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

To introduce non-negative slack variables x4 and x5 and write down the corresponding linear system, we can rewrite the given inequalities as equations by adding the slack variables:

x1 + 2x2 - x3 + x4 = 3,

x2 + 3x3 + x5 = 2,

with the additional constraints x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0, and x5 ≥ 0.

This linear system represents the feasible region in R3 defined by the given inequalities and non-negativity constraints.

Next, let's consider the basic solution obtained by making x1 and x2 basic variables. In this case, we set x4 and x5 to 0, and solve the resulting system of equations:

x1 + 2x2 - x3 + 0 = 3,

x2 + 3x3 + 0 = 2.

Simplifying, we have:

x1 + 2x2 - x3 = 3,

x2 + 3x3 = 2.

To find the values of the basic variables, we can arbitrarily set x1 = 0 and x2 = 0, and solve for x3:

0 + 2(0) - x3 = 3,

0 + 3x3 = 2.

From the second equation, we find x3 = 2/3.

Therefore, the basic solution obtained by making x1 and x2 basic variables is (x1, x2, x3, x4, x5) = (0, 0, 2/3, 0, 0).

To determine if this basic solution is feasible, we need to check if it satisfies all the constraints. In this case, the non-negativity constraints are already satisfied since x1 = 0, x2 = 0, x3 = 2/3, x4 = 0, and x5 = 0 are all non-negative.

Now, let's find the basic solution obtained by making x1 and x3 basic variables. Setting x4 and x5 to 0, we have:

x1 + 2x2 - x3 + 0 = 3,

x2 + 3x3 + 0 = 2.

Simplifying, we get:

x1 + 2x2 - x3 = 3,

x2 + 3x3 = 2.

Setting x1 = 0 and x3 = 0, we can solve for x2:

0 + 2x2 - 0 = 3,

x2 + 3(0) = 2.

From the first equation, we find x2 = 3/2. From the second equation, we find x2 = 2.

Since we have conflicting values for x2, this basic solution is not feasible.

Please note that the values of x1, x2, and x3 in the basic solutions are arbitrary choices made for solving the system, and other values may also yield feasible or infeasible solutions.