Consider the feasible region in R3 defined by the inequalities

Question

Consider the feasible region in R3 defined by the inequalities

x1 + 2x2 − x3 ≥ 3,

x2 + 3x3 ≤ 2, and

x1 ≥ 0, x2 ≥ 0, x3 ≥ 0.

(a) Introduce non-negative slack variables x4 and x5 and write down the corresponding linear system.

(b) Find the basic solution obtained by making x1 and x2 basic variables. Is this basic solution feasible?

(c) Find the basic solution obtained by making x1 and x3 basic variables. Is this basic solution feasible?

Answer 1

In this problem, we are dealing with a feasible region in three-dimensional space (R3) defined by a set of linear inequalities. To convert this problem into a linear system with non-negative slack variables, we introduce two non-negative slack variables, x4 and x5, representing the slack in the inequalities. We then create a linear system with seven equations and variables (x1, x2, x3, x4, x5) to represent the original inequalities and the non-negativity constraints.

We then find two different basic solutions:

(b) By making x1 and x2 basic variables, we set x3, x4, and x5 to 0 and obtain the solution (3, 0, 0, 0, 0). This basic solution is feasible as it satisfies all the constraints.

(c) By making x1 and x3 basic variables, we set x2, x4, and x5 to 0 and obtain the same solution (3, 0, 0, 0, 0). This basic solution is also feasible as it satisfies all the constraints.

In summary, both basic solutions (from part b and part c) are feasible within the constraints of the original system of inequalities.

(a) To convert the given system of linear inequalities into a linear system with non-negative slack variables, you can add two non-negative slack variables, x4 and x5, to represent the slack in the inequalities. The system becomes:

1. x1 + 2x2 - x3 + x4 = 3

2. x2 + 3x3 + x5 = 2

3. x1 ≥ 0

4. x2 ≥ 0

5. x3 ≥ 0

6. x4 ≥ 0

7. x5 ≥ 0

This linear system incorporates the original inequalities along with the non-negativity constraints for x1, x2, x3, x4, and x5.

(b) To find a basic solution by making x1 and x2 basic variables, we set x3, x4, and x5 to 0. This results in the following system:

1. x1 + 2x2 = 3

2. x2 = 0

3. x1 ≥ 0

4. x2 ≥ 0

5. x3 = 0

6. x4 = 0

7. x5 = 0

Solving this system, you find x1 = 3 and x2 = 0. The solution is (x1, x2, x3, x4, x5) = (3, 0, 0, 0, 0).

This basic solution is feasible because it satisfies all the constraints:

1. x1 + 2x2 = 3 ≥ 3

2. x2 = 0 (satisfies x2 ≥ 0)

3. x1 = 3 ≥ 0

4. x2 = 0 ≥ 0

5. x3 = 0 (satisfies x3 ≥ 0)

6. x4 = 0 (satisfies x4 ≥ 0)

7. x5 = 0 (satisfies x5 ≥ 0)

(c) To find a basic solution by making x1 and x3 basic variables, we set x2, x4, and x5 to 0. This results in the following system:

1. x1 - x3 = 3

2. x2 = 0

3. x1 ≥ 0

4. x2 ≥ 0

5. x3 = 0

6. x4 = 0

7. x5 = 0

Solving this system, you find x1 = 3 and x3 = 0. The solution is (x1, x2, x3, x4, x5) = (3, 0, 0, 0, 0).

This basic solution is feasible because it satisfies all the constraints:

1. x1 - x3 = 3 ≥ 3

2. x2 = 0 (satisfies x2 ≥ 0)

3. x1 = 3 ≥ 0

4. x2 = 0 ≥ 0

5. x3 = 0 (satisfies x3 ≥ 0)

6. x4 = 0 (satisfies x4 ≥ 0)

7. x5 = 0 (satisfies x5 ≥ 0)

Both basic solutions obtained in (b) and (c) are feasible within the constraints of the original system of inequalities.