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  • Chapter 5Linear Programming: A Geometric Approach

    Chapter Overview

    In this chapter, you will learn to:

    1. Solve linear programming problems that maximize the objective function.

    2. Solve linear programming problems that minimize the objective function.

    Maximization Applications

    Application problems in business, economics, and social and life sciences often ask us to make decisions on the basis of certain conditions. These conditions or constraints often take the form of inequalities. In this section, we will look at such problems.

    A typical linear programming problem consists of finding an extreme value of a linear function subject to certain constraints. We are either trying to maximize or minimize our function. That is why these linear programming problems are classified as maximization or minimization problems, or just optimization problems. The function we are trying to optimize is called an objective function, and the conditions that must be satisfied are called constraints. In this chapter, we will do problems that involve only two variables, and therefore, can be solved by graphing. We begin by solving a maximization problem.

    Example 5.1

    Niki holds two part-time jobs, Job I and Job II. She never wants to work more than a total of 12 hours a week. She has determined that for every hour she works at Job I, she needs 2 hours of preparation time, and for every hour she works at Job II, she needs one hour of preparation time, and she cannot spend more than 16 hours for preparation. If she makes $40 an hour at Job I, and $30 an hour at Job II, how many hours should she work per week at each job to maximize her income?

    We start by choosing our variables.

    Let x=The number of hours per week Niki will work at Job I.

    and y=The number of hours per week Niki will work at Job II.

    Now we write the objective function. Since Niki gets paid $40 an hour at Job I, and $30 an hour at Job II, her total income I is given by the following equation.

    (5.1)I=40x+30y

    Our next task is to find the constraints. The second sentence in the problem states, "She never wants to work more than a total of 12 hours a week." This translates into the following constraint:

    (5.2)x+y≤12

    The third sentence states, "For every hour she works at Job I, she needs 2 hours of preparation time, and for every hour she works at Job II, she needs one hour of preparation time, and she cannot spend more than 16 hours for preparation." The translation follows.

    (5.3)2x+y≤16

    The fact that x and y can never be negative is represented by the following two constraints:

    x≥0, and y≥0.

    Well, good news! We have formulated the problem. We restate it as

    Maximize I = 40 x + 30 y

    Subject to: x + y ≤ 12

    (5.4)2x+y≤16
    (5.5)
    _autogen-svg2png-0013.png

    In order to solve the problem, we graph the constraints as follows.

    graphics1.png
    Figure 5.0

    Observe that we have shaded the region where all conditions are satisfied. This region is called the feasibility region or the feasibility polygon.

    The Fundamental Theorem of Linear Programming states that the maximum (or minimum) value of the objective function always takes place at the vertices of the feasibility region.

    Therefore, we will identify all the vertices of the feasibility region. We call these points critical points. They are listed as (0, 0), (0, 12), (4, 8), (8, 0). To maximize Niki's income, we will substitute these points in the objective function to see which point gives us the highest income per week. We list the results below.

    Table 5.1.
    Critical PointsIncome
    (0,0) 40 ( 0 ) + 30 ( 0 ) = $ 0
    (0.12) 40 ( 0 ) + 30 ( 12 ) = $ 360
    (4,8) 40 ( 4 ) + 30 ( 8 ) = $ 400
    (8,0) 40 ( 8 ) + 30 ( 0 ) = $ 320

    Clearly, the point (4, 8) gives the most profit: $400.

    Therefore, we conclude that Niki should work 4 hours at Job I, and 8 hours at Job II.

    Example 5.2

    A factory manufactures two types of gadgets, regular and premium. Each gadget requires the use of two operations, assembly and finishing, and there are at most 12 hours available for each operation. A regular gadget requires 1 hour of assembly and 2 hours of finishing, while a premium gadget needs 2 hours of assembly and 1 hour of finishing. Due to other restrictions, the company can make at most 7 gadgets a day. If a profit of $20 is realized for each regular gadget and $30 for a premium gadget, how many of each should be manufactured to maximize profit?

    We choose our variables.

    Let x=The number of regular gadgets manufactured each day.

    and y=The number of premium gadgets manufactured each day.

    The objective function is

    (5.6)P=20x+30y

    We now write the constraints. The fourth sentence states that the company can make at most 7 gadgets a day. This translates as

    (5.7)x+y≤7

    Since the regular gadget requires one hour of assembly and the premium gadget requires two hours of assembly, and there are at most 12 hours available for this operation, we get

    (5.8)x+2y≤12

    Similarly, the regular gadget requires two hours of finishing and the premium gadget one hour. Again, there are at most 12 hours available for finishing. This gives us the following constraint.

    (5.9)2x+y≤12

    The fact that x and y can never be negative is represented by the following two constraints:

    x≥0, and y≥0.

    We have formulated the problem as follows:

    Maximize P = 20 x + 30 y

    Subject to: x + y ≤ 7

    (5.10)x+2y≤12
    (5.11)2x+y≤12
    (5.12)x≥0;y≥0

    In order to solve the problem, we graph the constraints as follows:

    graphics2.png
    Figure 5.0

    Again, we have shaded the feasibility region, where all constraints are satisfied.

    Since the extreme value of the objective function always takes place at the vertices of the feasibility region, we identify all the critical points. They are listed as (0, 0), (0, 6), (2, 5), (5, 2), and (6, 0). To maximize profit, we will substitute these points in the objective function to see which point gives us the maximum profit each day. The results are listed below.

    Table 5.2.
    Critical pointIncome
    (0,0) 20 ( 0 ) + 30 ( 0 ) = $ 0
    (0,6) 20 ( 0 ) + 30 ( 6 ) = $ 180
    (2,5) 20 ( 2 ) + 30 ( 5 ) = $ 190
    (5,2) 20 ( 5 ) + 30 ( 2 ) = $ 160
    (6,0) 20 ( 6 ) + 30 ( 0 ) = $ 120

    The point (2, 5) gives the most profit, and that profit is $190. Therefore, we conclude that we should manufacture 2 regular gadgets and 5 premium gadgets daily for a profit of $190.

    Although we are mostly focusing on the standard maximization problems where all constraints are of the form _autogen-svg2png-0038.png, we will now consider an example where that is not the case.

    Example 5.3

    Solve the following maximization problem graphically.

    Maximize P = 10 x + 15 y

    Subject to: x + y ≥ 1

    (5.13)x+2y≤6
    (5.14)2x+y≤6
    (5.15)x≥0;y≥0

    The graph is shown below.

    graphics3.png
    Figure 5.0

    The five critical points are listed in the above figure. Clearly, the point (2, 2) maximizes the objective function to a maximum value of 50. The reader should observe that the first constraint x+y≥1 requires that feasibility region must be bounded below by the line x+y=1.

    Finally, we address an important question. Is it possible to determine the point that gives the maximum value without calculating the value at each critical point?

    The answer is yes.

    For example, in the above problem, we substituted the points (0, 0), (0, 6), (2, 5), (5, 2), and (6, 0), in the objective function P=20x+30y, and we got the values $0, $180, $190, $160, $120, respectively. Sometimes that is not the most efficient way of finding the optimum solution.

    To determine the largest P, we graph P=20x+30y for any value P of our choice. Let us say, we choose P=60. We graph 20x+30y=60. Now we move the line parallel to itself, that is, keeping the same slope at all times. Since we are moving the line parallel to itself, the slope is kept the same, and the only thing that is changing is the P. As we move away from the origin, the value of P increases. The largest value of P is realized when the line touches the last corner point. The figure below shows the movements of the line, and the optimum solution is achieved at the point (2, 5). In maximization problems, as the line is being moved away from the origin, this optimum point is the farthest critical point.

    graphics4.png
    Figure 5.1

    We summarize:

    Example 5.4The Maximization Linear Programming Problems
    1. Write the objective function.

    2. Write the constraints.

      1. For the standard maximization linear programming problems, constraints are of the form: _autogen-svg2png-0055.png

      2. Since the variables are non-negative, we include the constraints: x≥0;y≥0.

    3. Graph the constraints.

    4. Shade the feasibility region.

    5. Find the corner points.

    6. Determine the corner point that gives the maximum value.

      1. This is done by finding the value of the objective function at each corner point.

      2. This can also be done by moving the line associated with the objective function.

    Minimization Applications

    Minimization linear programming problems are solved in much the same way as the maximization problems. For the standard minimization linear programming problem, the constraints are of the form _autogen-svg2png-0057.png, as opposed to the form _autogen-svg2png-0058.png for the standard maximization problem. As a result, the feasible solution extends indefinitely to the upper right of the first quadrant, and is unbounded. But that is not a concern, since in order to minimize the objective function, the line associated with the objective function is moved towards the origin, and the critical point that minimizes the function is closest to the origin.

    However, one should be aware that in the case of an unbounded feasibility region, the possibility of no optimal solution exists.

    Example 5.5

    Professor Symons wishes to employ two students, John and Mary, to grade the homework papers for his classes. John can mark 20 papers per hour and charges $5 per hour, and Mary can mark 30 papers per hour and charges $8 per hour. Each student must be employed at least one hour a week to justify their employment. If Mr. Symons has at least 110 homework papers to be marked each week, how many hours per week should he employ each student to minimize his cost?

    We choose the variables as follows:

    Let x=The number of hours per week John is employed.

    and y=The number of hours per week Mary is employed.

    The objective function is

    (5.16)C=5x+8y

    The fact that each student must work at least one hour each week results in the following two constraints:

    (5.17)x≥1
    (5.18)y≥1

    Since John can grade 20 papers per hour and Mary 30 papers per hour, and there are at least 110 papers to be graded per week, we get

    (5.19)20x+30y≥110

    The fact that x and y are non-negative, we get

    x≥0, and y≥0.

    The problem has been formulated as follows.

    Minimize C = 5x + 8y

    Subject to: x ≥ 1

    (5.20)y≥1
    (5.21)20x+30y≥110
    (5.22)x≥0;y≥0

    To solve the problem, we graph the constraints as follows:

    graphics5.png
    Figure 5.1

    Again, we have shaded the feasibility region, where all constraints are satisfied.

    Since the extreme value of the objective function always takes place at the vertices of the feasibility region, we identify the two critical points, (1, 3) and (4, 1). To minimize cost, we will substitute these points in the objective function to see which point gives us the minimum cost each week. The results are listed below.

    Table 5.3.
    Critical PointsIncome
    (1,3) 5 ( 1 ) + 8 ( 3 ) = $ 29
    (4,1) 5 ( 4 ) + 8 ( 1 ) = $ 28

    The point (4, 1) gives the least cost, and that cost is $28. Therefore, we conclude that Professor Symons should employ John 4 hours a week, and Mary 1 hour a week at a cost of $28 per week.

    Example 5.6

    Professor Hamer is on a low cholesterol diet. During lunch at the college cafeteria, he always chooses between two meals, Pasta or Tofu. The table below lists the amount of protein, carbohydrates, and vitamins each meal provides along with the amount of cholesterol he is trying to minimize. Mr. Hamer needs at least 200 grams of protein, 960 grams of carbohydrates, and 40 grams of vitamins for lunch each month. Over this time period, how many days should he have the Pasta meal, and how many days the Tofu meal so that he gets the adequate amount of protein, carbohydrates, and vitamins and at the same time minimizes his cholesterol intake?

    Table 5.4.
     PastaTofu
    Protein8g16g
    Carbohydrates60g40g
    Vitamin C2g2g
    Cholesterol60mg50mg

    We choose the variables as follows.

    Let x = The number of days Mr. Hamer eats Pasta.

    and y = The number of days Mr. Hamer eats Tofu.

    Since he is trying to minimize his cholesterol intake, our objective function represents the total amount of cholesterol C provided by both meals.

    (5.23)C=60x+50y

    The constraint associated with the total amount of protein provided by both meals is as follows:

    (5.24)8x+16y≥200

    Similarly, the two constraints associated with the total amount of carbohydrates and vitamins are obtained, and they are

    (5.25)60x+40y≥960
    (5.26)2x+2y≥40

    The constraints that state that x and y are non-negative are

    (5.27)
    _autogen-svg2png-0078.png

    We summarize all information as follows:

    Minimize C=60x+50y

    Subject to: 8x+16y≥200

    (5.28)60x+40y≥960
    (5.29)2x+2y≥40
    (5.30)x≥0;y0

    To solve the problem, we graph the constraints as follows.

    graphics6.png
    Figure 5.1

    Again, we have shaded the unbounded feasibility region, where all constraints are satisfied.

    To minimize the objective function, we find the vertices of the feasibility region. These vertices are (0, 24), (8, 12), (15, 5) and (25, 0). To minimize cholesterol, we will substitute these points in the objective function to see which point gives us the smallest value. The results are listed below.

    Table 5.5.
    Critical PointsIncome
    (0,24) 60 ( 0 ) + 50 ( 24 ) = 1200
    (8,12) 60 ( 8 ) + 50 ( 12 ) = 1080
    (15,5) 60 ( 15 ) + 50 ( 5 ) = 1150
    (25,0) 60 ( 25 ) + 50 ( 0 ) = 1500

    The point (8, 12) gives the least cholesterol, which is 1080 mg. This states that for every 20 meals, Professor Hamer should eat Pasta 8 days, and Tofu 12 days.

    Although the method of solving minimization problems is similar to that of the maximization problems, we still feel that we should summarize the steps involved.

    Example 5.7Minimization Linear Programming Problems
    1. Write the objective function.

    2. Write the constraints.

      1. For standard minimization linear programming problems, constraints are of the form: _autogen-svg2png-0088.png

      2. Since the variables are non-negative, include the constraints: x≥0; y≥0.

    3. Graph the constraints.

    4. Shade the feasibility region.

    5. Find the corner points.

    6. Determine the corner point that gives the minimum value.

      1. This can be done by finding the value of the objective function at each corner point.

      2. This can also be done by moving the line associated with the objective function.

      3. There is the possibility that the problem has no solution.

    Solutions