SOLUTION: solve using simplex method max z=3x1+X2+3x3 for the following constraints 2x1+x2+x3≤2, x1+2x2+3x3≤5, 2x1+2x2+x3≤6

Maximize:

subject to the constraints:


We add non-negative slack variables to "take up the slack" in the inequalities
to change them into equations. We rearrange the equation of the variable z to
maximize and put it at the bottom:





We form the initial tableau:



The most negative number on the bottom row is -3. We find it twice at the bottom
of columns 1 and 3. So we could either pick column 1 or column 3.
We'll pick column 1, so we call column 1 "the pivot column".
We divide each positive number in the pivot column INTO the number
at the far right to see which gives the smallest positive answer.

  1         5         3
2)2       1)5       2)6

We get the smallest value when we divide 2 into the number at the far
right, so 2 in column 1 row 1 is the pivot element.  Its row, the 1st row,
is the pivot element'

Now we make the pivot element become 1, by dividing the pivot row
through by 2.  Then we make 0's elsewhere in the pivot column, using
the pivot row. Making an element of a column become 1 and all the other
elements in the column become 0 is called "pivoting" on the element
that we caused to become 1.  So here is the 2nd tableau:



It is not the final tableau, because there is still a negative number
on the bottom row.

The only negative number on the bottom row is -1.5. It is in 
column 3, so now column 3 is the pivot column. We divide each 
positive number in the pivot column INTO the number
at the far right to see which gives the smallest positive answer.

    2          1.6
0.5)1       2.5)4

We get the smallest value when we divide 2.5 into the number at the far
right, so 2.5 is the pivot element.

Now we make the pivot element become 1, by dividing the pivot row
through by 2.5.  Then we use it to make 0's elsewhere in the pivot column,
which is column 3, using the pivot element.



There are no more negative numbers on the bottom row, so
we have reached the final tableau:
 
Now we turn the final tableau back into a system of equations:



We solve the last equation for z:



Now we see that the maximum value of z will be 5.4 if we don't
subtract anything from it, so we choose x2=0, s1=0, s2=0.

Then the system above becomes:



Or z has maximum value of 5.4 when x1=0.2, x2=0, and x3=1.6

Edwin