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Standard  MM3A6. Students will solve linear programming problems in two variables.  a. Solve systems of inequalities in two variables, showing the solutions.

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Presentation on theme: "Standard  MM3A6. Students will solve linear programming problems in two variables.  a. Solve systems of inequalities in two variables, showing the solutions."— Presentation transcript:

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2 Standard  MM3A6. Students will solve linear programming problems in two variables.  a. Solve systems of inequalities in two variables, showing the solutions graphically. b. Represent and solve realistic problems using linear programming.

3 Linear programming is a method of using systems of inequalities to find the maximum or minimum of some function of interest (like cost or revenue or profit).

4 Vocabulary   Constraints: System of inequalities   Feasible region: Solution to the system of inequalities   Objective function: Quantity to be minimized or maximized   Vertices: Corner points of the feasible region where the inequalities intersect

5 Steps to solving a linear programming problem   1. Write a system of inequalities, and graph the feasible region.

6 Steps to solving a linear programming problem   1. Write a system of inequalities, and graph the feasible region.   2. Write the objective function to be maximized or minimized.

7 Steps to solving a linear programming problem   1. Write a system of inequalities, and graph the feasible region.   2. Write the objective function to be maximized or minimized.   3. Find the coordinates of the vertices of the feasible region. (May need to solve a system of equations.)

8 Steps to solving a linear programming problem   1. Write a system of inequalities, and graph the feasible region.   2. Write the objective function to be maximized or minimized.   3. Find the coordinates of the vertices of the feasible region. (May need to solve a system of equations.)   4. Evaluate the objective function (make a table) for each of the vertices in the feasible region. Identify the coordinates that give the required maximum or minimum.

9 Steps to solving a linear programming problem   1. Write a system of inequalities, and graph the feasible region.   2. Write the objective function to be maximized or minimized.   3. Find the coordinates of the vertices of the feasible region. (May need to solve a system of equations.)   4. Evaluate the objective function (make a table) for each of the vertices in the feasible region. Identify the   coordinates that give the required maximum or minimum. 5. Be sure to state your answer in the context of   the problem, if appropriate.

10 Practice Graphing

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12 The campus store sells stadium cushions and caps. The cushions cost $1.90; the caps cost $2.25. They sell the cushions for $5.00 and the caps for $6.00. They can obtain no more than 100 cushions and 75 caps per week. To meet demands, they have to sell a total of at least 120 of the two together. They cannot package more than 150 per week. What quantities will give the minimum cost? What quantities will give the maximum revenue? What quantities will give the maximum profit?

13 Variables  x = the number of cushions  y = the number of caps

14 Constraints  x < 100  y < 75  x + y > 120 y > -x  x + y < 150 y < -x  x > 0  y > 0

15 Graph  x < 100  y < 75  y > -x  y < -x  x > 0  y > 0 (0,150) (0,120) (0,75) (0,0) (100,0) (120,0)(150,0) (45,75) (75,75)(100,75) (100,50) (100,20) X-axis Y-axis

16 Vertices  (45,75)  (75,75)  (100,20)  (100,50)

17 Objective Functions  Cost: c(x,y) = 1.90x y  Revenue: c(x,y) = 5.00x y  Profit: c(x,y) = (5.00x – 1.90x) + (6.00y – 2.25y)

18 Objective Functions with Vertices  Costs: (45,75) = 1.90(45) (75) = $ (75,75) = 1.90(75) (75) = $ (100,20) = 1.90(100) (20) = $ (100,50) = 1.90(100) (50) = $  Revenues: (45,75) = 5.00(45) (75) = $ (75,75) = 5.00(75) (75) = $ (100,20) = 5.00(100) (20) = $ (100,50) = 5.00(100) (50) = $ Profits: (45,75) = (5.00(45) – 1.90(45)) + (6.00(75) – 2.25(75)) = $ (75,75) = (5.00(75) – 1.90(75)) + (6.00(75) – 2.25(75)) = $ (100,20) = (5.00(100) – 1.90(100)) + (6.00(20) – 2.25(20)) = $ (100,50) = (5.00(100) – 1.90(100)) + (6.00(50) – 2.25(50)) = $497.50

19 Conclusion  The campus store should buy 100 cushions and 20 caps for the minimum cost of $  They should sell 75 cushions and 75 caps for a maximum revenue of $  They should sell 75 cushions and 75 caps for a maximum profit of $

20 Tyler, Julian, and Nadia have been earning money designing and producing brochures and flyers for small businesses in their area. Over spring break they plan to use paper and ink they have on hand to fill some orders they have for publications. Unfortunately, they have more work than they can complete during this time period and would like to know how many of each type of publication they should manufacture to maximize their profits. They have 500 sheets of producing paper. Single sided flyers are sold in 20 copy packets and need 20 sheets of paper per pack. Double sided brochures come in 10 packs and would need 10 sheets of paper. They have 60 units of ink, and on average use 1 unit for a pack of flyers and 3 per pack of brochures. All of the projects they have are new orders and will need to be both designed and printed. On average, it takes 2 hours to design and print a packet of flyers and 3 for a brochure. They feel they can allocate a total of 72 hours over the break. If their profit on a flyer pack is $10, and $20 on a brochure pack, how many of each should they make to maximize their profit? What is the maximum profit?

21 Variables  x = the number packets of flyers  y = the number packets of brochures

22 Constraints  2x + 3y < 72 y < -2/3x + 24  20x + 10y < 500 y < -2x + 50  x + 3y < 60 y < -1/3x + 20

23 Graph  y < -2/3x + 24  y < -2x + 50  y < -1/3x + 20 (0,50) (0,24) (0,20) (0,0) (25,0)(36,0) (60,0) (18,14) (12,16) (19.5, 11) X-axis Y-axis

24 Vertices  (0,0)  (25,0)  (19.5,0)  (12,16)  (0,20)

25 Objective Function  Profit: c(x,y) = 10x + 20y

26 Objective Functions with Vertices  Profit: (0,0) = 10(0) + 20(0) = $0.00 (25,0) = 10(25) + 20(0) = $ (19.5,11) = 10(19.5) + 20(11) = $ (12,16) = 10(12) + 20(16) = $ (0,20) = 10(0) + 20(20) = $400.00

27 Conclusion  Tyler, Julian, and Nadia should produce 12 packets of flyers and 16 packets of brochures to make a maximum profit of $440.00


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