§ 4.5

Linear Programming

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 4.5

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 #3 Section 4.5

Linear Programming

Basic Linear ProgrammingObjective 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 ProblemLet 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 #4 Section 4.5

Linear Programming

EXAMPLE - Writing an objective functionEXAMPLE - 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 andLet y = # hours spent each week working at the newspaper office

Write an objective function that models total weekly earnings.

We can represent this statement by:z = 10x + 8y

Total dollars earned in a week =( x hours) times (10 dollars) +( y hours) times (8 dollars)

Blitzer, Intermediate Algebra, 5e – Slide #5 Section 4.5

Linear Programming

Continued

The student decides that he should work no more than 15 hours per weekso as to save time for his studies.

He must spend at least 3 hours per week tutoring since the tutoring centerrequires that.

He can not spend more than 8 hours per week tutoring since that is all thetutoring center will allow.

What system of inequalities models these three constraints?

8

3

15

x

x

yx

Blitzer, Intermediate Algebra, 5e – Slide #6 Section 4.5

Linear Programming

EXAMPLEEXAMPLE

(a) Graph the system of inequalities representing the constraints.

(b) Find the value of the objective function at each corner of the graphed region.

(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.

Objective Function50 x30 y

z = 5x – 2yConstraints

2 yx

Blitzer, Intermediate Algebra, 5e – Slide #7 Section 4.5

Linear Programming

(a) 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.

SOLUTIONSOLUTION

CONTINUECONTINUEDD

2 yx

Blitzer, Intermediate Algebra, 5e – Slide #8 Section 4.5

Linear Programming

2 yx

CONTINUECONTINUEDD Now we find the intercepts for the equation.

x + y = 2

We set y = 0 to find the x-intercept:

We set x = 0 to find the y-intercept:

x + y = 2x + 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 #9 Section 4.5

Linear Programming

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.

50 x

30 y

2 yx

CONTINUECONTINUEDD

Blitzer, Intermediate Algebra, 5e – Slide #10 Section 4.5

Linear Programming

Therefore, the shaded region below represents the constraints.CONTINUECONTINUE

DD

50 x

30 y

2 yx

Blitzer, Intermediate Algebra, 5e – Slide #11 Section 4.5

Linear Programming

CONTINUECONTINUEDD (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).

50 x

30 y

2 yx

Blitzer, Intermediate Algebra, 5e – Slide #12 Section 4.5

Linear Programming

CONTINUECONTINUEDD The next corner is where the lines x = 5 and y = 3 meet.

Therefore the coordinates of this point are (5,3).

50 x

30 y

2 yx

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 #13 Section 4.5

Linear Programming

CONTINUECONTINUEDD 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.

50 x

30 y

2 yx

x + y = 2

x + 0 = 2

x = 2

Therefore, the intersection point is (2,0).

Blitzer, Intermediate Algebra, 5e – Slide #14 Section 4.5

Linear Programming

CONTINUECONTINUEDD 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.

50 x

30 y

2 yx

x + y = 2

0 + y = 2

y = 2

Therefore, the intersection point is (0,2).

Blitzer, Intermediate Algebra, 5e – Slide #15 Section 4.5

Linear Programming

CONTINUECONTINUEDD 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 #16 Section 4.5

Linear Programming

CONTINUECONTINUEDD

The maximum value of the objective function is 25. This occurs when (x,y) = (5,0).

(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.

Blitzer, Intermediate Algebra, 5e – Slide #17 Section 4.5

Linear Programming

EXAMPLEEXAMPLE

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 #18 Section 4.5

Linear Programming

•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.

CONTINUECONTINUEDD

Blitzer, Intermediate Algebra, 5e – Slide #19 Section 4.5

Linear Programming

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.

CONTINUECONTINUEDD

SOLUTIONSOLUTION

Blitzer, Intermediate Algebra, 5e – Slide #20 Section 4.5

Linear Programming

Now we will determine the inequality constraints.

Since the number of planes cannot exceed 44 we get:

CONTINUECONTINUEDD

44 yx

512816 yx

000,300000,5000,9 yx

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 #21 Section 4.5

Linear Programming

1) Graph the system of inequalities representing the constraints.

CONTINUECONTINUEDD

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,

x + y = 44

We set y = 0 to find the x-intercept:

We set x = 0 to find the y-intercept:

x + y = 44x + 0 = 44 0 + y = 44

x = 44 y = 44Therefore, the two intercepts for the equation are: (44,0) and (0,44).

Blitzer, Intermediate Algebra, 5e – Slide #22 Section 4.5

Linear Programming

CONTINUECONTINUEDD Now we will determine the intercepts of the second

inequality. Upon rewriting it with an = symbol, we get 16x + 8y = 512. Thus,

16x + 8y = 512

We set y = 0 to find the x-intercept:

We set x = 0 to find the y-intercept:

16x + 8y = 51216x + 8(0) = 512 16(0) + 8y = 512

16x + 0 = 512 0 + 8y = 512

Therefore, the two intercepts for the equation are: (32,0) and (0,64).

16x = 512 8y = 512x = 32 y = 64

Blitzer, Intermediate Algebra, 5e – Slide #23 Section 4.5

Linear Programming

CONTINUECONTINUEDD 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,

9,000x + 5,000y = 300,000

We set y = 0 to find the x-intercept:

We set x = 0 to find the y-intercept:

9,000x + 5,000(0) = 300,000 9,000(0) + 5,000y = 300,0009,000x + 0 = 300,000 0 + 5,000y = 300,000

Therefore, the two intercepts for the equation are: (33.3,0) and (0,60).

9,000x = 300,000 5,000y = 300,000x = 33.3 y = 60

9,000x + 5,000y = 300,000

Blitzer, Intermediate Algebra, 5e – Slide #24 Section 4.5

Linear Programming

CONTINUECONTINUEDD 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 #25 Section 4.5

Linear Programming

CONTINUECONTINUEDD

44 yx

512816 yx

000,300000,5000,9 yx

On this graph, each tick-mark represents 10 units.

Blitzer, Intermediate Algebra, 5e – Slide #26 Section 4.5

Linear Programming

CONTINUECONTINUEDD

44 yx

512816 yx

000,300000,5000,9 yx

Therefore, the shaded region below represents the constraints.

Blitzer, Intermediate Algebra, 5e – Slide #27 Section 4.5

Linear Programming

CONTINUECONTINUEDD

44 yx

512816 yx

000,300000,5000,9 yx

The first point is clearly at the origin. So its coordinates are (0,0).

Blitzer, Intermediate Algebra, 5e – Slide #28 Section 4.5

Linear Programming

CONTINUECONTINUEDD

44 yx

512816 yx

000,300000,5000,9 yx

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 #29 Section 4.5

Linear Programming

CONTINUECONTINUEDD

44 yx

512816 yx

000,300000,5000,9 yx

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

Therefore, the intersection point is (32,0).

16x + 0 = 512

16x + 8(0) = 512

16x = 512x = 32

Blitzer, Intermediate Algebra, 5e – Slide #30 Section 4.5

Linear Programming

CONTINUECONTINUEDD

44 yx

512816 yx

000,300000,5000,9 yx

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

Upon solving this system, we find the intersection point is (20,24).

x + y = 44

Blitzer, Intermediate Algebra, 5e – Slide #31 Section 4.5

Linear Programming

CONTINUECONTINUEDD 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 = 0

(0,44) z = 30,000(0) + 20,000(44) = 0 + 880,000 = 880,000

(32,0) z = 30,000(32) + 20,000(0) = 960,000 + 0 = 960,000

(20,24) z = 30,000(20) + 20,000(24) = 600,000 + 480,000 = 1,080,000

Blitzer, Intermediate Algebra, 5e – Slide #32 Section 4.5

Linear Programming

CONTINUECONTINUEDD

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.