LPP KEY

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Algebra 2 NAME______________________________HR____

Linear Programming Packet (LT H)

Checklist:

⃞ I can graph linear inequalities

⃞ I can graph many linear inequalities on the same axes and identify the feasible region

⃞ I can identify each vertex of the feasible region

⃞ I can identify the variables

⃞ I can write constraints

⃞ I can write an objective function

⃞ I can use the objective function and vertices of the feasible region to find a maximum and minimum

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I can graph linear inequalities

Use the slope and the y-intercept to graph the following.

1. 𝑦 >1

3𝑥 − 4 2. 𝑦 ≤ 2𝑥 + 3

Use the intercepts to graph the following.

3. 3𝑥 + 2𝑦 ≤ 18 4. 7𝑥 + 3𝑦 ≤ 21

Graph the following.

5. 120𝑥 + 100𝑦 ≤ 6800 6. 1.8𝑥 + 3𝑦 > 5.4

IMPORTANT THINGS TO REMEMBER

[for y = mx + b form] 1. Plot the y-intercept 2. Use slope to plot another point 3. Draw the line < or > dashed line ≤ or ≥ solid line 4. Test a point (usually the origin) 5. Shade True, shade that side False, shade the other side

[for Ax + By = C form] 1. Plug in 0 for x to find the

y-intercept 2. plug in 0 for y to find the x-intercept 3. Draw the line < or > dashed line ≤ or ≥ solid line 4. Test a point (usually the origin) 5. Shade True, shade that side False, shade the other side Don’t be bothered by larger numbers or decimals!

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Graph. You may need to put into y = mx + b or Ax + By = C form first.

7. 20𝑥 ≤ 15𝑦

8. 𝑦 − 25 < 𝑥

9. 10𝑦 ≤ 2𝑥 + 50

You’ll need to decide which form is easiest. *If using a graphing calculator, you’ll need to solve for y!

20𝑥 ≤ 15𝑦

15𝑦 ≥ 20𝑥

𝑦 ≥20

15𝑥

𝑦 ≥4

3𝑥

𝑦 − 25 < 𝑥

𝑦 < 𝑥 + 25

10𝑦 ≤ 2𝑥 + 50

𝑦 ≤1

5𝑥 + 5

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Graph. The lines will be either horizontal or vertical.

10. 𝑥 > 0 11. 𝑥 ≤ 2.25

12. 𝑦 ≤ 35 13. 𝑦 > 3

14. 5 ≤ 𝑦 ≤ 15 15. 0.25 ≤ 𝑥 ≤ 1.5

Remember, lines in the form x = k will always be vertical. So these will always look like something like this:

Remember, lines in the form y = k will always be horizontal. So these will always look like something like this:

Read these as “y is between 5 and 15” … and the graphs should look something like this:

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Practice. Graph the following inequalities.

16. 𝑥 + 𝑦 ≤ 1.75 𝑦 ≤ −𝑥 + 1.75 17. 0.5 ≤ 𝑦 ≤ 1.5

18. 𝑦 ≥ 0.5𝑥 19. 𝑥 ≥ 0

Checkpoint: If you feel comfortable with graphing linear inequalities, check off the box

“I can graph linear inequalities” on the front page.

↺

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I can graph linear inequalities on the same axes and identify the feasible region

Graph the feasible regions defined by the systems of inequalities.

20. 𝑥 + 𝑦 ≤ 1.75

0.5 ≤ 𝑦 ≤ 1.5 𝑦 ≥ 0.5𝑥 𝑥 ≥ 0

21. 𝑥 + 𝑦 ≤ 200

20𝑥 + 30𝑦 ≤ 5000

50 ≤ 𝑦 ≤ 120

𝑥 ≥ 10

Notice this first one is just the previous page all put into one. Tip: Instead of shading all of them, draw arrows in the direction…then it will be cleaner to shade at the end.

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Practice. Graph the feasible regions defined by the systems of inequalities.

22. 𝑦 ≤ −0.8𝑥 + 40

0 ≤ 𝑥 ≤ 30

0 ≤ 𝑦 ≤ 30

23. 𝑦 ≤ 0.5𝑥 + 10

500 ≤ 𝑥 ≤ 1500

5𝑥 + 12𝑦 ≥ 12000

𝑦 ≤ 1200

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24. 𝑥 + 𝑦 ≥ 10

8𝑥 + 12𝑦 ≤ 200

2 ≤ 𝑦 ≤ 10

𝑥 ≤ 15

Checkpoint: If you feel comfortable with graphing feasible regions, check off the box

“I can graph many linear inequalities on the same axes and identify the feasible region” on the

front page.

↺

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I can identify each vertex of the feasible region

Below are inequalities and their related feasible regions. Identify each

vertex of the region.

24. 𝑦 ≤ 2𝑥

100𝑥 + 300𝑦 ≤ 15000

0 ≤ 𝑦 ≤ 30

(0,0) (150,0)

To find the vertices of the feasible region, you must find the points of intersection of the lines. For example, to find this vertex

you must find the solution to the system y = 30 100x + 300y = 15000 Don’t do extra work! You don’t need to find this one

You have options! Remember, you can find these points using substitution, elimination, or using a graphing calculator.

𝑦 = 2𝑥

100𝑥 + 300𝑦 = 15000

𝑦 = 30

𝑦 = 0

𝑦 = 2𝑥

30 = 2𝑥

𝑥 = 15

Need to use

and

𝑦 = 30.

Substitute.

(15,30)

100𝑥 + 300𝑦 = 15000

100𝑥 + 300 30 = 15000

100𝑥 + 9000 = 15000

100𝑥 = 6000

𝑥 = 60

Need to use

and

𝑦 = 30.

Substitute.

(60,30)

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25. 3𝑥 + 5𝑦 ≤ 70

𝑥 + 𝑦 ≥ 10

𝑥 − 𝑦 ≥ 5

𝑥 ≥ 5

3𝑥 + 5𝑦 = 70

𝑥 − 𝑦 = 5

𝑥 + 𝑦 = 10

𝑥 = 5

5 + 𝑦 = 10

𝑦 = 5

(5,5)

Use 𝑥 + 𝑦 = 10 and 𝑥 = 5

3 5 + 5𝑦 = 70

5𝑦 = 55

𝑦 = 11

(5,11)

Use 3𝑥 + 5𝑦 = 70 and 𝑥 = 5

𝑦 + 5 + 𝑦 = 10

2𝑦 + 5 = 10

2𝑦 = 5

𝑦 = 2.5

𝑥 + 2.5 = 10

𝑥 = 7.5

(2.5,7.5)

Use 𝑥 + 𝑦 = 10 and 𝑥 − 𝑦 = 5

𝑥 = 𝑦 + 5

3 𝑦 + 5 + 5𝑦 = 70

3𝑦 + 15 + 5𝑦 = 70

8𝑦 = 55

𝑦 = 6.875

𝑥 − 6.875 = 5

𝑥 = 11.875

(11.875,6.875)

Use 3𝑥 + 5𝑦 = 70 and 𝑥 − 𝑦 = 5

𝑥 = 𝑦 + 5

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Practice. Below are inequalities and their related feasible regions. Identify each vertex of the region.

26. 10𝑥 + 22𝑦 ≤ 550

𝑥 + 𝑦 ≥ 30

0 ≤ 𝑦 ≤ 15

𝑥 ≤ 40

27. 0.1𝑥 + 0.5𝑦 ≤ 2

𝑥 + 𝑦 ≤ 16

0 ≤ 𝑦 ≤ 2

𝑥 ≥ 0

(15,15)

(30,0)

(40,0)

(22,15)

(40,6.18)

(0,0) (16,0)

(15,1)

(0,2) (10,2)

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28. 0.4𝑥 − 𝑦 + 2 ≥ 0

𝑦 ≤ −0.2𝑥 + 4

𝑦 ≥ 1

𝑥 ≥ 1

𝑥 + 𝑦 ≤ 10

29. 1500𝑥 + 2400𝑦 ≤ 400000

𝑥 + 𝑦 ≤ 200

0 ≤ 𝑦 ≤ 150

0 ≤ 𝑥 ≤ 150

Checkpoint: If you feel comfortable finding vertices of feasible regions, check off the box

“I can identify each vertex of the feasible region” on the front page. ↺

(1,1) (9,1)

(7.5,2.5)

(3.3,3.3)

(1,2.4)

(0,150)

(150,0)

(150,50)

(88.9,111.1)

(0,0)

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NOW! Put it all together. Graph the inequalities and find each vertex of the feasible region.

30. 𝑥 + 𝑦 ≤ 8

𝑦 ≤ −1

2𝑥 + 7

1 ≤ 𝑥 ≤ 5

2𝑥 + 𝑦 ≤ 12

𝑦 ≥ 1

Vertices:

(0,0)

(1,6.5)

(2,6)

(4,4)

(5,2)

(5,0)

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31. 25𝑥 + 20𝑦 ≤ 800

𝑦 ≤ 3𝑥

10 ≤ 𝑦 ≤ 20

Vertices:

(3.3,10)

(6.6,20)

(16,20)

(24,10)

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I can identify the variables

For each word problem, identify the variables.

32. A computer company produces a laptop and a desktop. There is an

expected demand of at least 100 laptops and 80 desktops each day.

Because of limitations on production capacity, no more than 200 laptops

and 170 desktops can be made daily. To satisfy a shipping contract, a total

of at least 200 computers much be shipped each day. If each laptop

computer sold results in a $2 loss, but each desktop produces a $20 profit,

how many of each type should be made daily to maximize net profits?

SOLUTION

x = number of laptop computers

y = number of desktop computers

33. Piñatas are made to sell at a craft fair. It takes 2 hours to make a mini

piñata and 3 hours to make a regular sized piñata. The owner of the craft

booth will make a profit of $12 for each mini piñata sold and $24 for each

regular-sized piñata sold. If the craft booth owner has no more than 30

hours available to make piñatas and wants to have at least 12 piñatas to

sell, how many of each size piñata should be made to maximize profit?

SOLUTION

x = number of mini piñatas

y = number of regular-sized piñatas

Look at the question being asked. In this case the question is “How many of each type should be made daily to maximize net profits?” In this case the question is “How many of each size piñata should be made to maximize profit?”

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Practice. For each word problem, identify the variables.

34. A company manufactures two types of printers, an inkjet printer and a laser printer. The company can make a

total of 60 printers per day, and it has 120 labor-hours per day available. It takes 1 labor-hour to make an inkjet

printer and 3 labor-hours to make a laser printer. The profit is $40 per inkjet printer and $60 per laser printer.

How many of each type of printer should the company make to maximize its daily profit?

x = number of inkjet printers

y = number of laser printers

35. You have 180 tomatoes and 15 onions left over from your garden. You want to use these to make jars of tomato

sauce and jars of salsa to sell at a farm stand. A jar of tomato sauce requires 10 tomatoes and 1 onion, and a jar

of salsa requires 5 tomatoes and ¼ onion. You will make a profit of $2 on every jar of tomato sauce sold and a

profit of $1.50 on every jar of salsa sold. The owner of the farm stand wants at least three times as many jars of

tomato sauce as jars of salsa. How many jars of each should you make to maximize profit?

x = number of jars of tomato sauce

y = number of jars of salsa

36. A farmer has 10 acres to plant in wheat and rye. He has to plant at least 7 acres. However, he has only $1200 to

spend and each acre of wheat costs $200 to plant and each acre of rye costs $100 to plant. Moreover, the

farmer has to get the planting done in 12 hours and it takes an hour to plant an acre of wheat and 2 hours to

plant an acre of rye. If the profit is $500 per acre of wheat and $300 per acre of rye how many acres of each

should be planted to maximize profits?

x = number of wheat acres

y = number of rye acres

Checkpoint: If you feel comfortable IDENTIFYING VARIABLES, check off the box

“I can identify the variables” on the front page. ↺

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I can write constraints

For each word problem, write the constraints.














You’ve identified the variable already: use that information x = # of laptops y = # of desktops Take the information PIECE BY PIECE. “There is an expected demand of at least 100 laptops and 80 desktops each day.” (This should give you two inequalities) “no more than 200 laptops and 170 desktops can be made daily” (This should give you two more inequalities) “a total of at least 200 computers much be shipped each day” (This should give you another one)

You’ll get constraints from the following pieces: “It takes 2 hours to make a mini piñata and 3 hours to make a regular sized piñata.”

WITH “the craft booth owner has no more than 30 hours available to make piñatas” Also, “wants to have at least 12 piñatas to sell”

So what about PROFIT? You’ll deal with maximizing profit and the prices later when we write an objective function. For now, we just need to write as many INEQUALITIES as we can.

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Practice. For each word problem, write the constraints. You may want to refer back to the variables you already

identified in #34-36.















Checkpoint: If you feel comfortable WRITING CONSTRAINTS, check off the box

“I can write constraints” on the front page. ↺

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I can write an objective function

For each word problem, write the objective function.








P = -2x + 20y







P = 12x + 24y

The objective function deals with what you’re trying to maximize or minimize. Often, you’ll want to look at the last sentence: “If each laptop computer sold results in a $2 loss, but each desktop produces a $20 profit, how many of each type should be made daily to maximize net profits?” For this one, some of the information is in the middle: “The owner of the craft booth will make a profit of $12 for each mini piñata sold and $24 for each regular-sized piñata sold.”

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Practice. For each word problem, write the objective function.





P = 40x + 60y






P = 2x + 1.50y






P = 500x + 300y

Checkpoint: If you feel comfortable WRITING OBJECTIVE FUNCTIONS, check off the box

“I can write objective functions” on the front page. ↺

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I can use the objective function and vertices of the feasible region to

find a maximum and minimum

Given the vertices of the feasible region, find the maximum and minimum

values of the objective function.

47. Vertices: (0,0), (15,30), (60,30), (150,0)

𝑃 = 20𝑥 + 30𝑦

𝑃 = 20(0)+30(0)=0 *minimum

𝑃 = 20(15)+30(30)=1200

𝑃 = 20(60)+30(30)=2100

𝑃 = 20(150)+30(0)=3000 *maximum

48. Vertices: (5,5), (5,11), (11.875,6.875), (7.5,2.5)

𝑃 = 2.5𝑥 + 3𝑦

𝑃 = 2.5(5)+3(5)=27.5

𝑃 = 2.5(5)+3(11)=45.5

𝑃 = 2.5(11.875)+3(6.875)=50.3125 *maximum

𝑃 = 2.5(7.5)+3(2.5)=26.25 *minimum

Plug in each ordered pair. The biggest value of P is the maximum, the smallest value is the minimum.

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Practice. Given the vertices of the feasible region, find the maximum and minimum values of the objective

function.

49. Vertices:

(15,15)

(22,15)

(40,6.81)

(40,0)

(30,0)

P = 1000x - 200y

P = 1000(15) - 200(15) = 12000 *min

P = 1000(22) - 200(15) = 19000

P = 1000(40) - 200(6.81) = 38638

P = 1000(40) - 200(0) = 40000 *max

P = 1000(30) - 200(0) = 30000

50. (0,0)

(16,0)

(15,1)

(10,2)

(0,2)

P = 5x + 8y

P = 5(0) + 8(0) = 0 *min

P = 5(16) + 8(0) = 80

P = 5(15) + 8(1) = 83 *max

P = 5(10) + 8(2) =66

P = 5(0) + 8(2) = 16

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51. (3.3,3.3)

(7.5,2.5)

(9,1)

(1,1)

(1,2.4)

P = 100x + 200y

P = 100(3.3) + 200(3.3) = 990

P = 100(7.5) + 200(2.5) = 1250 *max

P = 100(9) + 200(1) = 1100

P = 100(1) + 200(1) = 300 *min

P = 100(1) + 200(2.4) = 580

52. (0,0)

(0,150)

(50,150)

(88.9,111.1)

(150,50)

(150,0)

P = 12x + 18y

P = 12(0) + 18(0) = 0 *min

P = 12(0) + 18(150) = 2700

P = 12(50) + 18(150) = 3300 *max

P = 12(88.9) + 18(111.1) = 3066.6

P = 12(150) + 18(50) = 2700

P = 12(150) + 18(0) = 1800

Checkpoint: If you feel comfortable finding a maximum and minimum using the objective function, check off

the box “I can use the objective function and vertices of the feasible region to find a maximum and

minimum” on the front page.

↺

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Now, put it all together!

53. A certain refinery is trying to decide how much fuel oil and gasoline to produce.

I. The refining process requires the production of at least two gallons of gasoline for each gallon of

fuel oil.

II. To meet the anticipated demands of winter, at least three million gallons of fuel oil a day will need

to be produced.

III. The demand for gasoline, on the other hand, is not more than 6.4 million gallons a day.

Objective: If gasoline is selling for $1.90 per gallon and fuel oil sells for $1.50/gal, how much of each should

be produced in order to maximize revenue?

A. Identify the variables.

x = gallons of gasoline

y = gallons of fuel oil

B. Write three constraints that come from

I,II, and III above.

x ≥ 2y

y ≥ 3

x ≤ 6.4

C. Graph your inequalities on the

coordinate plane at right. (In millions; that

is, 2 represents 2 million)

D. Write an objective function.

P = 1.90x + 1.50y

E. How much of each should be produced in order to maximize revenue?

Vertices:

(6.4,3)

(6.4,3.2) → 1.90(6.4) + 1.50(3.2) = 16.96

(6,3)

6.4 gallons of gasoline, 3.2 gallons of fuel oil

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54. A gold processor is trying to keep his plant running and extract as much gold as possible. He has two

sources of gold ore, source A and source B.

I. At least three tons of gold ore must be processed each day.

II. Ore from source A costs $20 per ton to process and ore from source B costs $10 per ton to process.

Costs must be kept under $80 per day.

III. The amount of ore from source B cannot exceed twice the amount of ore from source A.

Objective: Source A yields 2 oz of gold per ton, source B yields 3 oz of gold per ton. How many tons of ore

from both sources must be processed each day to maximize the amount of gold extracted?


x = tons from source A

y = tons from source B

B. Write three constraints that come from

I,II, and III above.

x + y ≥ 3

20x + 10y ≤ 80

y ≤ 2x


coordinate plane at right.


P = 2x + 3y

E. How many tons of ore from both sources must be processed each day to maximize the amount of gold

extracted?

(1,2)

(2,4) → 2(2) + 3(4) = 16

(4,0)

(3,0)

2 tons from source A and 4 tons from source B

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55. Mary works at a very popular fresh-squeezed lemonade stand on Myrtle Beach that sells regular

lemonade and strawberry lemonade. To make a 20 oz. regular lemonade it takes 4 whole lemons. To

make a 20 oz. strawberry lemonade it takes 3 whole lemons and 2 average sized strawberries.

I. The owners figured they have a budget so that each morning a truck delivers 900 lemons to the

stand.

II. That same truck brings 408 strawberries to the stand.

III. The owners want Mary to sell at least twice the amount of strawberry lemonade each day as

regular lemonade because more profit is made on the strawberry lemonade.

IV. They have also asked that at least 50 strawberry lemonades get sold each day because the

strawberries spoil more quickly than the lemons.

Objective: On each 20 oz. glass of regular lemonade a profit of $.95 is made. The stand is able to charge

more for strawberry lemonades and a profit of $1.10 is made on each 20 oz. glass of that. How many

regular lemonades and how many strawberry lemonades should Mary sell each day to maximize profit?


x = glasses of regular lemonade

y = glasses of strawberry lemonade

B. Write four constraints that come from

I,II, III, and IV above.

4x + 3y ≤ 900

2y ≤ 408

y ≥ 2x

y ≥ 50


coordinate plane at right.


P = 0.95x + 1.10y

E. How many regular lemonades and how many strawberry lemonades should Mary sell each day to

maximize profit?

(72,204) → 0.95(72) + 1.1(204) = 292.8

72 glasses of regular lemonade, 204 glasses of strawberry lemonade

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Finish #32-36.

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