StandardStandard

MM3A6. Students will solve linear MM3A6. Students will solve linear programming problems in two programming problems in two variables. variables.

a. Solve systems of inequalities in two a. Solve systems of inequalities in two variables, showing the solutions variables, showing the solutions graphically.graphically.b. Represent and solve realistic problems b. Represent and solve realistic problems using linear programming.using linear programming.

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).

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

Steps to solving a linear programming problem

1. Write a system of inequalities, and graph the feasible region.

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.

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.)

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.

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.

Practice GraphingPractice Graphing

2 8

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C x y

Practice GraphingPractice Graphing

3

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23 2 12

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x y

x y

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

VariablesVariables

x = the number of cushionsx = the number of cushions y = the number of capsy = the number of caps

ConstraintsConstraints

x < 100x < 100 y < 75y < 75 x + y > 120 y > -x + 120x + y > 120 y > -x + 120 x + y < 150 y < -x + 150x + y < 150 y < -x + 150 x > 0x > 0 y > 0y > 0

GraphGraph

x < 100x < 100 y < 75y < 75 y > -x + 120y > -x + 120 y < -x + 150y < -x + 150 x > 0x > 0 y > 0y > 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

VerticesVertices

(45,75)(45,75) (75,75)(75,75) (100,20)(100,20) (100,50)(100,50)

Objective FunctionsObjective Functions

Cost:Cost:c(x,y) = 1.90x + 2.25yc(x,y) = 1.90x + 2.25y

Revenue:Revenue:c(x,y) = 5.00x + 6.00yc(x,y) = 5.00x + 6.00y

Profit:Profit:c(x,y) = (5.00x – 1.90x) + (6.00y – 2.25y)c(x,y) = (5.00x – 1.90x) + (6.00y – 2.25y)

Objective Functions with VerticesObjective Functions with Vertices

Costs:Costs:(45,75) = 1.90(45) + 2.25(75) = $254.25(45,75) = 1.90(45) + 2.25(75) = $254.25

(75,75) = 1.90(75) + 2.25(75) = (75,75) = 1.90(75) + 2.25(75) = $311.25$311.25

(100,20) = 1.90(100) + 2.25(20) = (100,20) = 1.90(100) + 2.25(20) = $235.00$235.00

(100,50) = 1.90(100) + 2.25(50) = $302.50(100,50) = 1.90(100) + 2.25(50) = $302.50

Revenues:Revenues:(45,75) = 5.00(45) + 6.00(75) = $675.00(45,75) = 5.00(45) + 6.00(75) = $675.00

(75,75) = 5.00(75) + 6.00(75) = (75,75) = 5.00(75) + 6.00(75) = $825.00$825.00

(100,20) = 5.00(100) + 6.00(20) = (100,20) = 5.00(100) + 6.00(20) = $620.00$620.00

(100,50) = 5.00(100) + 6.00(50) = $800.00(100,50) = 5.00(100) + 6.00(50) = $800.00

• Profits:(45,75) = (5.00(45) – 1.90(45)) + (6.00(75) – 2.25(75)) = $420.75

(75,75) = (5.00(75) – 1.90(75)) + (6.00(75) – 2.25(75)) = $513.75

(100,20) = (5.00(100) – 1.90(100)) + (6.00(20) – 2.25(20)) = $385.00

(100,50) = (5.00(100) – 1.90(100)) + (6.00(50) – 2.25(50)) = $497.50

ConclusionConclusion

The campus store should buy 100 The campus store should buy 100 cushions and 20 caps for the minimum cushions and 20 caps for the minimum cost of $235.00.cost of $235.00.

They should sell 75 cushions and 75 caps They should sell 75 cushions and 75 caps for a maximum revenue of $825.00.for a maximum revenue of $825.00.

They should sell 75 cushions and 75 caps They should sell 75 cushions and 75 caps for a maximum profit of $513.75.for a maximum profit of $513.75.

Tyler, Julian, and Nadia have been earning money designing and Tyler, Julian, and Nadia have been earning money designing and producing brochures and flyers for small businesses in their area. producing brochures and flyers for small businesses in their area. Over spring break they plan to use paper and ink they have on hand Over spring break they plan to use paper and ink they have on hand to fill some orders they have for publications. Unfortunately, they to fill some orders they have for publications. Unfortunately, they have more work than they can complete during this time period and 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 would like to know how many of each type of publication they should manufacture to maximize their profits.manufacture to maximize their profits.

They have 500 sheets of producing paper. Single sided flyers are sold 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 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 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 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 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. 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 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 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 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 brochure pack, how many of each should they make to maximize their profit? What is the maximum profit?their profit? What is the maximum profit?

VariablesVariables

x = the number packets of flyersx = the number packets of flyers y = the number packets of brochuresy = the number packets of brochures

ConstraintsConstraints

2x + 3y < 72 y < -2/3x + 242x + 3y < 72 y < -2/3x + 24 20x + 10y < 500 y < -2x + 5020x + 10y < 500 y < -2x + 50 x + 3y < 60 y < -1/3x + 20x + 3y < 60 y < -1/3x + 20

GraphGraph y < -2/3x + 24y < -2/3x + 24 y < -2x + 50y < -2x + 50 y < -1/3x + 20y < -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

VerticesVertices

(0,0)(0,0) (25,0)(25,0) (19.5,0)(19.5,0) (12,16)(12,16) (0,20)(0,20)

Objective FunctionObjective Function

Profit:Profit:

c(x,y) = 10x + 20yc(x,y) = 10x + 20y

Objective Functions with VerticesObjective Functions with Vertices

Profit:Profit:

(0,0) = 10(0) + 20(0) = (0,0) = 10(0) + 20(0) = $0.00$0.00

(25,0) = 10(25) + 20(0) = $250.00(25,0) = 10(25) + 20(0) = $250.00

(19.5,11) = 10(19.5) + 20(11) = $415.00(19.5,11) = 10(19.5) + 20(11) = $415.00

(12,16) = 10(12) + 20(16) = (12,16) = 10(12) + 20(16) = $440.00$440.00

(0,20) = 10(0) + 20(20) = $400.00(0,20) = 10(0) + 20(20) = $400.00

ConclusionConclusion

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