# Service for Solving Linear Programming Problems

and other interesting typical problems
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## Example №5. Solving a Linear Programming Problem Using a Graphical Method. Function has a Maximum Value on the Ray

This solution has been made using the calculator presented on the site.
Problem:
Find the maximum value of the function

F = - x1 + x2

subject to the constraints: x1 - 2 x2 ≤ 4 x1 - x2 ≥ 1 x1 + x2 ≥ 2

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 3)

The last two steps are necessary to get the 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.

See the plan for solving this problem in pictures

By the condition of the problem: x1 ≥ 0     x2 ≥ 0.

Now we have the region of feasible solutions shown in the picture. Big pictures
Step №1

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

x1 - 2 x2  ≤  4

We need to plot a straight line: x1 - 2 x2 = 4

Let x1 =0 => - 2 x2 = 4 => x2 = -2

Let x2 =0 => x1 = 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.

x1 - 2 x2  ≤  4

We need to transform the inequality so that only x2 is on the left side.

- 2 x2  ≤  - x1 + 4

x2  ≥  1/2 x1 - 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. Big pictures
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. Big pictures
Step №3

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

x1 + x2  ≥  2

We need to plot a straight line: x1 + x2 = 2

Let x1 =0 => x2 = 2

Let x2 =0 => x1 = 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.

x1 + x2  ≥  2

We need to transform the inequality so that only x2 is on the left side.

x2  ≥  - x1 + 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. Big pictures
Step №4

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. Big pictures
Step №5

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 on the straight line (3) at the same time. x1 - x2 = 1 => x1 = 3/2 x1 + x2 = 2 x2 = 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:

x1 = 3/2 + 1 * t

x2 = 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.

x1 = 3/2 + 1 * 1 = 5/2

x2 = 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. Big pictures
Result: