1. Linear Programming Warm Up page 12- write the question and answer it you have 10 mins to complete it. See coaching on page 85

2. Linear Programming Problems
A linear programming problem is made up of an objective function and a system of constraints.
The objective function is an algebraic expression in two or more variables describing a quantity that must be maximized or minimized. It has the form z = Ax + By
The system of constraints are made up of a system of linear inequalities. In applications, the variables are also limited to positive values.
The solution to the linear programming problem, if it exists, will occur at the vertices (corner points) of the solution region obtained from the system of constraints.

3. Part 1 Writing the objective function
Bottled water and medical supplies are to be shipped to victims of an earth-quake by plane. Each container of bottled water will serve 10 people and each medical kit will aid 6 people. If x represents the number of bottles of water to be shipped and y represents the number of medical kits, write the objective function that describes the number of people that can be helped.
Solution Because each bottle of water serves 10 people and each medical kit aids 6 people, we have
z = x y.
Using z to represent the objective function, we have
z = 10x + 6y.
Unlike the functions that we have seen so far, the objective function is an equation in three variables. For a value of x and a value of y, there is one and only one value of z. Thus, z is a function of x and y.
6 times the number
of medical kits.
10 times the number
of bottles of water
plus is
The number of
People helped

4. Example Writing the objective function
Each plane can carry no more than 80,000 pounds. The bottled water weighs 20 pounds per container and each medical kit weighs 10 pounds. If x represents the number of bottles of water to be shipped and y represents the number of medical kits, write an inequality that describes this constraint .
Solution Because each plane can carry no more than 80,000 pounds,
The total weight of
the water bottles
The total weight of
the medical kits
must be less than or equal to
80,000 pounds
plus
20x + 10y < 80,000.
Each bottle is
20 pounds.
Each kit is
10 pounds.
The plane's volume constraint is described by the inequality 20x + 10y < 80,000.

5. must be less than or equal to
Example
Writing the constraint
Planes can carry a total volume for supplies that does not exceed 6000 cubic feet. Each water bottle is 1 cubic foot and each medical kit also has a volume of 1 cubic foot. With x still representing the number of water bottles and y the number of medical kits, write an inequality that describes this constraint.
Solution Because each plane can carry a volume of supplies that does not exceed 6000 cubic feet, we have
The total volume of
the water bottles
The total volume of
the medical kits
must be less than or equal to
6000
cubic feet.
plus
lx + ly < 6000.
Each bottle is
1 cubic foot.
Each kit is
1 cubic foot.
The plane's volume constraint is described by the inequality x + y < 6000.

6. Solution of Linear Inequalities
Part 2 Linear Inequalities (Constraints)
Expressions of the type x + 2y ≤ 8 and 3x – y > 6 are called linear inequalities in two variables (x and y) or constraints.
The sign (< or < or > or >) tells they are inequalities and the variables are x and y
A solution of a linear inequality in two variables is an ordered pair (x, y) which makes the inequality true.
Example: (1, 3) is a solution to x + 2y ≤ 8
since (1) + 2(3) = 7 ≤ 8.
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Solution of Linear Inequalities

7. Before we can find the maximum or minimum values(solutions) of any system we must first find the objective function and the inequalities (constraints) Problem 1
A farmer can plant up to 8 acres of land with wheat and barley. He can earn $5,000 for every acre he plants with wheat and $3,000 for every acre he plants with barley. His use of a necessary pesticide is limited by federal regulations to 10 gallons for his entire 8 acres. Wheat requires 2 gallons of pesticide for every acre planted and barley requires just 1 gallon per acre. What is the maximum profit he can make?

8. Objective Function (aim or purpose or goal)
let x = the number of acres of wheat let y = the number of acres of barley. since the farmer earns $5,000 for each acre of wheat and $3,000 for each acre of barley, then the total profit the farmer can earn is 5000*x *y. let p = total profit that can be earned. your equation for profit becomes: p = 5000x y that's your objective function. It's what you want to maximize/minimize or to use to gain the maximum/minimum profit.

9. The constraints are:
number of acres has to be greater than or equal to 0.
1) x > 0
2) y > 0
number of acres has to be less than or equal to 8.
3) x + y < 8
amount of pesticide has to be less than or equal to 10.
4)2x + y < 10
We can use a table to find the constraints x y
WHEAT
BARLEY
TOTAL
> 0
Acres
< 8
Pesticide 2x
< 10

10. Your constraint inequalities are:
x > 0
y > 0
x + y < 8
2x + y < 10

11. PROBLEM NUMBER 2
A painter has exactly 32 units of yellow dye and 54 units of green dye. He plans to mix as many gallons as possible of color A and color B. Each gallon of color A requires 4 units of yellow dye and 1 unit of green dye. Each gallon of color B requires 1 unit of yellow dye and 6 units of green dye. Find the maximum number of gallons he can mix.

12. Objective Function (this is not an inequality/constraint)
let x = the number of gallons of color A. let y = the number of gallons of color B. if we let g = the maximum gallons the painter can make, then the objective function becomes: g = x + y

13. Your Constraints make a table for color A and color B
to determine the amount of
each dye required.
Each gallon (color A/B)
has to be greater than or
equal to 0 or x and y
have to each be greater
than or equal to 0 because
the number of gallons
can't be negative.
1) x > 0
2) y > 0
each gallon of color A or B will require:
total units of yellow dye available are 32
3) 4x + y < 32 total units of green dye available are 54
4) x + 6y < 54 x y
COLOR A
COLOR B
TOTAL X
> 0
YELLOW 4x
< 32
GREEN 6y
< 54

14. Your Constraints/ Inequalities:
1) x > 0
2) y > 0
x >= 0
y >= 0
4x + y <= 32
3) 4x + y < 32
4) x + 6y < 54
allons can't be negative.

15. PROBLEM NUMBER 3
The Bead Store sells material for customers to make their own jewelry. Customer can select beads from various bins. Grace wants to design her own Halloween necklace from orange and black beads. She wants to make a necklace that is at least 12 inches long, but no more than 24 inches long. Grace also wants her necklace to contain black beads that are at least twice the length of orange beads. Finally, she wants her necklace to have at least 5 inches of black beads.

16. Objective Function
let x = the number of inches of black beads. let y = the number of inches of orange beads. the objective function is the length of the necklace there is a maximum length and a minimum length. if you let n equal the length of the necklace, then the objective function becomes: n = x + y

17. The constraints are:
x > 0 is there because the number of inches of black beads can't be negative.
y > 0 is there because the number of inches of orange beads can't be negative.
x + y > 12 is there because the total length of the necklace has to be greater than or equal to 12 inches.
x + y < 24 is there because the total length of the necklace has to be less than or equal to 24 inches.
x > 2y is there because the length of the black beads has to be greater than or equal to twice the length of the orange beads.
x > 5 is there because the number of inches of black beads has to be greater than or equal to 5.

18. GRACE NECKLACE CHOICES
On a table:
GRACE NECKLACE CHOICES x y
BLACK BEADS
ORANGE BEADS
TOTAL
1st CHOICE
> 0
2nd CHOICE
> 0
3RD CHOICE
> 12
4TH CHOICE
< 24
5TH CHOICE
x > 2y
6TH CHOICE
x > 5

19. Constraints
Since the problem is looking for the number of inches of black beads and the number of inches of orange beads, we will let: the constraint equations for this problem are: 1) x > 0 2) y > 0 3) x + y >12 4) x + y < 24 5) x > 2y 6) x > 5

20. PROBLEM NUMBER 4
A garden shop wishes to prepare a supply of special fertilizer at a minimal cost by mixing two fertilizers, A and B. The mixture is to contain: at least 45 units of phosphate at least 36 units of nitrate at least 40 units of ammonium Fertilizer A costs the shop $.97 per pound. Fertilizer B costs the shop $1.89 per pound. fertilizer A contains 5 units of phosphate and 2 units of nitrate and 2 units of ammonium. fertilizer B contains 3 units of phosphate and 3 units of nitrate and 5 units of ammonium. how many pounds of each fertilizer should the shop use in order to minimize their cost.

21. Objective Equation
let x = the number of pounds of fertilizer A. let y = the number of pounds of fertilizer B. the objective function is to minimize the cost. the objective function becomes: c = .97x y

22. The constraints are:
since the number of pounds of each fertilizer can't be negative, 2 of the constraint equations become: x >= 0 y >= 0 since the number of units of phosphate has to be at least 45, the constraint equation for phosphate becomes: 5x + 3y >= 45 since the number of units of nitrate must be at least 36, the constraint equation for nitrates becomes: 2x + 3y >= 36 since the number of units of ammonium must be at least 40, the constraint equation for ammonium becomes: 2x + 5y >= 40

23. Constraints Equations:
x >= 0
y >= 0
5x + 3y >= 45
2x + 3y >= 36
2x + 5y >= 40

24. Let’s graph :
x + 2y ≤ 8

25. Part 3:To graph any inequality:
YOU DO NOT NEED TO WRITE THE STEPS I WILL GIVE IT TO YOU AT THE END
Make x = 0 and solve for y (y intercept)
Make y = 0 and solve for x ( x intercept)
Plot both intercepts and draw the line
< or > tells you that the line should be broken(dotted line)
< or > tells you that the line should not be broken
< or < shade up on the y axis and to the right on the x axis
> or > shade down on the y axis and to the left of the x axis

26. Example: The solution set for x + 2y ≤ 8 is the shaded region.
The solution set, or feasible set, of a linear inequality is the set of all solutions (the region (shaded) that satisfies the inequality)
Example: The solution set for x + 2y ≤ 8 is the shaded region. x y 2
The solution set is a half-plane. It consists of the line
x + 2y ≤ 8 and all the points below and to its left.
The line is called the boundary line of the half-plane.
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Feasible Set

27. The solution set is a half-plane.
A test point can be selected to determine which side of the half-plane to shade. x y
Example: For 2x – 3y ≤ 18 graph the boundary line. Use the previous steps to guide you.
(0, 0) 2 -2
The solution set is a half-plane.
Use (0, 0) as a test point. This is where x > 0and y > 0
(0, 0) is a solution. So all points on the (0, 0) side of the boundary line are also solutions.
Shade above and to the left of the line.
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Test Point

28. Example: Graph an Inequality
Example: Graph the solution set for x – y > 2.
1. Graph the boundary line x – y = 2 as a dotted line. x y
(0, 0)
2. Select a test point not on the line, say (0, 0).
(2, 0)
(0, -2)
(0) – 0 = 0 > 2 is false.
3. Since this is a not a solution, shade in the half-plane not containing (0, 0).
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Example: Graph an Inequality

29. Inequalities in One Variable
Solution sets for inequalities with only one variable can be graphed in the same way. x y 4
- 4
Example: Graph the solution set for x < - 2. x y 4
- 4
Example: Graph the solution set for x ≥ 4.
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Inequalities in One Variable

30. Solution of a System of Linear Inequalities
A solution of a system of linear inequalities is an ordered pair that satisfies all the inequalities.
Example: Find a solution for the system
(5, 4) is a solution of x + y > 8. (5, 4) is also a solution of 2x – y ≤ 7.
Since (5, 4) is a solution of both inequalities in the system, it is a solution of the system.
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Solution of a System of Linear Inequalities

31. 2. Shade in the intersection of the half-planes.
The set of all solutions of a system of linear inequalities is called its solution set.
To graph the solution set for a system of linear inequalities in two variables:
1. Shade the half-plane of solutions for each inequality in the system.
2. Shade in the intersection of the half-planes.
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Solution Set

32. Example: Graph a System of Two Inequalities
Example: Graph the solution set for the system x y
Graph the solution set for x + y > 8.
Graph the solution set for 2x – y ≤ 7. 2
The solution set is the intersection of these two half-planes. This is the wedge-shaped region at the top of the diagram.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
Example: Graph a System of Two Inequalities

33. Example: Graph a System of Two Inequalities
Example: Graph the solution set for the system of linear inequalities: x y
-2x + 3y ≥ 6
Graph the two half-planes. 2
The two half-planes do not intersect; therefore, the solution set is the empty set.
2x – 3y ≥ 12
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
Example: Graph a System of Two Inequalities

34. Linear Programming Warm Up page 12- write the question and answer it you have 10 mins to complete it. See coaching on page 85

35. Example: Graph a System of Four Inequalities
Try this: Graph the solution set for the linear system.
(2) x y 4
- 4
(1)
(3)
(4)
Graph each linear inequality.
The solution set is the intersection of all the half-planes.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
Example: Graph a System of Four Inequalities

36. Solving a Linear Programming Problem Review
If a maximum or minimum value of
z = Ax + By exists, it can be determined as follows:
Find the constraints (inequalities)
Graph the system of inequalities representing the constraints.
Find the value of the objective function at each corner of the graphed region.
Use the values that you found in the prior step to determine the maximum or minimum value of the objective function and the values of x and y for which the maximum or minimum occurs.

37. Given the following constraints and an objective function of z = 10x + 6y find the maximum and minimum values. Give reasons for your answers. y> 0 x >0

38. Let’s practice
Work in pairs

39. The solution set for x + 2y ≤ 8 (the shaded region).
GRAPH THE FOLLOWING INEQUALITIES AND SHADE THE
APPROPRIATE REGION. Sit and work individually. All solutions will be reviewed at the end of the warm up. label and number all axes accurately.
The solution set for x + 2y ≤ 8 (the shaded region).
The boundary line of the solution set of 3x – y ≥ 2
The boundary line of the solution set of x + y < 2
For 2x – 3y ≤ 18 graph the boundary line.
Graph the solution set for x – y > 2
Graph the solution set for x < - 2
Graph the solution set for x ≥ 4.

40. Solving a Linear Programming Problem Review
If a maximum or minimum value of
z = Ax + By exists, it can be determined as follows:
Find the constraints (inequalities)
Graph the system of inequalities representing the constraints.
Find the value of the objective function at each corner of the graphed region.
Use the values that you found in the prior step to determine the maximum or minimum value of the objective function and the values of x and y for which the maximum or minimum occurs.

41. Example Solve: Steps: Write/Identify the constraints
Graph the system of inequalities representing the constraints.
Find the value of the objective function at each corner of the graphed region.
Use the values that you found in the prior step to determine the maximum value of the objective function and the values of x and y for which the maximum occurs.

42. Example cont.
Graph the constraints:

43. Example cont.
x > 0
y > 0

44. Example cont.
2x+y<8

45. Example cont.
x+y>4

46. Example cont.
Corners:
(4,0), (0,4), (0,8)

47. Example cont. Objective function z = 3x+2y Corners Objective function
The maximum value of the objective function is 16 and it occurs when x = 0 and y = 8

48. Example
Find the maximum value of the objective function z = 2x + y subject to the constraints: x > 0, y > 0, x + 2y < 5, x – y < 2. -1 -3 -2 1 3 4 5 6 7 2 -4 -5
x – y = 2
x + 2y = 5
(0, 2.5)
(0, 0)
(3, 1)
(2, 0)
Solution We begin by graphing the region in quadrant I (x > 0, y > 0) formed by the constraints.
Now we evaluate the objective function at the four vertices of this region.
Corners Obj. Func.: z = 2x + y
(0, 0) z = 2 • = 0
(2, 0) z = 2 • = 4
(3, 1) z = 2 • = 7
(0, 2.5) z = 2 • = 2.5
The maximum value of z.
Thus, the maximum value of z is 7, and this occurs when x = 3 and y = 1.

49. NO MARK WILL BE AWARDED IF AXES ARE NOT LABELLED AND NUMBERED
5 MINS
NO MARK WILL BE AWARDED IF AXES ARE NOT LABELLED AND NUMBERED
USING ONE GRAPH SHEET DRWA UP FOUR DIFFERENT PAIRS OF AXES
LABEL EACH AXIS USING “x” AND “y”
3. NUMBER EACH AXIS AS FOLLOWS AND DRAW THE FOLLOWING LINES AND SHADE THEM.
FIRST PAIR OF AXES MUST BE 4 POINTS TO ONE UNIT. DRAW AND SHADE x > -8
SECOND PAIR OF AXES MUST BE 100 POINTS TO ONE UNIT ON BOTH AXES. DRAW AND SHADE y < 400
THIRD PAIR OF AXES MUST BE 500 POINTS TO ONE UNIT ON BOTH AXES. DRAW AND SHADE 1000 < x > 4000
FOURTH AXES MUST BE 1000 POINTS TO ONE UNIT ON THE “x” AXIS AND 2 POINTS TO ONE UNIT ON THE “y” AXIS.
Draw the line for the following points ( 1500, 4) and (3000, 3.5)

50. Lets listen: