1. § 4.5
Linear Programming

2. Linear Programming
Linear programming is a method for solving problems in which a particular quantity that must be maximized or minimized is limited by other factors.
In this section, you will learn how to:
(1) Write an objective function modeling a quantity that must be maximized or minimized.
(2) Use inequalities to model the limitations in the problem.
(3) Use linear programming to solve the problem.
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 4.5

3. Basic Linear Programming Solving a Linear Programming Problem
Objective Function: An algebraic expression in two or more variables describing a quantity that must be maximized or minimized.
Constraint: A restriction, based on a situation, that can be expressed as an inequality.
Solving a Linear Programming Problem
Let z = ax + by be an objective function that depends on x and y. Furthermore, z is subject to a number of constraints on x and y. If a maximum or minimum value of z exists, it can be determined as follows:
1) Graph the system of inequalities representing the constraints.
2) Find the value of the objective function at each corner, or vertex, of the graphed region. The maximum and minimum of the objective function occur at one or more of the corner points.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 4.5

4. Linear Programming
EXAMPLE - Writing an objective function
Suppose a student earns $10 per hour for tutoring and $8 per hour working at a newspaper office.
Let x = # hours spent each week tutoring and
Let y = # hours spent each week working at the newspaper office
Write an objective function that models total weekly earnings.
Total dollars earned in a week =( x hours) times (10 dollars) +( y hours) times (8 dollars)
We can represent this statement by:
z = 10x + 8y
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 4.5

5. Linear Programming
Continued
The student decides that he should work no more than 15 hours per week
so as to save time for his studies.
He must spend at least 3 hours per week tutoring since the tutoring center
requires that.
He can not spend more than 8 hours per week tutoring since that is all the
tutoring center will allow.
What system of inequalities models these three constraints?
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 4.5

6. Linear Programming
EXAMPLE
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 in part (b) to determine the maximum value of the objective function and the values of x and y for which the maximum occurs.
Objective Function
z = 5x – 2y
Constraints
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 4.5

7. Linear Programming
CONTINUED
SOLUTION
Graph the system of inequalities representing the constraints.
We must first determine what the graph of each constraint looks like. The graphs of the first two constraints are relatively simple and it is left to you to verify that their graphs as contained here are correct. Let’s determine the graph of the third constraint.
We first rewrite the inequality as x + y = 2.
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 4.5

8. Linear Programming Now we find the intercepts for the equation.
CONTINUED
Now we find the intercepts for the equation.
We set y = 0 to find the x-intercept:
We set x = 0 to find the y-intercept:
x + y = 2
x + y = 2
x + 0 = 2
0 + y = 2
x = 2
y = 2
Therefore, the two intercepts for the equation are: (2,0) and (0,2).
Upon plugging the test-point (0,0) into the equation , we get a false statement. Therefore, the region not containing the test point (above the line) is the region to shade.
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 4.5

9. Linear Programming
CONTINUED
We are now ready to graph all three constraints. We will graph the first constraint in yellow, the second constraint in red, and the third constraint in blue.
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 4.5

10. Linear Programming
CONTINUED
Therefore, the shaded region below represents the constraints.
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 4.5

11. Linear Programming
CONTINUED
(b) Find the value of the objective function at each corner of the graphed region. First we have to find the coordinates of each of the five corners.
The first corner is where the lines x = 0 and y = 3 meet. Therefore the coordinates of the first point are (0,3).
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 4.5

12. Linear Programming
CONTINUED
The next corner is where the lines x = 5 and y = 3 meet. Therefore the coordinates of this point are (5,3).
The next corner is where the lines x = 5 and y = 0 meet. Therefore the coordinates of this point are (5,0).
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 4.5

13. Linear Programming
CONTINUED
The next corner is where the lines x + y = 2 and y = 0 meet. Therefore the y-coordinate of the intersection point is 0. To find the x-coordinate we do the following.
x + y = 2
x + 0 = 2
x = 2
Therefore, the intersection point is (2,0).
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 4.5

14. Linear Programming
CONTINUED
The next corner is where the lines x + y = 2 and x = 0 meet. Therefore the x-coordinate of the intersection point is 0. To find the y-coordinate we do the following.
x + y = 2
0 + y = 2
y = 2
Therefore, the intersection point is (0,2).
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 4.5

15. Linear Programming Corner (x,y) Objective Function z = 5x – 2y
CONTINUED
Now we can determine the value of the objective function at each corner of the graphed region.
Corner
(x,y)
Objective Function
z = 5x – 2y
(0,3)
z = 5(0) – 2(3) = 0 – 6 = -6
(5,3)
z = 5(5) – 2(3) = 25 – 6 = 19
(5,0)
z = 5(5) – 2(0) = 25 – 0 = 25
(2,0)
z = 5(2) – 2(0) = 10 – 0 = 10
(0,2)
z = 5(0) – 2(2) = 0 – 4 = -4
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 4.5

16. Linear Programming
CONTINUED
(c) Use the values in part (b) to determine the maximum value of the objective function and the values of x and y for which the maximum occurs.
The maximum value of the objective function is 25. This occurs when (x,y) = (5,0).
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 4.5

17. Linear Programming
EXAMPLE
On June 24, 1948 the former Soviet Union blocked all land and water routes through East Germany to Berlin. A gigantic airlift was organized using American and British planes to bring food, clothing, and other supplies to the more than 2 million people in West Berlin. The cargo capacity was 30,000 cubic feet for an American plane and 20,000 cubic feet for a British plane. To break the Soviet blockade, the Western Allies had to maximize cargo capacity, but were subject to the following restrictions:
No more than 44 planes could be used.
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 4.5

18. Linear Programming
CONTINUED
The larger American planes required 16 personnel per flight, double that of the requirement for the British planes. The total number of personnel available could not exceed 512.
The cost of an American flight was $9,000 and the cost of a British flight was $5,000. Total weekly costs could not exceed $300,000.
Find the number of American planes and the number of British planes that were used to maximize cargo capacity.
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 4.5

19. Linear Programming
CONTINUED
SOLUTION
First we need to define a pair of variables. Let x = the number of American planes and let y = the number of British planes.
Now we can determine the objective function and the inequalities that define the constraints.
Since the total cargo capacity for an American plane is 30,000 cubic feet and the total cargo capacity for a British plane is 20,000 cubic feet, let z = the total cargo being transported. This is represented by the objective function:
z = 30,000x + 20,000y.
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 4.5

20. Linear Programming Now we will determine the inequality constraints.
CONTINUED
Now we will determine the inequality constraints.
Since the number of planes cannot exceed 44 we get:
Since the number of personnel cannot exceed 512 and the number of personnel an American plane requires is 16 and the number of personnel a British plane requires is 8:
Since the cost cannot exceed $300,000 and the cost of an American flight costs $9,000 and the cost of a British flight costs $5,000:
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 4.5

21. Linear Programming
CONTINUED
1) Graph the system of inequalities representing the constraints.
First, we need to determine the intercepts of each of the inequalities. Upon rewriting the first inequality with an = symbol, we get x + y = 44. Thus,
We set y = 0 to find the x-intercept:
We set x = 0 to find the y-intercept:
x + y = 44
x + y = 44
x + 0 = 44
0 + y = 44
x = 44
y = 44
Therefore, the two intercepts for the equation are: (44,0) and (0,44).
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 4.5

22. Linear Programming
CONTINUED
Now we will determine the intercepts of the second inequality. Upon rewriting it with an = symbol, we get x + 8y = Thus,
We set y = 0 to find the x-intercept:
We set x = 0 to find the y-intercept:
16x + 8y = 512
16x + 8y = 512
16x + 8(0) = 512
16(0) + 8y = 512
16x + 0 = 512
0 + 8y = 512
16x = 512
8y = 512
x = 32
y = 64
Therefore, the two intercepts for the equation are: (32,0) and (0,64).
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 4.5

23. Linear Programming
CONTINUED
Now we will determine the intercepts of the third inequality. Upon rewriting it with an = symbol, we get 9,000x + 5,000y = 300,000. Thus,
We set y = 0 to find the x-intercept:
We set x = 0 to find the y-intercept:
9,000x + 5,000y = 300,000
9,000x + 5,000y = 300,000
9,000x + 5,000(0) = 300,000
9,000(0) + 5,000y = 300,000
9,000x + 0 = 300,000
0 + 5,000y = 300,000
9,000x = 300,000
5,000y = 300,000
x = 33.3
y = 60
Therefore, the two intercepts for the equation are: (33.3,0) and (0,60).
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 4.5

24. Linear Programming
CONTINUED
Upon plugging the test-point (0,0) into each of the inequalities, we get true statements for all three. Therefore, the regions containing the test point (below the lines) are the regions to shade.
We are now ready to graph all three constraints. We will graph the first constraint in yellow, the second constraint in red, and the third constraint in blue.
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 4.5

25. Linear Programming On this graph, each tick-mark represents 10 units.
CONTINUED
On this graph, each tick-mark represents 10 units.
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 4.5

26. Linear Programming
CONTINUED
Therefore, the shaded region below represents the constraints.
Blitzer, Intermediate Algebra, 5e – Slide #26 Section 4.5

27. Linear Programming
CONTINUED
The first point is clearly at the origin. So its coordinates are (0,0).
Blitzer, Intermediate Algebra, 5e – Slide #27 Section 4.5

28. Linear Programming
CONTINUED
The next corner is where the lines x + y = 44 and x = 0 meet. Therefore the x-coordinate of the intersection point is 0. To find the y-coordinate we do the following.
x + y = 44
0 + y = 44
y = 44
Therefore, the intersection point is (0,44).
Blitzer, Intermediate Algebra, 5e – Slide #28 Section 4.5

29. Linear Programming
CONTINUED
The next corner is where the lines 16x + 8y = 512 and y = 0 meet. Therefore the y-coordinate of the intersection point is 0. To find the x-coordinate we do the following.
16x + 8y = 512
16x + 8(0) = 512
16x + 0 = 512
16x = 512
x = 32
Therefore, the intersection point is (32,0).
Blitzer, Intermediate Algebra, 5e – Slide #29 Section 4.5

30. Linear Programming
CONTINUED
The next corner is where the lines x + y = 44 and 16x + 8y = 512 meet. To find the intersection point we must solve the system of equations:
16x + 8y = 512
x + y = 44
Upon solving this system, we find the intersection point is (20,24).
Blitzer, Intermediate Algebra, 5e – Slide #30 Section 4.5

31. Linear Programming Corner (x,y) Objective Function
CONTINUED
Now we can determine the value of the objective function at each corner of the graphed region.
Corner
(x,y)
Objective Function
z = 30,000x + 20,000y
(0,0)
z = 30,000(0) + 20,000(0) = = 0
(0,44)
z = 30,000(0) + 20,000(44) = ,000 = 880,000
(32,0)
z = 30,000(32) + 20,000(0) = 960, = 960,000
(20,24)
z = 30,000(20) + 20,000(24) = 600, ,000 = 1,080,000
Blitzer, Intermediate Algebra, 5e – Slide #31 Section 4.5

32. Linear Programming
CONTINUED
Therefore, the maximum value of the objective function is 1,080,000. This occurs when (x,y) = (20,24). That means that the ally’s efforts were maximized when they used 20 American planes and 24 British planes.
Blitzer, Intermediate Algebra, 5e – Slide #32 Section 4.5