Big-M method max:P=2x₁+x₂+3x₃

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

• Constraint 1 is ≤: add slack variable s₁.
• Constraint 2 is an equation: 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 P = 2x₁ + x₂ + 3x₃ − M·A₁
subject to
  x₁ + x₂ + 2x₃ + s₁ = 5
  2x₁ + 3x₂ + 4x₃ + A₁ = 12

Step 2: Initial solution. Basic variables: the slack variables (for ≤ rows) and the artificial variables (for ≥ and = rows).

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    3  0 -M           
Basis     CB   x₁   x₂   x₃ s₁ A₁    b Ratio
--------------------------------------------
s₁         0    1    1    2  1  0    5   5/2
A₁        -M    2    3    4  0  1   12     3
--------------------------------------------
          Zj  -2M  -3M  -4M  0 -M -12M      
       Cj-Zj 2+2M 1+3M 3+4M  0  0           
Entering: x₃   Leaving: s₁   Pivot = 2

Table 2

          Cj   2      1  3       0 -M              
Basis     CB  x₁     x₂ x₃      s₁ A₁       b Ratio
---------------------------------------------------
x₃         3 1/2    1/2  1     1/2  0     5/2     5
A₁        -M   0      1  0      -2  1       2     2
---------------------------------------------------
          Zj 3/2  3/2-M  3  3/2+2M -M 15/2-2M      
       Cj-Zj 1/2 -1/2+M  0 -3/2-2M  0              
Entering: x₂   Leaving: A₁   Pivot = 1

Table 3

          Cj   2  1  3    0           
Basis     CB  x₁ x₂ x₃   s₁    b Ratio
--------------------------------------
x₃         3 1/2  0  1  3/2  3/2     3
x₂         1   0  1  0   -2    2     -
--------------------------------------
          Zj 3/2  1  3  5/2 13/2      
       Cj-Zj 1/2  0  0 -5/2           
Entering: x₁   Leaving: x₃   Pivot = 1/2

Table 4

          Cj  2  1  3  0  
Basis     CB x₁ x₂ x₃ s₁ b
--------------------------
x₁         2  1  0  2  3 3
x₂         1  0  1  0 -2 2
--------------------------
          Zj  2  1  4  4 8
       Cj-Zj  0  0 -1 -4  

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

Answer: x₁ = 3, x₂ = 2, x₃ = 0, and Maximum P = 8.