# Service for Solving Linear Programming Problems

and other interesting typical problems
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## Example ¹5. Solving a Linear Programming Problem Using a Graphical Method. Function has a Maximum Value on the Ray

This solution was 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.

Step ¹1

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.

Step ¹2

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.

Step ¹3

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.

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 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.

Result: