Duality LPP min:Z=3x₁+2x₂

Step 1: Write the dual. For the primal (minimisation, all constraints ≥), the dual is:

Primal: Minimize Z = 3x₁ + 2x₂
  subject to  3x₁ − x₂ ≥ −5
              −x₁ + 4x₂ ≥ 1
              x₁ + 9x₂ ≥ 6
              x₁, x₂ ≥ 0

Dual:   Maximize W = −5y₁ + y₂ + 6y₃
  subject to  3y₁ − y₂ + y₃ ≤ 3
              −y₁ + 4y₂ + 9y₃ ≤ 2
              y₁, y₂, y₃ ≥ 0

Here y₁, y₂, y₃ are the dual variables, one for each primal constraint.

Step 2: Standard form of the dual. Add slack variables s₁, s₂ (one per dual constraint). Since all right sides are positive, the slack variables give the starting basis.

Maximize W = −5y₁ + y₂ + 6y₃
subject to
  3y₁ − y₂ + y₃ + s₁ = 3
  −y₁ + 4y₂ + 9y₃ + s₂ = 2

Step 3: Simplex tables for the dual

Table 1

          Cj -5  1  6  0  0        
Basis     CB y₁ y₂ y₃ s₁ s₂ b Ratio
-----------------------------------
s₁         0  3 -1  1  1  0 3     3
s₂         0 -1  4  9  0  1 2   2/9
-----------------------------------
          Zj  0  0  0  0  0 0      
       Cj-Zj -5  1  6  0  0        
Entering: y₃   Leaving: s₂   Pivot = 9

Table 2

          Cj    -5     1  6  0    0     
Basis     CB    y₁    y₂ y₃ s₁   s₂    b
----------------------------------------
s₁         0  28/9 -13/9  0  1 -1/9 25/9
y₃         6  -1/9   4/9  1  0  1/9  2/9
----------------------------------------
          Zj  -2/3   8/3  6  0  2/3  4/3
       Cj-Zj -13/3  -5/3  0  0 -2/3     

Step 4: Read the answer. All Cj − Zj ≤ 0, so the dual is solved: y₁ = 0, y₂ = 0, y₃ = 2/9, Max W = 4/3.

By the duality theorem, Min Z = Max W. The values of the primal variables are the numbers (without sign) in the Cj − Zj row under the slack columns: x₁ = 0 (under s₁), x₂ = 2/3 (under s₂).

Check: Z = 3(0) + 2(2/3) = 4/3.

Answer: x₁ = 0, x₂ = 2/3, Minimum Z = 4/3.