Duality LPP min:Z=21x+50y

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

Primal: Minimize Z = 21x + 50y
  subject to  2x + 5y ≥ 12
              3x + 7y ≥ 17
              x, y ≥ 0

Dual:   Maximize W = 12y₁ + 17y₂
  subject to  2y₁ + 3y₂ ≤ 21
              5y₁ + 7y₂ ≤ 50
              y₁, y₂ ≥ 0

Here y₁, y₂ 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 = 12y₁ + 17y₂
subject to
  2y₁ + 3y₂ + s₁ = 21
  5y₁ + 7y₂ + s₂ = 50

Step 3: Simplex tables for the dual

Table 1

          Cj 12 17  0  0         
Basis     CB y₁ y₂ s₁ s₂  b Ratio
---------------------------------
s₁         0  2  3  1  0 21     7
s₂         0  5  7  0  1 50  50/7
---------------------------------
          Zj  0  0  0  0  0      
       Cj-Zj 12 17  0  0         
Entering: y₂   Leaving: s₁   Pivot = 3

Table 2

          Cj   12 17     0  0          
Basis     CB   y₁ y₂    s₁ s₂   b Ratio
---------------------------------------
y₂        17  2/3  1   1/3  0   7  21/2
s₂         0  1/3  0  -7/3  1   1     3
---------------------------------------
          Zj 34/3 17  17/3  0 119      
       Cj-Zj  2/3  0 -17/3  0          
Entering: y₁   Leaving: s₂   Pivot = 1/3

Table 3

          Cj 12 17  0  0    
Basis     CB y₁ y₂ s₁ s₂   b
----------------------------
y₂        17  0  1  5 -2   5
y₁        12  1  0 -7  3   3
----------------------------
          Zj 12 17  1  2 121
       Cj-Zj  0  0 -1 -2    

Step 4: Read the answer. All Cj − Zj ≤ 0, so the dual is solved: y₁ = 3, y₂ = 5, Max W = 121.

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: x = 1 (under s₁), y = 2 (under s₂).

Check: Z = 21(1) + 50(2) = 121.

Answer: x = 1, y = 2, Minimum Z = 121.