




Setting y = 0 and z = 0 leads to s = 1, x = 1, which point obeys all the constraints.
We may therefore choose y and z as basis variables and solve for x and s.
We get
x = 2  y  z, s = 1 2y z
We also write the objective function in terms of the basis variables:
OF = x + 2z = 2  y + z
The value of the objective function at the origin in this basis is 2.
If we take z out of the basis, and put s in instead, the objective function
at the origin increases;
We get
z = 1  2y  s, x = 2  y  z = 1 + y + s
and the objective function is
33ys
Obviously this cannot exceed 3 if y and s are to be positive, so that 3 is the solution for the objective function, and the solution point is z = 1, x = 1, y = s = 0.