Big-M method max:Z=2x₁+x₂

Step 1: Make the problem ready for the simplex method.

• Constraint 1 is ≤: add slack variable s₁.
• Constraint 2 is ≥: subtract surplus variable s₂ and add artificial variable A₁.

Artificial variables carry a cost of −M in the objective, where M is a very large positive number, so that the simplex method drives them out of the basis.

Maximize Z = 2x₁ + x₂ − M·A₁
subject to
  x₁ + x₂ + s₁ = 10
  −x₁ + x₂ − s₂ + A₁ = 2

Step 2: Initial solution. Put x₁ = x₂ = s₂ = 0. The basic variables are the slack s₁ = 10 (from the ≤ row) and the artificial A₁ = 2 (from the ≥ row).

Step 3: Simplex tables. An artificial variable's column is dropped as soon as it leaves the basis. Cj − Zj entries contain M, and the large-M part decides which variable enters.

Table 1

          Cj   2   1  0  0 -M          
Basis     CB  x₁  x₂ s₁ s₂ A₁   b Ratio
---------------------------------------
s₁         0   1   1  1  0  0  10    10
A₁        -M  -1   1  0 -1  1   2     2
---------------------------------------
          Zj   M  -M  0  M -M -2M      
       Cj-Zj 2-M 1+M  0 -M  0          
Entering: x₂   Leaving: A₁   Pivot = 1

Table 2

          Cj  2  1  0  0        
Basis     CB x₁ x₂ s₁ s₂ b Ratio
--------------------------------
s₁         0  2  0  1  1 8     4
x₂         1 -1  1  0 -1 2     -
--------------------------------
          Zj -1  1  0 -1 2      
       Cj-Zj  3  0  0  1        
Entering: x₁   Leaving: s₁   Pivot = 2

Table 3

          Cj  2  1    0    0   
Basis     CB x₁ x₂   s₁   s₂  b
-------------------------------
x₁         2  1  0  1/2  1/2  4
x₂         1  0  1  1/2 -1/2  6
-------------------------------
          Zj  2  1  3/2  1/2 14
       Cj-Zj  0  0 -3/2 -1/2   

Step 4: Optimality. All Cj − Zj ≤ 0 and no artificial variable remains in the basis, so the solution is optimal and feasible.

Answer: x₁ = 4, x₂ = 6, and Maximum Z = 14.