Use the simplex method to maximize the objective function z
Solution
Using Simplex method, rewriting the given equations in tableau form, we get :
Tableau #1
x1 x2 s1 s2 s3 s4 z
2 10 1 0 0 0 0 50
2 1 0 1 0 0 0 20
1 0 0 0 -1 0 0 0
0 1 0 0 0 -1 0 0
-9 -30 0 0 0 0 1 0
Tableau #2
x1 x2 s1 s2 s3 s4 z
2 10 1 0 0 0 0 50
2 1 0 1 0 0 0 20
-1 0 0 0 1 0 0 0
0 1 0 0 0 -1 0 0
-9 -30 0 0 0 0 1 0
Tableau #3
x1 x2 s1 s2 s3 s4 z
2 10 1 0 0 0 0 50
2 1 0 1 0 0 0 20
-1 0 0 0 1 0 0 0
0 -1 0 0 0 1 0 0
-9 -30 0 0 0 0 1 0
Tableau #4
x1 x2 s1 s2 s3 s4 z
1/5 1 1/10 0 0 0 0 5
9/5 0 -1/10 1 0 0 0 15
-1 0 0 0 1 0 0 0
1/5 0 1/10 0 0 1 0 5
-3 0 3 0 0 0 1 150
Tableau #5
x1 x2 s1 s2 s3 s4 z
0 1 1/9 -1/9 0 0 0 10/3
1 0 -1/18 5/9 0 0 0 25/3
0 0 -1/18 5/9 1 0 0 25/3
0 0 1/9 -1/9 0 1 0 10/3
0 0 17/6 5/3 0 0 1 175
So the Optimal Solution is:
z = 175;
x1 = 25/3,
x2 = 10/3
