Simplex method max:P=50x₁+80x₂

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 P = 50x₁ + 80x₂
subject to
  x₁ + 2x₂ + s₁ = 32
  3x₁ + 4x₂ + s₂ = 84

Step 2: Initial basic feasible solution. Put x₁, x₂ = 0, so the slack variables are the basic variables: s₁ = 32, s₂ = 84. Then P = 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 50 80  0  0         
Basis     CB x₁ x₂ s₁ s₂  b Ratio
---------------------------------
s₁         0  1  2  1  0 32    16
s₂         0  3  4  0  1 84    21
---------------------------------
          Zj  0  0  0  0  0      
       Cj-Zj 50 80  0  0         
Entering: x₂   Leaving: s₁   Pivot = 2

Table 2

          Cj  50 80   0  0           
Basis     CB  x₁ x₂  s₁ s₂    b Ratio
-------------------------------------
x₂        80 1/2  1 1/2  0   16    32
s₂         0   1  0  -2  1   20    20
-------------------------------------
          Zj  40 80  40  0 1280      
       Cj-Zj  10  0 -40  0           
Entering: x₁   Leaving: s₂   Pivot = 1

Table 3

          Cj 50 80   0    0     
Basis     CB x₁ x₂  s₁   s₂    b
--------------------------------
x₂        80  0  1 3/2 -1/2    6
x₁        50  1  0  -2    1   20
--------------------------------
          Zj 50 80  20   10 1480
       Cj-Zj  0  0 -20  -10     

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

Answer: x₁ = 20, x₂ = 6, and Maximum P = 1480.