## Example ¹10. Solving the Linear Programming Problem Using the Graphical Method. |

3 x_{1} | + | 5 x_{2} | ≤ | 30 | ||

4 x_{1} | - | 3 x_{2} | ≤ | 12 | ||

x_{1} | - | 3 x_{2} | ≥ | 6 |

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 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: 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.

3 x_{1} + 5 x_{2} ≤ 30

We need to plot a straight line: 3 x_{1} + 5 x_{2} = 30

Let x_{1} =0 => 5 x_{2} = 30 => x_{2} = 6

Let x_{2} =0 => 3 x_{1} = 30 => x_{1} = 10

Two points were found: (0, 6) and (10 ,0)

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

Let's go back to the inequality.

3 x_{1} + 5 x_{2} ≤ 30

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

5 x_{2} ≤ - 3 x_{1} + 30

x_{2} ≤ - 3/5 x_{1} + 6

The inequality sign is ≤

Therefore, we must consider points below 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.

4 x_{1} - 3 x_{2} ≤ 12

We need to plot a straight line: 4 x_{1} - 3 x_{2} = 12

Let x_{1} =0 => - 3 x_{2} = 12 => x_{2} = -4

Let x_{2} =0 => 4 x_{1} = 12 => x_{1} = 3

Two points were found: (0, -4) and (3 ,0)

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

Let's go back to the inequality.

4 x_{1} - 3 x_{2} ≤ 12

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

- 3 x_{2} ≤ - 4 x_{1} + 12

x_{2} ≥ 4/3 x_{1} - 4

The inequality sign is ≥

Therefore, we must consider points above 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} - 3 x_{2} ≥ 6

We need to plot a straight line: x_{1} - 3 x_{2} = 6

Let x_{1} =0 => - 3 x_{2} = 6 => x_{2} = -2

Let x_{2} =0 => x_{1} = 6

Two points were found: (0, -2) and (6 ,0)

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

Let's go back to the inequality.

x_{1} - 3 x_{2} ≥ 6

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

- 3 x_{2} ≥ - x_{1} + 6

x_{2} ≤ 1/3 x_{1} - 2

The inequality sign is ≤

Therefore, we must consider points below the straight line (3).

This result does not have common points with the region of possible solutions of the previous step (see picture).

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