Simplex method max:P=15x+10y

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 = 15x + 10y
subject to
  2x + y + s₁ = 10
  x + 3y + s₂ = 10

Step 2: Initial basic feasible solution. Put x, y = 0, so the slack variables are the basic variables: s₁ = 10, s₂ = 10. 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 15 10  0  0         
Basis     CB  x  y s₁ s₂  b Ratio
---------------------------------
s₁         0  2  1  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: x   Leaving: s₁   Pivot = 2

Table 2

          Cj 15   10     0  0         
Basis     CB  x    y    s₁ s₂  b Ratio
--------------------------------------
x         15  1  1/2   1/2  0  5    10
s₂         0  0  5/2  -1/2  1  5     2
--------------------------------------
          Zj 15 15/2  15/2  0 75      
       Cj-Zj  0  5/2 -15/2  0         
Entering: y   Leaving: s₂   Pivot = 5/2

Table 3

          Cj 15 10    0    0   
Basis     CB  x  y   s₁   s₂  b
-------------------------------
x         15  1  0  3/5 -1/5  4
y         10  0  1 -1/5  2/5  2
-------------------------------
          Zj 15 10    7    1 80
       Cj-Zj  0  0   -7   -1   

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

Answer: x = 4, y = 2, and Maximum P = 80.