Consider the convex set given by

Question

Consider the convex set given by

3 x1+6 x2 ≤16

5 x1+2 x2 ≥18

x1≥0,x2≥0

(a) Introduce a slack variable x3≥0 to convert the first inequality to an equation.

(b) Introduce a slack variable x4≥0 to convert the second inequality to an equation.

(c) Enter the 2x5 augmented matrix that comes from your answers to (a) and (b), with pivots in the third and fourth columns

(d) Find a basic solution that is not feasible. It should have four components, x1,x2,x3,x4

(e) Find a feasible basic solution. It should have four components, x1,x2,x3,x4.

(a),(b),(c) I did it correctly but I have no idea how to do (d) and (e)

Answer 1

The set is the perimeter and inside of a triangle with vertices A = (18/5, 0), B = (16/3, 0), C = (19/6,13/12)

The 5×2 matrix I get in part (c) is

row 1: (3,6,1,0,16)

row 2: (5,2,0,-1,18)

I'm not entirely sure what parts (d) and (e) are asking, but...

...solving for x₁ and x₂ I get the matrix

row 1: (1,0,-1/12,-1/4,19/6)

row 2: (0,1,5/24,1/8,13/12)

meaning

x₁ – (1/12)x₃ – (1/4)x₄ = 19/6

x₂ + (5/24)x₃ + (1/8)x₄ = 13/12

Since x₃ and x₄ must both be ≥ 0 this ultimately leads to

x₁ ≥ 19/6

x₂ ≤ 13/12

Since (19/6,13/12) is in the set (point C of the triangle), this contains that point as a solution.

So I guess the only feasible solution [for part (e)] is

x₁ = 19/6

x₂ = 13/12

x₃ = x₄ = 0

and the sets of solutions that are not feasible [for part (d)] are

x₁ = 19/6

x₂ = 13/12

x₃ > 0

x₄ ≥ 0

and

x₁ = 19/6

x₂ = 13/12

x₃ ≥ 0

x₄ > 0