1. 24 Case Study Inequalities and Linear Programming 24.1Compound Linear Inequalities in One Unknown 24.2Quadratic Inequalities in One Unknown 24.3Linear Inequalities in Two Unknowns Chapter Summary 24.4Linear Programming 24.5Applications of Linear Programming

2. P. 2 In an art lesson, students are asked to make lantern frames using sticks and plasticine. Each group of students is given 100 sticks and 70 pieces of plasticine. There is also a requirement that the difference between the number of rectangular prisms and the number of pentagonal pyramids should not be more than 3. To make the frame of a rectangular prism, we need 12 sticks and 8 pieces of plasticine, while 10 sticks and 6 pieces of plasticine are needed to make the frame of a pentagonal pyramid. It is not difficult. Let me show you. What is the maximum number of lantern frames we can make if we have 100 sticks and 70 pieces of plasticine? Case Study

3. P. 3 The following table shows some combinations of the lantern frames that can be made. Case Study Number of rectangular prisms Number of pentagonal pyramids Number of sticks used Number of pieces of plasticine used 369660 539058 5410064 It is not easy to list all possible combinations. In fact, we can use linear programming to find the answer.

4. P. 4 For example: (a)x > 4(b)x  6 24.1 Compound Linear Inequalities in One Unknown One Unknown When we solve these kinds of compound inequalities, we have to find the values of x that can satisfy all the given inequalities. In junior forms, we learnt that the solutions of a linear inequality in one unknown can be represented on a number line. Now, we are going to investigate the solution of compound inequalities connected by the word ‘and’. A. Solving Compound Inequalities with ‘and’ If we want to represent the solution of compound inequalities on a number line, we can shade the region represented by each inequality Then the overlapping region is the solution: x > 4 and x  6, that is, 4 < x  6.

5. P. 5 Example 24.1T Solution: Solve 2x + 3  7x – 2 and 3x + 1 < 4x + 2, and represent the solution on the number line. Solving 2x + 3  7x – 2, we have 2x – 7x  –2 – 3 24.1 Compound Linear Inequalities in One Unknown One Unknown A. Solving Compound Inequalities with ‘and’ –5x  –5 x  1………..(1) Solving 3x + 1 < 4x + 2, we have 3x – 4x < 2 – 1 –x  x  –1……..(2) Combining (1) and (2), we have Therefore, the required solution is –1 < x  1.

6. P. 6 Example 24.2T Solve the compound inequality and –2(x – 1) + 3 < x – 7, and represent the solution on the number line. Solution: 24.1 Compound Linear Inequalities in One Unknown One Unknown A. Solving Compound Inequalities with ‘and’ Solving, we have 6 – x – 3  2x –x – 2x < 3 – 6 x  1………..(1) Solving –2(x – 1) + 3 < x – 7, we have –2x + 2 + 3 < x – 7 –2x – x < –7 – 2 – 3 x  4………..(2) Therefore, the required solution is x > 4. –3x < –3–3x < –12

7. P. 7 Example 24.3T Fanny has $200 and she wants to buy two types of gifts, which are pencil cases and mouse pads, for her students. The total number of gifts is 20 and the number of pencil cases should be at least three. If the prices of a pencil case and a mouse pad are $14 and $8 respectively, find the possible number of pencil cases that she can buy. Solution: Let x be the number of pencil cases, then the number of mouse pads is (20  x). We have 24.1 Compound Linear Inequalities in One Unknown One Unknown A. Solving Compound Inequalities with ‘and’ 14x + 8(20  x)  200 and x  3 14x + 160  8x  200 and x  3 160 + 6x  200 and x  3 6x  40 and x  3 and x  3 Combining the above results, we have  The possible number of pencil cases are 3, 4, 5, 6.

8. P. 8 Now, we are going to study another kind of compound inequalities connected by the word ‘or’. For example, 3x – 4 > 5 or 2x + 5 < 2. The solution is the whole shaded region on the number line. Solving compound inequalities with ‘or’, we need to solve each inequality separately and shade the region represented by each inequality. 24.1 Compound Linear Inequalities in One Unknown One Unknown B. Solving Compound Inequalities with ‘or’

9. P. 9 Example 24.4T Solution: Solve or, and represent the solution on the number line. 24.1 Compound Linear Inequalities in One Unknown One Unknown B. Solving Compound Inequalities with ‘or’ Solving, we have x + 4  6 x  2………..(1) Solving, we have 2x  24 x  12………..(2) Therefore, the required solution is x > 2.

10. P. 10 Example 24.5T Solution: Solve or, and represent the solution on the number line. 24.1 Compound Linear Inequalities in One Unknown One Unknown B. Solving Compound Inequalities with ‘or’ Solving, we have Solving, we have Therefore, the required solution is x  –17 or x  –11.

11. P. 11 Consider the graph of a quadratic function y  ax 2 + bx + c, where a > 0. 24.2 Quadratic Inequalities in One Unknown Unknown If the graph meets the x-axis at x  p and x  q, the x-axis can be divided into three intervals: A. Solving Quadratic Inequalities in One Unknown by the Graphical Method Unknown by the Graphical Method Then we can solve the inequality by considering the value of y in each interval. For example: (i)ax 2 + bx + c > 0, where a > 0 : The solution is x q. (ii)ax 2 + bx + c 0 : The solution is p < x < q. (iii)ax 2 + bx + c  0, where a > 0 : The solution is x  p or x  q. (iv)ax 2 + bx + c  0, where a > 0 : The solution is p  x  q.

12. P. 12 Example 24.6T Solution: Solve the inequality 6x – x 2  0 graphically and represent the solution on the number line. 6x – x 2  0 x 2 – 6x  0 x(x – 6)  0 24.2 Quadratic Inequalities in One Unknown Unknown A. Solving Quadratic Inequalities in One Unknown by the Graphical Method Unknown by the Graphical Method  The solution of the inequality is 0  x  6.

13. P. 13 In addition to using the graphical method, a quadratic inequality can also be solved by the algebraic method. Before solving, it is necessary to factorize the quadratic equation to get the root(s) or the boundary point(s). 1. If ab > 0, then either a > 0 and b > 0 or a < 0 and b < 0. That is, the product of two numbers that have the same sign is positive. (1)Sign Testing Method The basic multiplication sign rules state that: 24.2 Quadratic Inequalities in One Unknown Unknown B. Solving by the Algebraic Method 2. If ab 0 and b 0. That is, the product of two numbers that have the same sign is negative.

14. P. 14 Example 24.7T Solution: Solve the following quadratic inequalities algebraically. (a)x 2 – 6x + 8 < 0(b)x(3x – 8)  –4 (a) x 2 – 6x + 8 < 0 (x – 4)(x – 2) < 0 24.2 Quadratic Inequalities in One Unknown Unknown B. Solving by the Algebraic Method x – 4 0 or x – 4 > 0 and x – 2 < 0 x 2 or x > 4 and x < 2 There is no solution for x > 4 and x < 2.  The solution is 2 < x < 4.

15. P. 15 Example 24.7T Solution: (b) x(3x – 8)  –4 3x 2 – 8x + 4  0 24.2 Quadratic Inequalities in One Unknown Unknown B. Solving by the Algebraic Method x – 2  0 and 3x – 2  0 or x – 2  0 and 3x – 2  0 x  2 and x  or x  2 and x   The solution is x  or x  2. (x – 2)(3x – 2)  0 Solve the following quadratic inequalities algebraically. (a)x 2 – 6x + 8 < 0(b)x(3x – 8)  –4

16. P. 16 (2)Test Value Method To solve a quadratic inequality by the test value method, Remarks: We can choose any values of x in an interval. Remember that we are only interested in the sign (positive or negative) that we get after substituting the value x into the quadratic expression. 24.2 Quadratic Inequalities in One Unknown Unknown B. Solving by the Algebraic Method Boundary numbers are found by replacing the inequality symbol in the quadratic inequality with the sign ‘  ’, and then solving the equation.  first find the boundary numbers (the roots) and divide the number line into intervals by the boundary numbers,  then take a value of x from each interval and check whether it satisfies the inequality or not.

17. P. 17 Example 24.8T Solution: Solve the inequality by the test value method and represent the solution on the number line. When x   2, (  2  4)[1(  2) + 3]  6 > 0. 24.2 Quadratic Inequalities in One Unknown Unknown B. Solving by the Algebraic Method When x  0, (0  4)[1(0) + 3]   12 < 0. When x  5, (5  4)[1(5) + 3]  13 > 0.  The solution is or x  4. –2 0 5

18. P. 18 Example 24.9T Solution: If the equation 2x 2 + k(x – 1) + 4(1 – x)  0 has two distinct real roots, find the range of values of k. 2x 2 + k(x – 1) + 4(1 – x)  0 24.2 Quadratic Inequalities in One Unknown Unknown B. Solving by the Algebraic Method 2x 2 + kx – k + 4 – 4x  0 2x 2 + (k – 4)x + (4 – k)  0 The discriminant of the equation  (k – 4) 2 – 4(2)(4 – k)  k 2 – 8k + 16 – 32 + 8k  k 2 – 16 Since the equation has two distinct real roots, we have  > 0, k 2 – 16> 0 (k + 4)(k – 4)> 0  k 4

19. P. 19 24.3 Linear Inequalities in Two Unknowns A. Linear Inequalities in Two Unknowns Linear inequalities in two unknowns are inequalities which can be represented in the form Ax + By + C > 0 or Ax + By + C  0, where A, B and C are real numbers, with A and B not both equal to zero. Consider the graph of the straight line x + y  1. The straight line divides the coordinate plane into two parts, one above the line and the other below the line. Each region is called a half-plane. The region above the line is called the upper half-plane, and that below the line is called the lower half-plane. The straight line itself is called the boundary line of each half-plane.

20. P. 20 A. Linear Inequalities in Two Unknowns A dashed line is used to indicate that all points on the line do not satisfy the inequality x + y > 1. It means that all the points on the line are not included in the solution of the inequality x + y > 1. If we want to represent the solution of the inequality x + y  1, we have to draw a solid line. In this case, the points on the line x + y  1 are also included in the solution of x + y  1. 24.3 Linear Inequalities in Two Unknowns Every point in the upper half-plane satisfies the inequality x + y > 1. Conversely, every point in the lower half-plane satisfies the inequality x + y < 1.

21. P. 21 A. Linear Inequalities in Two Unknowns We can summarize the steps of finding the solution of a linear inequality in two unknowns as follows: 1.Draw the line for the corresponding equality first. Use a solid line if the straight line is a part of the solution and a dashed line if the line is not a part of the solution. 2.Choose a point not on the line and substitute its coordinates into the inequality. Check whether the point satisfies the inequality. 3.The solution of the inequality is (a)the half-plane containing the test point if the inequality is satisfied by the coordinates of the point; or (b)the half-plane not containing the test point if the inequality is not satisfied by the coordinates of the point. 24.3 Linear Inequalities in Two Unknowns

22. P. 22 Example 24.10T Solution: Solve 2x + y < 10 graphically. A. Linear Inequalities in Two Unknowns 24.3 Linear Inequalities in Two Unknowns x01234 y108642 Step 1:Plot the graph of 2x + y  10 with a dashed line. Step 2:Choose (0, 0) as a test point and substitute its coordinates into 2x + y (the L.H.S. of the inequality): 2(0) + 0  0 < R.H.S. which satisfies the given inequality. Step 3:Since the test point satisfies the inequality, the graph of 2x + y < 10 is the half-plane containing (0, 0).

23. P. 23 B. System of Linear Inequalities in Two Unknowns Unknowns We have just studied the method to represent the solution of a linear inequality in two unknowns graphically. 24.3 Linear Inequalities in Two Unknowns We can use the same technique to solve two or more linear inequalities in two unknowns graphically.

24. P. 24 B. System of Linear Inequalities in Two Unknowns Unknowns Combining the two graphs of the inequalities, the overlapping region is the solution of the system of inequalities. Remarks: Alternatively, we can use arrows to indicate the solution of each inequality. 24.3 Linear Inequalities in Two Unknowns For example, to solve the system of inequalities : Step 1: Consider the graph of x + y  8, and choose (0, 0) as the test point. ∵ 0 + 0  0 < 8 ∴ The lower half-plane represents the solution of x + y  8. Step 2: Consider the graph of x – y  6 and (0, 0). ∵ 0 + 0  0 < 6 ∴ The upper half-plane represents the solution of x – y  6.

25. P. 25 Example 24.11T Solution: (a)Plot the straight lines y – x  0 and x + y + 2  0 on the same coordinate plane. (b)Hence solve the system of inequalities. (a)For y  x  0, we have For x + y + 2  0, we have (b)The shaded region represents the solution of the system of inequalities. x 33 22 023 y 33 22 023 x 33 22 023 y10 22 44 55 B. System of Linear Inequalities in Two Unknowns Unknowns 24.3 Linear Inequalities in Two Unknowns

26. P. 26 Example 24.12T Solution: (a)Solve the system of inequalities graphically. (b)How many integral pairs of (x, y) satisfy the above system of inequalities? (a)For 2x + y  12, B. System of Linear Inequalities in Two Unknowns Unknowns x3456 y6420 For x  y  10, x46810 y 66 44 22 0 24.3 Linear Inequalities in Two Unknowns The shaded region represents the solution of the system of inequalities. (b)44 pairs

27. P. 27 A linear function in two variables x and y is a function in the form C  ax + by, where a and b are constants. 24.4 Linear Programming Usually, we want to find the optimum values (maximum or minimum) of a linear function C, where x and y are under some restrictions. A. Optimum Values of a Linear Function in Two Variables Two Variables For each pair of x and y, there is a corresponding value of the function. Consider a linear function C  x + 2y. For different values of C, the graphs of the linear function C  x + 2y are a group of parallel lines and the value of C increases as the line shifts to the right. x + 2y = 2 x + 2y = 4 x + 2y = 6

28. P. 28 24.4 Linear Programming A. Optimum Values of a Linear Function in Two Variables Two Variables The above results demonstrate a fundamental concept of a new branch of mathematics, which is called linear programming. Linear programming is the method of finding the optimum values of a linear function under some restrictions, which are called constraints. The linear functions we want to optimize are called the objective functions.

29. P. 29 There are two methods to find the optimum values of a linear function under given constraints. 24.4 Linear Programming B. Methods of Finding the Optimum Values of a Linear Function a Linear Function Method I: Method of Sliding Line  First draw the line x + y  0 which is parallel to the graph of the objective function C  x + y. Then move the line in a parallel direction to obtain the optimum values of C.  The value of C increases as the line shifts to the right, so the maximum value of C will be obtained at the extreme right vertex of the solution region, while the minimum value of C will be obtained at the extreme left vertex of the solution region.  From the graph, C attains its maximum and minimum values at R(8, –1) and Q(0, –1) respectively. Maximum value of C  8 + (–1)  7 Minimum value of C  0 + (–1)  –1

30. P. 30 24.4 Linear Programming B. Methods of Finding the Optimum Values of a Linear Function a Linear Function Method II: Method of Testing Vertices When solving problems of linear programming, the objective function always attains its maximum and minimum values at the vertices of the shaded region. We can find the optimum values of C by substituting the coordinates of all the vertices into the objective function. At P(2, 3), C  2 + 3  5 At Q(0, –1), C  0 + (–1)  –1  Maximum value of C  7 minimum value of C  –1 At R(8, –1), C  8 + (–1)  7 Sometimes the objective function attains its optimum value at more than one point.

31. P. 31 Example 24.13T Solution: Find the maximum and minimum values of the linear function C  x – 2y, subject to the constraints: At A(0, 15), C  0  2(15)   30 B. Methods of Finding the Optimum Values of a Linear Function a Linear Function At B(0,  1), C  0  2(  1)  2 At C(12, 3), C  12  2(3)  6  Maximum value minimum value 24.4 Linear Programming Solve the system of inequalities graphically. By the method of testing vertices,

32. P. 32 Example 24.14T Solution: Find the maximum and minimum values of the linear function C  x – 2y, where x and y are integers, subject to the constraints x + y  2, 3y – 2x  10 and y  4x – 19. Using the method of sliding line, P attains its maximum value at (4,  2) and minimum value at (4, 6) and (6, 7).  Maximum value of C  4 – 2(–2) B. Methods of Finding the Optimum Values of a Linear Function a Linear Function Minimum value of C  6 – 2(7) 24.4 Linear Programming Solve the system of inequalities graphically.

33. P. 33 24.5 Applications of Linear Programming There are many applications of linear programming in daily life. To solve a real-life problem of linear programming, we can follow the steps below. Step 1:Identify the variables x and y. Step 2:Set up the system of inequalities for the constraints. Step 3:Solve the system of inequalities graphically and shade the region of feasible solution. Step 4:Write down the objective function C. Step 5:Find the optimal solution of the objective function by the methods we learnt in the previous section. People can use it to allocate resources (such as capital, materials, labour force, land, etc.) in order to achieve the best performance.

34. P. 34 Example 24.15T The table below shows the ingredients per unit of product A and product B. Suppose there are 60 units of chemical R and 25 units of chemical S. Let x units and y units be the amounts of product A and product B that are produced respectively. (a)Set up the system of constraints in terms of x and y. (b)Draw and shade the region of feasible solution that represents the constraints in (a). (c)If the profit on selling product A is $100 and the profit on selling product B is $120, find the maximum profit. Chemical RChemical S Product A2 units1 unit Product B3 units1 unit 24.5 Applications of Linear Programming (a) The constraints are: Solution:

35. P. 35 Example 24.15T Solution: Let x units and y units be the amounts of product A and product B that are produced respectively. (a) Set up the system of constraints in terms of x and y. (b) Draw and shade the region of feasible solution that represents the constraints in (a). (c) If the profit on selling product A is $100 and the profit on selling product B is $120, find the maximum profit. (b) Refer to the figure. (c) The objective function is P  100x + 120y. Draw a straight line 100x + 120y  0. By the method of sliding line, we observe P attains its maximum value at (15, 10).  The maximum profit  $[100(15) + 120(10)] 24.5 Applications of Linear Programming

36. P. 36 Example 24.16T Solution: In a candy shop, there are two packets of candies. Packet A contains 5 pieces of milk candies and 20 pieces of fruit candies. Packet B contains 25 pieces of milk candies and 25 pieces of fruit candies. Peter wants to buy at least 140 pieces of milk candies and 375 pieces of fruit candies. If the prices of packets A and B are $15 and $20 respectively, find the minimum cost that Peter needs to pay. Let x be the number of packets A and y be the number of packets B. The constraints are: The objective function is C  15x + 20y. 24.5 Applications of Linear Programming

37. P. 37 Example 24.16T Solution: In a candy shop, there are two packets of candies. Packet A contains 5 pieces of milk candies and 20 pieces of fruit candies. Packet B contains 25 pieces of milk candies and 25 pieces of fruit candies. Peter wants to buy at least 140 pieces of milk candies and 375 pieces of fruit candies. If the prices of packets A and B are $15 and $20 respectively, find the minimum cost that Peter needs to pay. As x, y are non-negative integers, C attains its minimum value at (15, 3).  The minimum cost  $[15(15) + 20(3)] 24.5 Applications of Linear Programming In the figure, we draw a line 15x + 20y  0 and move it to the right in a parallel direction.

38. P. 38 Example 24.17T Solution: David buys x toy cars and y toy robots from a factory. The cost of a car and a robot is $35 and $15 respectively and David’s budget is $900. He uses a plastic bag to carry the toys. The capacity of the bag is 50 items. The weight of a toy car and a toy robot are 300 g and 550 g respectively and David can carry a load of 20 kg. If the profit from selling a car and a robot is $15 and $10 respectively, find the maximum profit he can make and the number of toy cars and robots he should buy. The constraints are: The objective function is P  15x + 10y. 24.5 Applications of Linear Programming

39. P. 39 Example 24.17T Solution: David buys x toy cars and y toy robots from a factory. The cost of a car and a robot is $35 and $15 respectively and David’s budget is $900. He uses a plastic bag to carry the toys. The capacity of the bag is 50 items. The weight of a toy car and a toy robot are 300 g and 550 g respectively and David can carry a load of 20 kg. If the profit from selling a car and a robot is $15 and $10 respectively, find the maximum profit he can make and the number of toy cars and robots he should buy. Draw a line 15x + 10y  0 and move it to the right in a parallel direction. We observe that P attains its maximum value at (13, 29).  The maximum profit Number of toy cars  13; number of robots  29. 24.5 Applications of Linear Programming

40. P. 40 24.1 Compound Linear Inequalities in One Unknown Compound inequalities that contain the word ‘and’ are the set of values which satisfy both inequalities. Chapter Summary Compound inequalities that contain the word ‘or’ are the set of values which satisfy one or both of the inequalities.

41. P. 41 A quadratic inequality can be solved by the: Chapter Summary 24.2 Quadratic Inequalities in One Unknown 1.graphical method 2.algebraic method (a)Sign testing method (b)Test value method

42. P. 42 Chapter Summary 24.3 Linear Inequalities in Two Unknowns 1. Linear inequalities in two unknowns can be represented on the coordinate plane. 2. The line y  mx + c divides the plane into two parts, the upper half-plane and the lower half-plane. (a)All the points in the upper half-plane satisfy the inequality y > mx + c. (b)All the points in the lower half-plane satisfy the inequality y < mx + c. 3. The plane which satisfies the inequality is called the solution region. 4. The common region of all the inequalities represents the feasible solution of the system of inequalities. Notes:The straight line y  mx + c is included as part of the solution for inequalities with signs ‘  ’ or ‘  ’.

43. P. 43 Chapter Summary 24.4 Linear Programming Linear programming is a method for finding the optimum values of a linear function under given constraints.

44. P. 44 Chapter Summary 24.5 Applications of Linear Programming Steps in solving problems of linear programming: (i)Identify the decision variables x and y. (ii)Set up the system of inequalities for the constraints. (iii)Set up the system of inequalities for the constraints. (iv)Solve the system of inequalities graphically and shade the region of feasible solution. (v)Write down the objective function C. (vi)Find the optimum solution of the objective function.

45. Follow-up 24.1 Solution: Solve 2x – 1 < 9 and 2(x – 3) < 12, and represent the solution on the number line. Solving 2x – 1  9, we have 2x <  10 x  5………(1) Solving 2(x – 3) < 12, we have x – 3 < 6 x  ………(2) Combining (1) and (2), we have Therefore, the required solution is x  5. 24.1 Compound Linear Inequalities in One Unknown One Unknown A. Solving Compound Inequalities with ‘and’

46. Follow-up 24.2 (a)Solve the compound inequality 2(x – 3) – 5(x – 1)  11 and (b) Represent the solution in (a) on the number line. Solution: (a)Solving 2(x – 3) – 5(x – 1)  11, we have 2x – 6 – 5x + 5  11 x  –4……..(1) Solving, we have x  …..(2) Combining (1) and (2), we have –4  x  2. (b) –3x – 1  11 –3x  12x + 10  12 24.1 Compound Linear Inequalities in One Unknown One Unknown A. Solving Compound Inequalities with ‘and’

47. Follow-up 24.3 There are some $50 notes and $100 notes in a purse. The number of $50 notes is three times that of the $100 notes. If the total value of the notes is at most 1000 dollars and is greater than 500 dollars, find the possible numbers of $100 notes. Solution: Let x be the number of $100 notes, then the number of $50 notes is 3x. we have 500 < 50(3x) + 100x  1000 The inequality can be rewritten as 500 < 150x + 100x and 150x + 100x  1000 500 < 250x and 250x  1000 2 < x and x  4 Combining the above results, we have: Therefore, the possible numbers of $100 notes are 3 or 4. 24.1 Compound Linear Inequalities in One Unknown One Unknown A. Solving Compound Inequalities with ‘and’

48. Follow-up 24.4 Solution: Solve 5(x + 1)  2x – 1 or –2x  x – 12, and represent the solution on the number line. Solving 5(x + 1)  2x – 1, we have 5x + 5  2x – 1 x  –2………(1) Solving –2x  x – 12, we have –2x – x  –12 x  4………..(2) Therefore, the required solution is x  4. 5x – 2x  –1 – 5 3x  –6 –3x  –12 24.1 Compound Linear Inequalities in One Unknown One Unknown B. Solving Compound Inequalities with ‘or’

49. Follow-up 24.5 Solve or, and represent the solution on the number line. Solution: Solving, we have 8x – 9  15 x <  3……..(1) Solving we have Therefore, the required solution is x < 8. 8x  24 3(x – 2)  2(x + 1) x  8……..(2) 3x – 6  2x + 2 3x – 2x  6 + 2 24.1 Compound Linear Inequalities in One Unknown One Unknown B. Solving Compound Inequalities with ‘or’

50. Follow-up 24.6 Solve the inequality 15  x 2 – 2x graphically and represent the solution on the number line. Solution: 15  x 2 – 2x x 2 – 2x – 15  0 (x – 5)(x + 3)  0  The solution of the inequality is x  –3 or x  5. 24.2 Quadratic Inequalities in One Unknown Unknown A. Solving Quadratic Inequalities in One Unknown by the Graphical Method Unknown by the Graphical Method

51. Follow-up 24.7 Solve the following quadratic inequalities algebraically. (a)x 2 + 19x + 34 > 0(b)2 – 3x 2  5x Solution: (a) x 2 + 19x + 34 > 0 (x + 2)(x + 17) > 0 x + 2 > 0 and x + 17 > 0 or x + 2 < 0 and x + 17 < 0 x > –2 and x > –17 or x < –2 and x < –17  The solution is x –2. 24.2 Quadratic Inequalities in One Unknown Unknown B. Solving by the Algebraic Method

52. Follow-up 24.7 Solve the following quadratic inequalities algebraically. (a)x 2 + 19x + 34 > 0(b)2 – 3x 2  5x Solution: (b) 2 – 3x 2  5x 3x 2 + 5x – 2  0 3x – 1  0 and x + 2  0 or 3x – 1  0 and x + 2  0 There is no solution for and x  –2. x  and x  –2 or x  and x  –2  The solution is. 24.2 Quadratic Inequalities in One Unknown Unknown B. Solving by the Algebraic Method (3x – 1)(x + 2)  0

53. Follow-up 24.8 Solve the inequality x(3x + 5)  2 by the test value method and represent the solution on the number line. Solution: x(3x + 5)  2 3x 2 + 5x – 2  0 When x   3, [3(  3)  1](  3 + 2)  10 > 0. (3x – 1)(x + 2)  0 When x  0, [3(0)  1](0 + 2)   2 < 0. When x  1, [3(1)  1](1 + 2)  6 > 0.  The solution is x  –2 or. 24.2 Quadratic Inequalities in One Unknown Unknown B. Solving by the Algebraic Method

54. Follow-up 24.9 Given an equation x 2 + 2kx – x + 4  0. If the equation does not have any real root, find the range of values of k. Solution: x 2 + 2kx – x + 4  0 x 2 + (2k – 1)x + 4  0 The discriminant of the equation  (2k – 1) 2 – 4(1)(4)  4k 2 – 4k + 1 – 16  4k 2 – 4k – 15 Since the equation does not have any real root, we have  < 0, 4k 2 – 4k – 15< 0 (2k + 3)(2k – 5)< 0  –1.5 < k < 2.5 24.2 Quadratic Inequalities in One Unknown Unknown B. Solving by the Algebraic Method

55. Solve x – y  5 graphically. Solution: Follow-up 24.10 A. Linear Inequalities in Two Unknowns 24.3 Linear Inequalities in Two Unknowns x–20246 y–7–5–3–11 Step 1:Plot the graph of x – y  5 with a solid line. Step 2:Choose (0, 0) as a test point and substitute its coordinates into x – y (the L.H.S. of the inequality): 0 – 0  0 < R.H.S. which does NOT satisfy the given inequality. Step 3:Since the test point does NOT satisfy the inequality, the graph of x – y  5 is the half-plane NOT containing (0, 0).

56. (a)Plot the straight lines x + y  6 and 2x – y  –6 on the same coordinate plane. (b)Hence solve the system of inequalities. Solution: (a)For x + y  6, we have Follow-up 24.11 x 22 0246 y86420 For 2x – y  –6, we have x 44 22 024 y 22 261014 (b) The shaded region represents the solution of the system of inequalities. B. System of Linear Inequalities in Two Unknowns Unknowns 24.3 Linear Inequalities in Two Unknowns

57. Follow-up 24.12 Solution: B. System of Linear Inequalities in Two Unknowns Unknowns 24.3 Linear Inequalities in Two Unknowns (a)Solve the system of inequalities graphically. (b)Find the point(s) of (x, y), where x and y are integers, which satisfies/satisfy the above system of inequalities. (a)For x  y  6, x0246 y 66 44 22 0 For x  2y  8, x0286 y 44 33 22 11 The shaded region represents the solution of the system of inequalities. (b)The possible point is (7, 0) only.

58. Follow-up 24.13 Find the maximum and minimum values of the linear function C  x + y subject to the constraints. Solution: B. Methods of Finding the Optimum Values of a Linear Function a Linear Function 24.4 Linear Programming At P(0, 10), C  0 + 10  10 At Q(0,  8), C  0 + (  8)   8 At R(6,  2), C  6 + (  2)  4  Maximum value minimum value Solve the system of inequalities graphically. By the method of testing vertices,

59. Follow-up 24.14 (a)Find the minimum value of the linear function P  y – 2x subject to the constraints. (b)Repeat (a) if x and y are integers. Solution: (a)By the method of sliding lines, P is minimum at (1.5,  8).  Minimum value   8  2(1.5) (b)If x and y are integers, then P is minimum at (1,  7).  Minimum value   7  2(1) B. Methods of Finding the Optimum Values of a Linear Function a Linear Function 24.4 Linear Programming Solve the system of inequalities graphically.

60. Follow-up 24.15 A factory produces two types of television sets. The standard set requires two hours for assembly and one hour for testing and packaging. The deluxe set requires three hours for assembly and four hours for testing and packaging. There are a total of 150 hours and 160 hours available for assembly, testing and packaging process respectively. The profit on each standard set is $150 and the profit on each deluxe set is $250. (a)Let x be the number of standard sets and y be the number of deluxe sets produced. Set up the system of constraints in terms of x and y. (b)Draw and shade the region of feasible solution that represents the constraints in (a). (c)Set up an objective function P and find the maximum profit. 24.5 Applications of Linear Programming Solution: (a) The constraints are:

61. Follow-up 24.15 Solution: (b) Refer to the figure. (c) The objective function is P  150x + 250y. Draw a straight line 150x + 250y  0. By the method of sliding line, we observe P attains its maximum value at (24, 34).  The maximum profit  $(150  24 + 250  34) 24.5 Applications of Linear Programming The profit on each standard set is $150 and that on each deluxe set is $250. (a)Let x be the number of standard sets and y be the number of deluxe sets produced. Set up the system of constraints in terms of x and y. (b)Draw and shade the region of feasible solution that represents the constraints in (a). (c)Set up an objective function P and find the maximum profit.

62. Follow-up 24.16 The owner of an amusement park wants to install two types of new machines. Each small machine takes up a floor area of 4 m 2 and consumes 120 units of electricity per day. Each large machine takes up a floor area of 6 m 2 and consumes 480 units of electricity per day. It is expected that the total area used should be at least 54 m 2 and the total electricity consumption should be at least 4000 units per day. If the running cost of each small machine and large machine are $1000 and $2500 per day respectively, find the minimum cost of operating the machines each day. Solution: Let x be the number of small machines and y be the number of large machines. The constraints are: 24.5 Applications of Linear Programming The objective function is C  1000x + 2500y.

63. Follow-up 24.16 In the figure, we draw a line 2x + 5y  0 and move it to the right in a parallel direction such that C increases. As x, y are non-negative integers, C attains its minimum value at (2, 8).  The minimum cost  $[1000(2) + 2500(8)] The owner of an amusement park wants to install two types of new machines. Each small machine takes up a floor area of 4 m 2 and consumes 120 units of electricity per day. Each large machine takes up a floor area of 6 m 2 and consumes 480 units of electricity per day. It is expected that the total area used should be at least 54 m 2 and the total electricity consumption should be at least 4000 units per day. If the running cost of each small machine and large machine are $1000 and $2500 per day respectively, find the minimum cost of operating the machines each day. Solution: 24.5 Applications of Linear Programming

64. Follow-up 24.17 A company decides to employ x full-time workers and y part-time workers. Each full-time worker earns $300 per day and each part-time worker earns $200 per day. The company cannot afford to pay more than $15 000 a day. Also, the total number of workers should not exceed 70 and the number of part-time workers should not exceed the number of full-time workers by more than 10. (a)Set up the system of constraints in terms of x and y. (b)Draw and shade the feasible region that represents the constraints in (a). (c)If a full-time worker generates a profit of $2000 per day and a part-time worker generates a profit of $1500 per day, find the maximum profit of the company. 24.5 Applications of Linear Programming (a) The constraints are: Solution:

65. Follow-up 24.17 (b) Refer to the figure. (c) The objective function is P  2000x + 1500y.  P is maximum at (26, 36). Maximum profit  $[2000(26) + 1500(36)] 24.5 Applications of Linear Programming (a)Set up the system of constraints in terms of x and y. (b)Draw and shade the feasible region that represents the constraints in (a). (c)If a full-time worker generates a profit of $2000 per day and a part-time worker generates a profit of $1500 per day, find the maximum profit of the company. Solution: