Question

MAXIMIZATION BY THE SIMPLEX METHOD

Maximize z = x₁ + 2x₂ + x₃

subject to

x₁ + x₂ ≤ 3

x₂ + x₃ ≤ 4

x₁ + x3 ≤ 5

x₁, x₂, x₃ ≥0

Answer 1

Maximize Z = X₁ + 2X₂ + 3X₃

subject to

X₁ + X₂   <= 3

X₂ + X₃ <= 4

X₁ + X₃ <= 5

X₁ , X₂ , X₃ >= 0

So this is a standard linear programming and we can add slack variables S₁ ,S₂ and S₃ to get equations from inequalities.

X₁ + X₂ + S₁ = 3

X₂ + X₃ + S₂ = 4

X₁ + X₃ + S₃ = 5

And -X₁ - 2X₂ - 3X_3­­ + Z = 0

So the initial tableau for above linear programming

X₁ X₂ X₃     S₁ S₂     S₃      Z            

1       1      0       1        0      0      0      3     

0      1       1       0        1       0      0      4     

1       0       1       0        0       1      0      5     

-1     -2     -3        0        0       0      1      0     

Here the most negative element in the bottom row will indicates the pivot element so here -3 ,so I am taking 3^rd column as a pivot column and for pivot row the least positive result when last column divided by pivot column will indicates

i.e. +min (3/0 , 4/1, 5/1) = 4/1 so 2^nd row as a pivot row.

R₃-> R₃  - R₂              R₄-> R₄ + 3 R₂

X₁     X₂ X₃    S₁      S₂     S₃    Z            

1      1      0       1        0      0      0      3     

0        1       1      0        1      0      0      4     

1      -1       0       0      -1     1      0      1     

-1      1      0       0        3      0      1      12

Here the most negative element in the bottom row will indicates the pivot element so here -1 ,so I am taking 1^st  column as a pivot column and for pivot row the least positive result when last column divided by pivot column will indicates

i.e. +min (3/1 , 4/0, 1/1) = 1/1 so 3^rd  row as a pivot row.

R₁-> R₁  - R₃              R₄-> R₄ + R₃

X₁ X₂ X₃     S₁    S₂     S₃ Z            

0      2      0      1      1      -1     0      2     

0      1       1       0      1       0      0      4     

1      -1     0      0    -1       1      0      1     

0      0       0      0      2       1      1      13

So now we did not have any negative elements in bottom row so we can stop the iterations. Now the optimum solution is Maximum Z = 13   At    X₁ = 1 , X₂= 0 ,X₃ =4