes through (80, 0) and (0, 80) ≤ x ≤ 80 y can be any value ≤ y ≤ 30 x can be any value

48 4.5 Linear Programming 4. Graph the inequalities (80, 0) (0, 80) x + y ≤ ≤ x ≤ ≤ y ≤ 30 A D C B

49 4.5 Linear Programming 5. Determine the corners of the solution area To find A: x = 30 y = 30 (30, 30) To find B: y = 30 x + 30 = 80 x = 50 (50, 30) To find C: y = 10 x + 10 = 80 x = 70 (70, 10) To find D: (30, 10) (80, 0) (0, 80) x + y = 80 x = 80 y = 30A D C B y = 10 x = 30

50 4.5 Linear Programming 6. Check the objective equation with the corner points Corner (x, y) Objective Function z = 25x + 55y (30, 30) z = 25(30) + 55(30) = 2400 (50, 30) z = 25(50) + 55(30) = 2900 (70, 10) z = 25(70) + 55(10) = 2300 (30, 10) z = 25(30) + 55(10) = 1300 Solution: 50 bookcases, 30 desks, resulting in $2900 profit

51 Linear Programming (Another Example) Problem: Food and clothing are shipped to survivors of a hurricane. Each carton of food will feed 12 people, while each carton of clothing will help 5 people. Food and clothing are shipped to survivors of a hurricane. Each carton of food will feed 12 people, while each carton of clothing will help 5 people. Each 20 ft 3 box of food weights 50 lb, and each 10 ft 3 box of food will weight 20 lb Each 20 ft 3 box of food weights 50 lb, and each 10 ft 3 box of food will weight 20 lb Planes are bound by the following constraints Planes are bound by the following constraints Total weight per plane ≤ lb Total volume per plane ≤ 8000 ft 3 How many cartons of food and how many cartons of clothing should be sent with each plane to maximize the number of people who can be helped? How many cartons of food and how many cartons of clothing should be sent with each plane to maximize the number of people who can be helped?

52 Linear Programming Solution 1. Use variables to represent quantities x = cartons of food y = cartons of clothing z = number of people helped 2. Form objective function z = 12x + 5y 3. Write constraints as inequalities 50x + 20y ≤ 19,000 20x + 10y ≤ 8,000

53 Linear Programming 4. Graph the inequalities 1. 50x + 20y ≤ x + 20y = Find 2 points—e.g., y-intercept & x-intercept (0, 950) & (380, 0) 2. 20x + 10y ≤ x + 10y = 8000 Find 2 points (0, 800) & (400, 0) y can be any value

54 Linear Programming 4. Graph the inequalities (380, 0) (0, 950) 20x + 10y ≤ x + 20y ≤ A C B (400, 0) (0, 800)

55 Linear Programming 5. Determine the corners of the solution area To find A: (0, 800) To find B: 50x + 20y = x + 10y = x + 20y = x – 20y = x = 3000 x = 300 y = 200 (300, 200) To find C: (380, 0) (380, 0) (0, 950) 20x + 10y ≤ x + 20y ≤ A C B (400, 0) (0, 800)

56 Linear Programming 6. Check the objective equation with the corner points Corner (x, y) Objective Function z = 12x + 5y (0, 800) z = 12(0) + 5(800) = 4000 (300, 200) z = 12(300) + 5(200) = 4600 (380, 0) z = 12(380) + 5(0) = 4560 Solution: 300 food cartons, 200 clothing cartons, resulting in 4600 people helped

57 Your turn Problem: A theater is presenting a program on drinking and driving for students and their parents. A theater is presenting a program on drinking and driving for students and their parents. Admission $2.00 for parents $1.00 for students. Admission $2.00 for parents $1.00 for students. However, the situation has two constraints: However, the situation has two constraints: The theater can hold no more than 150 people, and every two parents must bring at least one students How many parents and students should attend to raise the maximum amount of money? How many parents and students should attend to raise the maximum amount of money?

58 Solution Write the objective function Write the objective function Write the constraints inequalities Write the constraints inequalities

59 Solutions Variables x = number of parents x = number of parents y = number of students y = number of students z = total amount of money z = total amount of money Objective Function z = 2x + y z = 2x + yConstraints x + y ≤ 150 x + y ≤ 150 x ≥ 2y x ≥ 2y

60 Solution x (parents) y (students) x ≤ 2y

61 Solution (0, 150) (150, 0) x + y = 150 x = 2y (100, 50)


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