Graphical method min:z=4x₁+3x₂+x₃

Reading of the question. The problem as printed has three variables (x₁, x₂, x₃), so its feasible region lies in three-dimensional space and cannot be drawn on a plane. It is solved here graphically through its dual, which has only two variables (one per constraint) and can be drawn. The solution is the one for the problem exactly as printed.

Step 1: Write the dual. Primal: Minimize z = 4x₁ + 3x₂ + x₃ subject to x₁ + 2x₂ + 4x₃ ≥ 12, 3x₁ + 2x₂ + x₃ ≥ 8, x₁, x₂, x₃ ≥ 0. The dual is:

Maximize W = 12w₁ + 8w₂
subject to   w₁ + 3w₂ ≤ 4
             2w₁ + 2w₂ ≤ 3
             4w₁ + w₂ ≤ 1
             w₁, w₂ ≥ 0

(Each primal variable gives one dual constraint: the coefficients of x₁ are (1, 3), of x₂ are (2, 2) and of x₃ are (4, 1), and the costs 4, 3, 1 become the right-hand sides.)

Step 2: Draw the dual constraint lines. Replace each inequality by an equation, find the intercepts and decide which side of the line is allowed (test the origin).

Also w₁ ≥ 0 and w₂ ≥ 0, so the region lies in the first quadrant.

Step 3: Feasible region of the dual. The region satisfying all constraints is the shaded polygon below (dots), with corner points marked A, B, C, ….

   2|                                            
    |                                            
    |                                            
    |                                            
    |                                            
    |                                            
    |                                            
    |                                            
    |C                                           
    |..                                          
    |...                                         
    |.....                                       
    |......                                      
    |.......                                     
    |.........                                   
    |..........                                  
   0|A..........B                                
    +--------------------------------------------
     0                                          1

Step 4: Corner points and the value of W. The optimum of a linear function over a convex polygon occurs at a corner point.

Step 5: Conclusion. The largest W is 8 at (w₁, w₂) = (0, 1). By the duality theorem, Min z = Max W = 8.

Step 6: Primal values (complementary slackness). At (0, 1) the first two dual constraints are not tight (0 + 3 = 3 < 4 and 0 + 2 = 2 < 3), so x₁ = 0 and x₂ = 0. Since w₂ = 1 > 0, the second primal constraint is tight: 3(0) + 2(0) + x₃ = 8, so x₃ = 8. The first primal constraint is then 4(8) = 32 ≥ 12, which holds with w₁ = 0.

Check: z = 4(0) + 3(0) + 1(8) = 8, and all primal constraints are satisfied.

Answer: Minimum z = 8 at x₁ = 0, x₂ = 0, x₃ = 8.