Simplex method max:Z=15x₁+10x₂

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 = 15x₁ + 10x₂
subject to
  2x₁ + 2x₂ + s₁ = 10
  x₁ + 3x₂ + s₂ = 10

Step 2: Initial basic feasible solution. Put x₁, x₂ = 0, so the slack variables are the basic variables: s₁ = 10, s₂ = 10. 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 15 10  0  0         
Basis     CB x₁ x₂ s₁ s₂  b Ratio
---------------------------------
s₁         0  2  2  1  0 10     5
s₂         0  1  3  0  1 10    10
---------------------------------
          Zj  0  0  0  0  0      
       Cj-Zj 15 10  0  0         
Entering: x₁   Leaving: s₁   Pivot = 2

Table 2

          Cj 15 10     0  0   
Basis     CB x₁ x₂    s₁ s₂  b
------------------------------
x₁        15  1  1   1/2  0  5
s₂         0  0  2  -1/2  1  5
------------------------------
          Zj 15 15  15/2  0 75
       Cj-Zj  0 -5 -15/2  0   

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

Answer: x₁ = 5, x₂ = 0, and Maximum Z = 75.