Presentation is loading. Please wait.

Presentation is loading. Please wait.

Systems of Equations and Inequalities

Similar presentations


Presentation on theme: "Systems of Equations and Inequalities"— Presentation transcript:

1 Systems of Equations and Inequalities
C-4 Linear Programming

2 ACT WARM-UP If x + 3y = 1 and 3x + y = 11, then 3x + 3y = ?
A) 3 B) C) 9 D) 18 E) 36 Using elimination, multiply the first equation by −3 to get −3x −9y = −3. Add to 3x + y = 11 to get −8y = 8, so y = −1. Substitute to get x = 4. 3(4) + 3(−1) = 12 − 3 = 9. The answer is C) 9.

3 Objectives Find the maximum and minimum values of a function over a region Solve real-world problems using linear programming

4 Essential Question What are the Linear Programming Procedures?

5 A system of linear inequalities has many solutions.
Linear programming is a method of finding the best solutions by determining a maximum or minimum value of a function that satisfies a given set of conditions called constraints. A constraint is one of the inequalities in a linear programming problem. The solution to the set of constraints can be graphed as a feasible region. This is located where the graphs intersect. The maximum or minimum value of a related function always occurs at one of the vertices of the feasible region.

6 When the graph of a system of constraints is a closed polygonal region we say that the region is bounded. Sometimes a system of inequalities forms a region that is open on one end. In this case, the region is said to be unbounded. Always test a point contained in the feasible region when the graph is unbounded. Do not assume that there is no minimum value if the feasible region is unbounded below the line, or that there is no maximum value if the feasible region is unbounded above the line.

7 Step 1 Find the vertices of the region. Graph the inequalities.
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 for this region. Step 1 Find the vertices of the region. Graph the inequalities. The polygon formed is a triangle with vertices at (–2, 4), (5, –3), and (5,4). Example 4-1a

8 (x, y) 3x – 2y f(x, y) (–2, 4) 3(– 2) – 2(4) (5, –3) 3(5) – 2(–3) 21
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. (x, y) 3x – 2y f(x, y) (–2, 4) 3(– 2) – 2(4) – 14 (5, –3) 3(5) – 2(–3) 21 (5, 4) 3(5) – 2(4) 7 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). Example 4-1a

9 Graph the following system of inequalities
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 for this region. Answer: vertices: (1, 5), (4, 5) (4, 2); maximum: f(4, 2) = 10, minimum: f(1, 5) = –11 Example 4-1b

10 Graph the following system of inequalities
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 for this region. Graph the system of inequalities. There are only two points of intersection, (–2, 0) and (0, –2). Example 4-2a

11 (x, y) 2x + 3y f(x, y) (–2, 0) 2(–2) + 3(0) –4 (0, –2) 2(0) + 3(–2) –6
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). Example 4-2a

12 Answer: vertices: (0, –3), (6, 0); maximum: f(6, 0) = 6; no minimum
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 for this region. Answer: vertices: (0, –3), (6, 0); maximum: f(6, 0) = 6; no minimum Example 4-2b

13 Real World Problems The process of finding maximum or minimum values of a function for a region defined by inequalities is called linear programming. The steps used to solve a problem using linear programming are listed on the next slide.

14 Linear Programming Procedures
Define the variables Write a system of inequalities Graph the system of inequalities Find the coordinates of the vertices of the feasible region Write a function to be maximized or minimized Substitute the coordinates of the vertices into the function Select the greatest or least result. Answer the problem.

15 m = the number of mowing jobs p = the number of pruning jobs
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 crew’s 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. Step 1 Define the variables. m = the number of mowing jobs p = the number of pruning jobs Example 4-3a

16 m  0, p  0 p  2 Step 2 Write 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. There are no more than 2 pruning jobs a day. p  2 Example 4-3a

17 Step 3 Graph the system of inequalities.
Example 4-3a

18 Step 4 Find 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 5 Write the function to be maximized. The function that describes the income is We want to find the maximum value for this function. Example 4-3a

19 (m, p) 40m + 120p f(m, p) (0, 2) 40(0) + 120(2) 240 (3, 2)
Step 6 Substitute the coordinates of the vertices into the function. (m, p) 40m + 120p f(m, p) (0, 2) 40(0) + 120(2) 240 (3, 2) 40(3) + 120(2) 360 (9, 0) 40(9) + 120(0) (0, 0) 40(0) + 120(0) Step 7 Select the greatest amount. Example 4-3a

20 Answer: The maximum values are 360 at (3, 2) and 360. at (9, 0)
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 mows and 2 prunings or 9 mows and 0 prunings. Example 4-3a

21 Landscaping A landscaping company has crews who rake leaves and mulch
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 crew’s 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. Find a combination of raking leaves and mulching that will maximize the income the company receives per day from one of its crews. Example 4-3b

22 0 raking jobs and 4 mulching jobs
Answer: 0 raking jobs and 4 mulching jobs Example 4-3b

23 Essential Question What are the Linear Programming Procedures?
1) Define the variables 2) Write a system of inequalities 3) Graph the system of inequalities 4) Find the coordinates of the vertices of the feasible region 5) Write a function to be maximized or minimized 6) Substitute the coordinates of the vertices into the function 7) Select the greatest or least result. Answer the problem.

24 What do you get when you cross a linebacker with a computer geek?
A linear programmer.


Download ppt "Systems of Equations and Inequalities"

Similar presentations


Ads by Google