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This chapter covers principles of a geometrical approach to linear programming. After completing this chapter students should be able to: solve linear programming problems that maximize the objective function and solve linear programming problems that minimize the objective function.

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.

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 . size 12{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 . size 12{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.

I = 40 x + 30 y size 12{I="40"x+"30"y} {}

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:

x + y 12 size 12{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.

2x + y 16 size 12{2x+y<= "16"} {}

The fact that x size 12{x} {} and y size 12{y} {} can never be negative is represented by the following two constraints:

x 0 size 12{x>= 0} {} , and y 0 size 12{y>= 0} {} .

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

Maximize I = 40 x + 30 y size 12{I="40"x+"30"y} {}

Subject to: x + y 12 size 12{x+y<= "12"} {}

2x + y 16 size 12{2x+y<= "16"} {}
x 0 ; y 0 size 12{x>= 0;``y>= 0} {}

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

 The graph shows that the lines 2x+y=16 and x+y+12 intersect at the point (4,8). The shaded region represents the area of the graph that meets the required conditions.

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.

Critical Points Income
(0,0) 40 ( 0 ) + 30 ( 0 ) = $ 0 size 12{"40" \( 0 \) +"30" \( 0 \) =$0} {}
(0.12) 40 ( 0 ) + 30 ( 12 ) = $ 360 size 12{"40" \( 0 \) +"30" \( "12" \) =$"360"} {}
(4,8) 40 ( 4 ) + 30 ( 8 ) = $ 400 size 12{"40" \( 4 \) +"30" \( 8 \) =$"400"} {}
(8,0) 40 ( 8 ) + 30 ( 0 ) = $ 320 size 12{"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.

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Source:  OpenStax, Applied finite mathematics. OpenStax CNX. Jul 16, 2011 Download for free at http://cnx.org/content/col10613/1.5
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