Duality LPP min:C=16x₁+45x₂

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

Primal: Minimize C = 16x₁ + 45x₂
  subject to  2x₁ + 5x₂ ≥ 50
              x₁ + 3x₂ ≥ 27
              x₁, x₂ ≥ 0

Dual:   Maximize W = 50y₁ + 27y₂
  subject to  2y₁ + y₂ ≤ 16
              5y₁ + 3y₂ ≤ 45
              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 = 50y₁ + 27y₂
subject to
  2y₁ + y₂ + s₁ = 16
  5y₁ + 3y₂ + s₂ = 45

Step 3: Simplex tables for the dual

Table 1

          Cj 50 27  0  0         
Basis     CB y₁ y₂ s₁ s₂  b Ratio
---------------------------------
s₁         0  2  1  1  0 16     8
s₂         0  5  3  0  1 45     9
---------------------------------
          Zj  0  0  0  0  0      
       Cj-Zj 50 27  0  0         
Entering: y₁   Leaving: s₁   Pivot = 2

Table 2

          Cj 50  27    0  0          
Basis     CB y₁  y₂   s₁ s₂   b Ratio
-------------------------------------
y₁        50  1 1/2  1/2  0   8    16
s₂         0  0 1/2 -5/2  1   5    10
-------------------------------------
          Zj 50  25   25  0 400      
       Cj-Zj  0   2  -25  0          
Entering: y₂   Leaving: s₂   Pivot = 1/2

Table 3

          Cj 50 27   0  0    
Basis     CB y₁ y₂  s₁ s₂   b
-----------------------------
y₁        50  1  0   3 -1   3
y₂        27  0  1  -5  2  10
-----------------------------
          Zj 50 27  15  4 420
       Cj-Zj  0  0 -15 -4    

Step 4: Read the answer. All Cj − Zj ≤ 0, so the dual is solved: y₁ = 3, y₂ = 10, Max W = 420.

By the duality theorem, Min C = Max W. The values of the primal variables are the numbers (without sign) in the Cj − Zj row under the slack columns: x₁ = 15 (under s₁), x₂ = 4 (under s₂).

Check: C = 16(15) + 45(4) = 420.

Answer: x₁ = 15, x₂ = 4, Minimum C = 420.