## Example ¹5. Solving the Linear Programming Problem Using the Graphical Method. |

x_{1} | - | 2 x_{2} | ≤ | 4 | ||

x_{1} | - | x_{2} | ≥ | 1 | ||

x_{1} | + | x_{2} | ≥ | 2 |

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.

x_{1} - 2 x_{2} ≤ 4

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

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

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

Two points were found: (0, -2) and (4 ,0)

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

Let's go back to the inequality.

x_{1} - 2 x_{2} ≤ 4

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

- 2 x_{2} ≤ - x_{1} + 4

x_{2} ≥ 1/2 x_{1} - 2

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} + x_{2} ≥ 2

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

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

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

Two points were found: (0, 2) and (2 ,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} ≥ 2

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

x_{2} ≥ - x_{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.

Step ¹4

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 ¹5

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

There is an assumption that the function F has a maximum value on the ray, which has its start at point A. (see picture)

Let's find the coordinates of point A.

Point A is on the straight line (2) and the straight line (3) at the same time.

x_{1} | - | x_{2} | = | 1 | => | x_{1} = 3/2 | ||

x_{1} | + | x_{2} | = | 2 | x_{2} = 1/2 |

Let's calculate the value of the function F at point A (3/2,1/2).

F (A) = -1 * 3/2 + 1 * 1/2 = -1

The coordinates of any point of the ray can be written as follows:

x_{1} = 3/2 + 1 * t

x_{2} = 1/2 + 1 * t

t ≥ 0

Suppose that if t = 1 then we have a point B. Let's find the coordinates of point B.

x_{1} = 3/2 + 1 * 1 = 5/2

x_{2} = 1/2 + 1 * 1 = 3/2

Let's calculate the value of the function F at point B (5/2,3/2).

F (B) = -1 * 5/2 + 1 * 3/2 = -1

F(A) = F(B).

Then we can conclude that the function F has a maximum value at any point on the ray which has its start at point A.

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