Service for Solving Linear Programming Problems

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Example 9. Solving the Linear Programming Problem Using the Graphical Method.
Region of Feasible Solutions is a Point

This solution has been done by the calculator presented on the site.
Problem:
Find the maximum value of the function

F = x1 + x2

subject to the constraints:

 x1 + x2 1
x1 - x2 1
x1 1
2 x1 + x2 1
x1 + 2 x2 7

x1 ≥ 0     x2 ≥ 0
Solution:

Points whose coordinates satisfy all the inequalities of the constraint system are called a region of feasible solutions.

It is necessary to solve each inequality of the constraint system to find the region of feasible solutions to this problem. (see step 1 - step 5)

The last two steps are necessary to get an answer. (see step 6 - step 7)

This is a standard solution plan. If the region of feasible solutions is a point or an empty set then the solution will be shorter.

By the condition of the problem: x1 ≥ 0     x2 ≥ 0.

Now we have the region of feasible solutions shown in the picture.

Picture 0. Graphical Method of Solving LPP.

Step 1

Let's solve 1 inequality of the system of constraints.

x1 + x2  ≥  1

We need to plot a straight line: x1 + x2 = 1

Let x1 =0 => x2 = 1

Let x2 =0 => x1 = 1

Two points were found: (0, 1) and (1 ,0)

Now we can plot the straight line (1) through the found two points.

Let's go back to the inequality.

x1 + x2  ≥  1

We need to transform the inequality so that only x2 is on the left side.

x2  ≥  - x1 + 1

The inequality sign is  ≥
Therefore, we must consider points above the straight line (1).

Let's combine this result with the previous picture.
Now we have the region of feasible solutions shown in the picture.

Picture 1. Graphical Method of Solving LPP.

Step 2

Let's solve 2 inequality of the system of constraints.

x1 - x2  ≥  1

We need to plot a straight line: x1 - x2 = 1

Let x1 =0 => - x2 = 1 => x2 = -1

Let x2 =0 => x1 = 1

Two points were found: (0, -1) and (1 ,0)

Now we can plot the straight line (2) through the found two points.

Let's go back to the inequality.

x1 - x2  ≥  1

We need to transform the inequality so that only x2 is on the left side.

- x2  ≥  - x1 + 1

x2  ≤  x1 - 1

The inequality sign is  ≤
Therefore, we must consider points below the straight line (2).

Let's combine this result with the previous picture.
Now we have the region of feasible solutions shown in the picture.

Picture 2. Graphical Method of Solving LPP.

Step 3

Let's solve 3 inequality of the system of constraints.

x1  ≤  1

We need to plot a straight line: x1 = 1

Now we can plot the straight line (3) through point (1,0) parallel to the OX2 axis

The inequality sign is  ≤
Therefore, we must consider points to the left of the straight line (3).

Let's combine this result with the previous picture.
Now the region of feasible solutions is a point A (see picture).

The coordinates of point A (1,0) are known. (see step 1)

Picture 3. Graphical Method of Solving LPP.

Step 4

We need to check whether the coordinates of point A (1,0) satisfy 4 inequality of the constraint system?

2 * 1 + 1 * 0  ≥  1

2  ≥  1

Yes, that's right.

Step 5

We need to check whether the coordinates of point A (1,0) satisfy 5 inequality of the constraint system?

1 * 1 + 2 * 0  ≤  7

1  ≤  7

Yes, that's right.

Let's calculate the value of the function F at point A (1, 0).

F (A) = 1 * 1 + 1 * 0 = 1

Result:

x1 = 1

x2 = 0

Fmax = 1









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