# 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. 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. 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. 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) 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: