solve the following linear programming problems using simple

solve the following linear programming problems using simplex method.

Maximize P=4x+5y+6z

subject to 2x+3y+z<=900

3x+y+z<=350

4x+2y+z<=400

x=>0 y=>0 z=>0

Solution

Tableau #1
x y z s1 s2 s3 s4 s5 s6 p   
2 3 1 1 0 0 0 0 0 0 900
3 1 1 0 1 0 0 0 0 0 350
4 2 1 0 0 1 0 0 0 0 400
1 0 0 0 0 0 -1 0 0 0 0
0 1 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 0 0 0 -1 0 0
-4 -5 -6 0 0 0 0 0 0 1 0

Tableau #2
x y z s1 s2 s3 s4 s5 s6 p   
2 3 1 1 0 0 0 0 0 0 900
3 1 1 0 1 0 0 0 0 0 350
4 2 1 0 0 1 0 0 0 0 400
-1 0 0 0 0 0 1 0 0 0 0
0 1 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 0 0 0 -1 0 0
-4 -5 -6 0 0 0 0 0 0 1 0

Tableau #3
x y z s1 s2 s3 s4 s5 s6 p   
2 3 1 1 0 0 0 0 0 0 900
3 1 1 0 1 0 0 0 0 0 350
4 2 1 0 0 1 0 0 0 0 400
-1 0 0 0 0 0 1 0 0 0 0
0 -1 0 0 0 0 0 1 0 0 0
0 0 1 0 0 0 0 0 -1 0 0
-4 -5 -6 0 0 0 0 0 0 1 0

Tableau #4
x y z s1 s2 s3 s4 s5 s6 p   
2 3 1 1 0 0 0 0 0 0 900
3 1 1 0 1 0 0 0 0 0 350
4 2 1 0 0 1 0 0 0 0 400
-1 0 0 0 0 0 1 0 0 0 0
0 -1 0 0 0 0 0 1 0 0 0
0 0 -1 0 0 0 0 0 1 0 0
-4 -5 -6 0 0 0 0 0 0 1 0

Tableau #5
x y z s1 s2 s3 s4 s5 s6 p   
-1 2 0 1 -1 0 0 0 0 0 550
3 1 1 0 1 0 0 0 0 0 350
1 1 0 0 -1 1 0 0 0 0 50   
-1 0 0 0 0 0 1 0 0 0 0
0 -1 0 0 0 0 0 1 0 0 0
3 1 0 0 1 0 0 0 1 0 350
14 1 0 0 6 0 0 0 0 1 2100   

Hence the maximum value for the constraint P is equal to 2100