Simplex method max:Z=10x₁+x₂+2x₃

Step 1: Standard form. Every constraint is of the ≤ type, so add one slack variable to each (s₁, s₂ ≥ 0) to turn it into an equation. Slack variables have zero cost.

Maximize Z = 10x₁ + x₂ + 2x₃
subject to
  x₁ + x₂ − 2x₃ + s₁ = 10
  4x₁ + x₂ + x₃ + s₂ = 20

Step 2: Initial basic feasible solution. Put x₁, x₂, x₃ = 0, so the slack variables are the basic variables: s₁ = 10, s₂ = 20. Then Z = 0.

Rules used: the entering variable is the one with the largest positive Cj − Zj; the leaving variable is the one with the smallest non-negative ratio b/column; the solution is optimal when every Cj − Zj ≤ 0.

Step 3: Simplex tables

Table 1

          Cj 10  1  2  0  0         
Basis     CB x₁ x₂ x₃ s₁ s₂  b Ratio
------------------------------------
s₁         0  1  1 -2  1  0 10    10
s₂         0  4  1  1  0  1 20     5
------------------------------------
          Zj  0  0  0  0  0  0      
       Cj-Zj 10  1  2  0  0         
Entering: x₁   Leaving: s₂   Pivot = 4

Table 2

          Cj 10    1    2  0    0   
Basis     CB x₁   x₂   x₃ s₁   s₂  b
------------------------------------
s₁         0  0  3/4 -9/4  1 -1/4  5
x₁        10  1  1/4  1/4  0  1/4  5
------------------------------------
          Zj 10  5/2  5/2  0  5/2 50
       Cj-Zj  0 -3/2 -1/2  0 -5/2   

Step 4: Optimality test. In the last table every Cj − Zj is ≤ 0, so the solution is optimal.

Answer: x₁ = 5, x₂ = 0, x₃ = 0, and Maximum Z = 50.