Question

Use the method of slack variables to find the vertices of the feasible region in R2 from Assignment 8, defined by the inequalities x + 2y ≤ 4, 3x + 2y ≤ 6, x, y ≥ 0

(a) Introduce slack variables and turn the system of inequalities into a linear system.

(b) Use Gauss-Jordan elimination to find the basic solution corresponding to the basic variables x1 and x4 and the basic solution corresponding to the basic variables x1 and x2.

(c) Which of the basic solutions that you found in part (b) are feasible?

Algebra Use the method of slack variables to find the vertices of the feasible region in R2 from Assignment 8, defined by the inequalities x2y < 4 Зх + 2у <6 X, y > 0 За + 2у — 6 4 3 (0, 2) (1,3 2 1 (2,0) (0, 0) 3 4 1 2 (a) Introduce slack variables and turn the system of inequalities into a linear system. (b) Use Gauss-Jordan elimination to find the basic solution corre- sponding to the basic variables x and x4 and the basic solution corresponding to the basic variables ri and x2. (c) Which of the basic solutions that you found in part (b) are feasi ble?

Answer 1

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Use the method of slack variables to find the vertices of the feasible region in R2 from Assignment 8, defined by the inequalities x + 2y ≤ 4, 3x + 2y ≤ 6, x, y ≥ 0

EXPLANATION ::-

Given system of inequalities is :

2y<4

3r2y6

T, y >0

(a) Introduce slack variables and turn the system of inequalities into a linear system.

SOL::-

ntroducing slack variables u, v in the given system we get,

(b) Use Gauss-Jordan elimination to find the basic solution corresponding to the basic variables x1 and x4 and the basic solution corresponding to the basic variables x1 and x2.

SOL ::-

Here computation table is :

x1	x2	x3	x4	B
1	2	1	0	4
3	2	0	1	6

Row operation : R₂-3R₁=R₂

x1	x2	x3	x4	B
1	2	1	0	4
0	-4	-3	1	-6

Therefore, the basic solution corresponding to the basic variables x₁ and x₄ is (4,0,0,-6).

Row operation : (-1/4)R₂=R₂

x1	x2	x3	x4	B
1	2	1	0	4
0	1	3/4	-1/4	3/2

Row operation : R₁-2R₂=R₁

x1	x2	x3	x4	B
1	0	-1/2	1/2	1
0	1	3/4	-1/4	3/2

Therefore, the basic solution corresponding to the basic variables x₁ and x₂ is (1,3/2,0,0).

(c) Which of the basic solutions that you found in part (b) are feasible

SOL ::-

We know that feasible solution must have non-negative values.

In the first solution, there is a negative value -6, but in the second solution, there exists only non-negative values.

Therefore, basic feasible solution is (1,3/2,0,0)

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