Service for Solving Linear Programming Problems

Let good people look good solutions

Example 7. Solving the Linear Programming Problem Using the Graphical Method.
Function Increases Unlimitedly

This solution is done by the program presented on the site.
Problem:
Find the maximum value of the function

F = x1 + x2

subject to the constraints:

 3 x1 + 2 x2 6
- x1 + x2 1
x1 - 2 x2 1

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 3)

The last two steps are necessary to get an answer. (see step 4 - step 5)

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.

3 x1 + 2 x2  ≥  6

We need to plot a straight line: 3 x1 + 2 x2 = 6

Let x1 =0 => 2 x2 = 6 => x2 = 3

Let x2 =0 => 3 x1 = 6 => x1 = 2

Two points were found: (0, 3) and (2 ,0)

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

Let's go back to the inequality.

3 x1 + 2 x2  ≥  6

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

2 x2  ≥  - 3 x1 + 6

x2  ≥  - 3/2 x1 + 3

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

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

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 - 2 x2  ≤  1

We need to plot a straight line: x1 - 2 x2 = 1

Let x1 =0 => - 2 x2 = 1 => x2 = -1/2

Let x2 =0 => x1 = 1

Two points were found: (0, -1/2) and (1 ,0)

Now we can plot the straight line (3) through the found two points.

Let's go back to the inequality.

x1 - 2 x2  ≤  1

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

- 2 x2  ≤  - x1 + 1

x2  ≥  1/2 x1 - 1/2

The inequality sign is  ≥
Therefore, we must consider points above the straight line (3).

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

Picture 3. Graphical Method of Solving LPP.

Step 4

We need to plot the vector C = (1, 1), whose coordinates are the coefficients of the function F.

Picture 4. Graphical Method of Solving LPP.

Step 5

We will move a "red" straight line perpendicular to vector C from the lower left corner to the upper right corner.

The function F has a minimum value at the point where the "red" straight line crosses the region of feasible solutions for the first time.

The function F has a maximum value at the point where the "red" straight line crosses the region of feasible solutions for the last time.

It is impossible to find the point where the "red" line crosses the region of feasible solutions for the last time, i.s. the function increases indefinitely (see picture).

Picture 5. Graphical Method of Solving LPP.

Result:

Fmax = + ∞









2010-2020 If you have comments, please write to matematika1974@yandex.ru