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Lesson 3-3 Ideas/Vocabulary Solve systems of inequalities by graphing. system of inequalities Determine the coordinates of the vertices of a region formed by the graph of a system of inequalities.

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Lesson 3-3 Example 1 A. Solve the system of inequalities by graphing. y 2x – 3 y < –x + 2 Intersecting Regions Solution of y 2x – 3 Regions 1 and 2 Solution of y < –x + 2 Regions 2 and 3

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Lesson 3-3 Example 1 Intersecting Regions Answer: The intersection of these regions is Region 2, which is the solution of the system of inequalities. Notice that the solution is a region containing an infinite number of ordered pairs. A.

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Lesson 3-3 Example 1 B. Solve the system of inequalities by graphing. y –x + 1 x + 1< 3 The inequality x + 1 –3, or x –4. Intersecting Regions

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Lesson 3-3 Example 1 Answer: Graph all of the inequalities on the same coordinate plane and shade the region or regions that are common to all. Intersecting Regions Animation: Use Graphing to Solve the System of Inequalities

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1.A 2.B 3.C 4.D Lesson 3-3 CYP 1 A. Solve the system of inequalities by graphing. What is the solution to the system of inequalities below? y 3x – 3 y > x + 1 A. B. C.D.

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1.A 2.B 3.C 4.D Lesson 3-3 CYP 1 B. Solve the system of inequalities by graphing. What is the solution to the system of inequalities below? y –2x – 3x + 2< 1 A. B. C.D.

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Lesson 3-3 Example 2 Graph both inequalities. Answer: The solution set is Ø. Separate Regions The graphs do not overlap, so the solutions have no points in common. Solve the system of inequalities by graphing.

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1.A 2.B 3.C 4.D Lesson 3-3 CYP 2 Solve the system of inequalities by graphing. A. B. C.D.

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Lesson 3-3 Example 3 MEDICINE Medical professionals recommend that patients have a cholesterol level below 200 milligrams per deciliter (mg/dL) of blood and a triglyceride level below 150 mg/dL. Write and graph a system of inequalities that represents the range of cholesterol levels and triglyceride levels for patients. Let c represent the cholesterol levels in mg/dL. It must be less than 200 mg/dL. Since cholesterol levels cannot be negative, we can write this as 0 c < 200. Let t represent the triglyceride levels in mg/dL. It must be less than 150 mg/dL. Since triglyceride levels also cannot be negative, we can write this as 0 t < 150. Write and Use a System of Inequalities

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Lesson 3-3 Example 3 Graph all of the inequalities. Any ordered pair in the intersection of the graphs is a solution of the system. Answer:0 c < t < 150 Write and Use a System of Inequalities

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Lesson 3-3 CYP 3 SAFETY The speed limits while driving on the highway are different for trucks and cars. Cars must drive between 45 and 65 miles per hour, inclusive. Trucks are required to drive between 40 and 55 miles per hour, inclusive. Let c represent the range of speeds for cars and t represent the range of speeds for trucks.

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1.A 2.B 3.C 4.D Lesson 3-3 CYP 3 Which graph represents this? A. B. C.D.

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Lesson 3-3 Example 4 Find the coordinates of the vertices of the figure formed by 2x – y –1, x + y 4, and x + 4y 4. Find Vertices Graph each inequality. The intersection of the graphs forms a triangle. Answer:The vertices of the triangle are at (0, 1), (4, 0), and (1, 3).

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Lesson 3-3 CYP 4 A. A B. B C. C D. D A.(–1, 0), (0, 3), and (5, –2) B.(–1, 0), (0, 3), and (4, –2) C.(–1, 1), (0, 3), and (5, –2) D.(0, 3), (5, –2), and (1, 0) Find the coordinates of the vertices of the figure formed by x + 2y 1, x + y 3, and –2x + y 3.

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End of Lesson 3-3

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Lesson 3-4 Menu Five-Minute CheckFive-Minute Check (over Lesson 3-3) Main Ideas and Vocabulary Example 1: Bounded Region Example 2: Unbounded Region Key Concept: Linear Programming Example 3: Real-World Example: Linear Programming

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constraints feasible region bounded vertex unbounded linear programming Lesson 3-4 Ideas/Vocabulary Find the maximum and minimum values of a function over a region. Solve real-world problems using linear programming.

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Lesson 3-4 Example 1 Graph the following system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the function f(x, y) = 3x – 2y for this region. x 5 y 4 x + y 2 Bounded Region Step 1 Graph the inequalities. The polygon formed is a triangle with vertices at (–2, 4), (5, –3), and (5, 4).

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Lesson 3-4 Example 1 Bounded Region Answer: The vertices of the feasible region are (–2, 4), (5, –3), and (5, 4). The maximum value is 21 at (5, –3). The minimum value is –14 at (–2, 4). Step 2 Use a table to find the maximum and minimum values of f(x, y). Substitute the coordinates of the vertices into the function. minimum maximum

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Lesson 3-4 CYP 1 A. A B. B C. C D. D A.maximum: f(4, 5) = 5 minimum: f(1, 5) = –11 B.maximum: f(4, 2) = 10 minimum: f(1, 5) = –11 C.maximum: f(4, 2) = 10 minimum: f(4, 5) = 5 D.maximum: f(1, 5) = –11 minimum: f(4, 2) = 10 Graph the following system of inequalities. What are the maximum and minimum values of the function f(x, y) = 4x – 3y for the feasible region of the graph? x 4y 5 x + y 6

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Lesson 3-4 Example 2 Graph the following system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the function f(x, y) = 2x + 3y for this region. –x + 2y 2 x – 2y 4 x + y –2 Graph the system of inequalities. There are only two points of intersection, (–2, 0) and (0, –2). Unbounded Region

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Lesson 3-4 Example 2 The minimum value is –6 at (0, –2). Although f(–2, 0) is –4, it is not the maximum value since there are other points that produce greater values. For example, f(2,1) is 7 and f(3, 1) is 10. It appears that because the region is unbounded, f(x, y) has no maximum value. Answer:The vertices are at (–2, 0) and (0, –2). There is no maximum value. The minimum value is –6 at (0, –2). Unbounded Region

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Lesson 3-4 CYP 2 A. A B. B C. C D. D Graph the following system of inequalities. What are the maximum and minimum values of the function f(x, y) = x + 2y for the feasible region of the graph? x + 3y 6–x – 3y 9 2y – x –6 A.maximum: no maximum minimum: f(6, 0) = 6 B.maximum: f(6, 0) = 6 minimum: f(0, –3) = –6 C.maximum: f(6, 0) = 6 minimum: no minimum D.maximum: no maximum minimum: f(0, –3) = –6

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Lesson 3-4 Key Concept 1 Linear Programming Procedure Step 1Define the variables. Step 2Write a system of inequalities. Step 3Graph the system of inequalities. Step 4Find the coordinates of the vertices of the feasible region. Step 5Write a function to be maximized or minimized. Step 6Substitute the coordinates of the vertices into the function. Step 7Select the greatest or least result. Answer the problem.

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Lesson 3-4 Example 3 LANDSCAPING A landscaping company has crews who mow lawns and prune shrubbery. The company schedules 1 hour for mowing jobs and 3 hours for pruning jobs. Each crew is scheduled for no more than 2 pruning jobs per day. Each crews schedule is set up for a maximum of 9 hours per day. On the average, the charge for mowing a lawn is $40 and the charge for pruning shrubbery is $120. Find a combination of mowing lawns and pruning shrubs that will maximize the income the company receives per day from one of its crews. Linear Programming

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Step 1Define the variables. Lesson 3-4 Example 3 Linear Programming m = the number of mowing jobs p = the number of pruning jobs

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Lesson 3-4 Example 3 Linear Programming Step 2Write a system of inequalities. Since the number of jobs cannot be negative, m and p must be nonnegative numbers. m 0, p 0 Mowing jobs take 1 hour. Pruning jobs take 3 hours. There are 9 hours to do the jobs. 1m + 3p 9 There are no more than 2 pruning jobs a day. p 2

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Lesson 3-4 Example 3 Linear Programming Step 3Graph the system of inequalities.

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Lesson 3-4 Example 3 Linear Programming Step 4Find the coordinates of the vertices of the feasible region. From the graph, the vertices are at (0, 2), (3, 2), (9, 0), and (0, 0). Step 5Write the function to be maximized. The function that describes the income is f(m, p) = 40m + 120p. We want to find the maximum value for this function.

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Lesson 3-4 Example 3 Linear Programming Step 6Substitute the coordinates of the vertices into the function. Step 7Select the greatest amount.

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Lesson 3-4 Example 3 Linear Programming Answer:The maximum values are 360 at (3, 2) and 360 at (9, 0). This means that the company receives the most money with 3 mowings and 2 prunings or 9 mowings and 0 prunings. Animation: Linear Programming

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Lesson 3-4 CYP 3 LANDSCAPING A landscaping company has crews who rake leaves and mulch. The company schedules 2 hours for mulching jobs and 4 hours for raking jobs. Each crew is scheduled for no more than 2 raking jobs per day. Each crews schedule is set up for a maximum of 8 hours per day. On the average, the charge for raking a lawn is $50 and the charge for mulching is $30.

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Lesson 3-4 CYP 2 A. A B. B C. C D. D What is a combination of raking leaves and mulching that will maximize the income the company receives per day from one of its crews? A.0 mulching; 2 raking B.4 mulching; 0 raking C.0 mulching; 4 raking D.2 mulching; 0 raking

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End of Lesson 3-4

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CIM 1

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CIM 2

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(over Chapter 2) 5 Min 1-1 A. A B. B C. C D. D Find the domain (D) and range (R) of the relation {(–4, 1), (0, 0), (1, –4), (2, 0), (–2, 0)}. Determine whether the relation is a function. A.D = {–4, –2, 0, 1, 2}, R = {–4, 0, 1}; no B.D = {–4, 0, 1}, R = {–4, –2, 0, 1, 2}; no C.D = {–4, –2, 0, 1, 2}, R = {–4, 0, 1}; yes D.D = {–4, 0, 1}, R = {–4, –2, 0, 1, 2}; yes

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(over Chapter 2) 5 Min 1-2 A. A B. B C. C D. D Find the value of f(4) for f(x) = 8 – x – x 2. A.20 B.12 C.–4 D.–12

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(over Chapter 2) 5 Min 1-3 A. A B. B C. C D. D A. B. C. D. Find the slope of the line that passes through (5, 7) and (–1, 0).

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(over Chapter 2) 5 Min 1-4 A. A B. B C. C D. D Write an equation in slope-intercept form for the line that has x-intercept –3 and y-intercept 6. A.y = 2x + 6 B.y = 3x – 6 C.y = –3x + 6 D.y = x + 3

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A. A B. B C. C D. D (over Chapter 2) 5 Min 1-5 A.12 B.15 C.16 D.19 The Math Club is using the prediction equation y = 1.25x + 10 to estimate the number of members it will have, where x represents the number of years the club has been in existence. About how many members does the club expect to have in its fifth year?

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(over Chapter 2) 5 Min 1-6 A. A B. B C. C D. D A.f(x) = x B.f(x) = [x] C.f(x) = |x| –1 D. f(x) = – |x| Which function has the greatest value for f(–2)?

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1.A 2.B 3.C 4.D 5 Min 2-1 Which choice shows a graph of the solution of this system? y = 3x – 2 y = –3x + 2 (over Lesson 3-1) A.B. C.D.

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5 Min 2-2 A. A B. B C. C D. D (over Lesson 3-1) A.consistent and independent B.consistent and dependent C.inconsistent D.none of the above State whether the graph of the system of equations is consistent and independent, consistent and dependent, or inconsistent? 2x + y = 6 3y = –6x + 6

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5 Min 2-3 A. A B. B C. C D. D Which equation is inconsistent with 4x + 5y = 30? (over Lesson 3-1) A. B.5y = –4x – 30 C. D.5y = 4x – 30

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(over Lesson 3-2) 5 Min 3-1 A. A B. B C. C D. D Solve the system of equations by using substitution. x – 4y = –12 3x + 2y = 20 A.(4, 4) B.(8, 5) C.no solution D.infinitely many solutions

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(over Lesson 3-2) 5 Min 3-2 A. A B. B C. C D. D Solve the system of equations by using substitution. x + 2y = 6 2x + 4y = 15 A.(3, 1.5) B.(3.5, 2) C.no solution D.infinitely many solutions

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(over Lesson 3-2) 5 Min 3-3 A. A B. B C. C D. D Solve the system of equations by using elimination. 2x + 5y = 9 –2x + 8y = 4 A.(–2, 0) B.(2, 1) C.no solution D.infinitely many solutions

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(over Lesson 3-2) 5 Min 3-4 A. A B. B C. C D. D Solve the system of equations by using elimination. x + 2y = 7 14 – 4y = 2x A.(0, 0) B.(–5, –6) C.no solution D.infinitely many solutions

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(over Lesson 3-2) 5 Min 3-5 A. A B. B C. C D. D Two times one number minus a second number is 7. The second number is three more than the first number. What are the two numbers? A.7 and 10 B.10 and 13 C.13 and 16 D.7 and 20

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1.A 2.B 3.C 4.D 5 Min 4-1 Which choice shows a graph of the solution of the system of inequalities? x –2 y > 3 (over Lesson 3-3) A.B. C.D.

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1.A 2.B 3.C 4.D 5 Min 4-2 (over Lesson 3-3) Which choice shows a graph of the solution of the system of inequalities? y 2x + 2 y > –x A.B. C.D.

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(over Lesson 3-3) 5 Min 4-3 A. A B. B C. C D. D Find the coordinates of the vertices of the figure formed by the system of inequalities. x 0 y 0 –3x + y = –6 A.(0, 0), (2, 0), (–6, 0) B.(0, 0), (0, 2), (0, 6) C.(0, 0), (2, 0), (0, –6) D.(0, 0), (0, –2), (0, –6)

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(over Lesson 3-3) 5 Min 4-4 A. A B. B C. C D. D Find the coordinates of the vertices of the figure formed by the system of inequalities. y 3 y 0 x 0 2y + 3x 12 A.(0, 0), (0, 6), (2, 3), (4, 0) B.(0, 0), (3, 0), (3, 2), (4, 0) C.(0, 0), (3, 0), (0, 6), (2, 3) D.(0, 0), (0, 3), (2, 3), (4, 0)

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(over Lesson 3-3) 5 Min 4-5 A. A B. B C. C D. D Which point is not the solution of the system? |x| 2 y 3 A.(0, 4) B.(–3, 5) C.(–2, 5) D.(1, 4)

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1.A 2.B 3.C 4.D 5 Min 5-1 Which choice shows a graph of the system of inequalities and the coordinates of the vertices of the feasible region? 1 x 4 y x y 2x + 3 (over Lesson 3-4) A.B. C.D.

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(over Lesson 3-4) 5 Min 5-2 A. A B. B C. C D. D A.min: f(0, 0) = 0 max: f(5, 5) = –10 B.min: f(0, 0) = 0 max: f(4, 2) = 2 C.min: f(0, 6) = –30 max: f(4, 2) = 2 D.min: f(0, 6) = –30 max: f(5, 5) = –10 The vertices of a feasible region are (0, 0), (0, 6), (4, 2), and (5, 6). What are the maximum and minimum values of the function f(x, y) = 3x – 5y over this region?

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