## Example ¹9. Solving the Linear Programming Problem Using the Graphical Method. |

x_{1} | + | x_{2} | ≥ | 1 | ||

x_{1} | - | x_{2} | ≥ | 1 | ||

x_{1} | ≤ | 1 | ||||

2 x_{1} | + | x_{2} | ≥ | 1 | ||

x_{1} | + | 2 x_{2} | ≤ | 7 |

x

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: x_{1} ≥ 0 x_{2} ≥ 0.

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

Step ¹1

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

x_{1} + x_{2} ≥ 1

We need to plot a straight line: x_{1} + x_{2} = 1

Let x_{1} =0 => x_{2} = 1

Let x_{2} =0 => x_{1} = 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.

x_{1} + x_{2} ≥ 1

We need to transform the inequality so that only x_{2} is on the left side.

x_{2} ≥ - x_{1} + 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 ¹2

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

x_{1} - x_{2} ≥ 1

We need to plot a straight line: x_{1} - x_{2} = 1

Let x_{1} =0 => - x_{2} = 1 => x_{2} = -1

Let x_{2} =0 => x_{1} = 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.

x_{1} - x_{2} ≥ 1

We need to transform the inequality so that only x_{2} is on the left side.

- x_{2} ≥ - x_{1} + 1

x_{2} ≤ x_{1} - 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 ¹3

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

x_{1} ≤ 1

We need to plot a straight line: x_{1} = 1

Now we can plot the straight line (3) through point (1,0) parallel to the OX_{2} 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 ¹4

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 ¹5

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

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