## Example ¹1. Solving a Linear Programming Problem Using a Graphical Method. |

x_{1} | + | x_{2} | ≤ | 7 | ||

x_{2} | ≤ | 5 | ||||

x_{1} | + | x_{2} | ≥ | 3 | ||

x_{2} | ≥ | 2 | ||||

x_{1} | ≤ | 4 |

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 the 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.

See the plan for solving this problem in pictures

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} ≤ 7

We need to plot a straight line: x_{1} + x_{2} = 7

Let x_{1} =0 => x_{2} = 7

Let x_{2} =0 => x_{1} = 7

Two points were found: (0, 7) and (7 ,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} ≤ 7

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

x_{2} ≤ - x_{1} + 7

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.

x_{2} ≤ 5

We need to plot a straight line: x_{2} = 5

Now we can plot the straight line (2) through point (0,5) parallel to the OX_{1} axis

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} + x_{2} ≥ 3

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

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

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

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

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

Let's go back to the inequality.

x_{1} + x_{2} ≥ 3

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

x_{2} ≥ - x_{1} + 3

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.

Step ¹4

Let's solve 4 inequality of the system of constraints.

x_{2} ≥ 2

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

Now we can plot the straight line (4) through point (0,2) parallel to the OX_{1} axis

The inequality sign is ≥

Therefore, we must consider points above the straight line (4).

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

Step ¹5

Let's solve 5 inequality of the system of constraints.

x_{1} ≤ 4

We need to plot a straight line: x_{1} = 4

Now we can plot the straight line (5) through point (4,0) parallel to the OX_{2} axis

The inequality sign is ≤

Therefore, we must consider points to the left of the straight line (5).

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

Step ¹6

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

Vector C is not shown at scale because it is too long.

Step ¹7

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

Function F has a maximum value at point A. (see picture)

Let's find the coordinates of point A.

Point A is on the straight line (4) and on the straight line (5) at the same time.

x_{2} | = | 2 | => | x_{1} = 4 | ||||

x_{1} | = | 4 | x_{2} = 2 |

Let's calculate the value of the function F at point A (4,2).

F (A) = 1 * 4 - 1 * 2 = 2

**Comment:** if there is any doubt that the function F has a maximum at point A, you should find the value of the function F at the point of interest and compare it to F (A).

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