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§ 4.5 Linear Programming

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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.

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Blitzer, Intermediate Algebra, 5e – Slide #3 Section 4.5 Linear Programming Basic Linear Programming 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.

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Blitzer, Intermediate Algebra, 5e – Slide #4 Section 4.5 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. 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)

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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 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?

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Blitzer, Intermediate Algebra, 5e – Slide #6 Section 4.5 Linear ProgrammingEXAMPLE (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 Functionz = 5x – 2y Constraints

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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. Lets determine the graph of the third constraint. We first rewrite the inequality as x + y = 2. SOLUTION CONTINUED

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Blitzer, Intermediate Algebra, 5e – Slide #8 Section 4.5 Linear ProgrammingCONTINUED 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 = 2 x + 0 = 20 + y = 2 x = 2y = 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.

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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. CONTINUED

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

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Blitzer, Intermediate Algebra, 5e – Slide #11 Section 4.5 Linear ProgrammingCONTINUED (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).

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Blitzer, Intermediate Algebra, 5e – Slide #12 Section 4.5 Linear ProgrammingCONTINUED 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).

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Blitzer, Intermediate Algebra, 5e – Slide #13 Section 4.5 Linear ProgrammingCONTINUED 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).

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Blitzer, Intermediate Algebra, 5e – Slide #14 Section 4.5 Linear ProgrammingCONTINUED 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).

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

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Blitzer, Intermediate Algebra, 5e – Slide #16 Section 4.5 Linear ProgrammingCONTINUED 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.

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

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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. CONTINUED

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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. CONTINUED SOLUTION

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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: CONTINUED 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:

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Blitzer, Intermediate Algebra, 5e – Slide #21 Section 4.5 Linear Programming 1) Graph the system of inequalities representing the constraints. CONTINUED 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 = 44 x + 0 = 440 + y = 44 x = 44y = 44 Therefore, the two intercepts for the equation are: (44,0) and (0,44).

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Blitzer, Intermediate Algebra, 5e – Slide #22 Section 4.5 Linear ProgrammingCONTINUED 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 = 512 16x + 8(0) = 51216(0) + 8y = 512 16x + 0 = 5120 + 8y = 512 Therefore, the two intercepts for the equation are: (32,0) and (0,64). 16x = 5128y = 512 x = 32 y = 64

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Blitzer, Intermediate Algebra, 5e – Slide #23 Section 4.5 Linear ProgrammingCONTINUED 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) = 300,0009,000(0) + 5,000y = 300,000 9,000x + 0 = 300,0000 + 5,000y = 300,000 Therefore, the two intercepts for the equation are: (33.3,0) and (0,60). 9,000x = 300,0005,000y = 300,000 x = 33.3 y = 60 9,000x + 5,000y = 300,000

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

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

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

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

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Blitzer, Intermediate Algebra, 5e – Slide #28 Section 4.5 Linear ProgrammingCONTINUED 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).

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Blitzer, Intermediate Algebra, 5e – Slide #29 Section 4.5 Linear ProgrammingCONTINUED 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 = 512 x = 32

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

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

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

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