An Introduction to Linear Programming
Introduction: In this lesson, a function in two variables will be either maximized or minimized according to given constraints. This lesson will require that you be able to graph both linear equations and linear inequalities. You will also need to be able to determine the intersection points where the lines in the system cross or share common values.

Additional lessons are available to assist you in how to calculate the solution to a linear system of equations, graph a system of linear inequalities, as well as how to use a graphing calculator to graph a linear system of equations.

The Lesson:

Suppose we are given the following information:

Based upon this information, we want to maximize the function P(x, y) = 400x + 500y. That is, we want to determine the maximum value of P(x, y).

To do this, we will use a technique known as linear programming, a graphical approach. Since there are only two variables, x and y, we can easily graph our initial given information.

Please examine and note the following features on our graph:

1. the graph is restricted to the first quadrant because
2. the graph of is delineated with vertical lines -- it has a y-intercept of 4 and an x-intercept of 6
3. the graph of  is delineated with horizontal lines -- it has a y-intercept of 3 and an x-intercept of 9
4. the doubly shaded region conforms to all of the constraints and has been shaded in yellow and labeled as region R in the second graph

The four original inequalities are called the constraints and P(x, y) is called the objective function.

Note in the Y= screen that we had to solve for y in order to graph the lines.

In the following graph, all four vertices have been identified and labeled. An additional line, having a slope of - 4/5, has also been included. The slope of this line matches the slope of our objective function P(x, y).

Our goal is to maximize the value of P represented by the additional line. Pwill be maximized or minimized when the line intersects a vertex of region defined by the constraints.

 Figure A Figure B Figure C

As our line is moved toward the origin point (0, 0), the first vertex it intersects in the vertex point (6, 0) [Figure A].

P(x, y) = 400x + 500y has a value of P(6, 0) = 400(6) + 500(0) = 2400.

As we continue to move this line closer to the origin [Figure B], we will next intersect the vertex point (3, 2).

P(x, y) = 400x + 500y has a value of P(3, 2) = 400(3) + 500(2) = 2200.

Continuing our journey [Figure C], we intersect the vertex  point (0, 3).

P(x, y) = 400x + 500y has a value of P(0, 3) = 400(0) + 500(3) = 1500.

Finally, when we arrive at (0, 0), P(x, y) has a value of  P(0, 0) = 0.

Clearly the maximum for P(x, y) is obtained at the point (6, 0) where the value of P is 2400. Similarly, the minimum value is obtained at the point (0,0) where P(0,0) = 0.

To summarize, the linear programming method is applied to an objective function, given constraint conditions. The maximum (and minimum) values of this function are found at the vertices of the region defined by the constraints.

Let's Practice:

Based on the following constraints, at what point would an objective function P(x, y) = 50x + 20y be maximized?

We begin by solving each equation for y and entering them into the calculator.

Please examine and note the following features of our graph:

1. the graph is restricted to the first quadrant because
2. the graph of is delineated with vertical lines -- it has a y-intercept of 18 and an x-intercept of 6
3. the graph of  is delineated with horizontal lines -- it has a y-intercept of 14 and an x-intercept of 7
4. the doubly shaded region conforms to all of the constraints

In the next graph, all four vertices have been identified and labeled. An additional line, having a slope of - 5/2, has also been included. The slope of this line matches the slope of our objective function P(x, y).

The objective function is P(x, y) = 50x + 20y. When evaluated at each vertex, P(x, y) yields the following values:

P(0, 14) = 50(0) + 20(14) = 280
P(4, 6) = 50(4) + 20(6) = 200 + 120 = 320
P(6, 0) = 50(6) + 20(0) = 300
P(0, 0) = 50(0) + 20(0) = 0

Clearly, the maximum value of P(x, y) is 320 at the point (4, 6) and the minimum value of P(x, y) is 0 at the point (0, 0).

Notice that in order to solve this problem, you must know the coordinates of all vertices, including the intersection of the two lines, the point (4, 6).

Example
 Minimize the function subject to the following constraints. What is your answer?

M Ransom

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