SOLUTION: Consider the following simplex tableau. x y z u v w P Constants 0 1/2 0 1 −1/2 0 0 9 1 1/2 1 0 1/2 0 0 0 2 1/2 0 0 −3/2 1 0 8 −1 3 0 0 1 0 1 3

There is one negative number on the bottom row, in the
1st column, so the first column is the pivot column. We
divide every POSITIVE number in the pivot column into the
rightmost number on its row to find the smallest quotient:

  4       0
2)8     1)0

0 is the smallest quotient so the 1 in the first column 2nd row
is the pivot element. It is already a 1, so we don't have to
make it 1.  We use it to make all the other elements on the pivot 
column 0.



We multiply the pivot row by -2 and add to the 3rd row to make the 2 in
the 3rd row become 0:

-2R2+R3->R3



We add the pivot row to the 4th row to make the -1 in
the 4th row become 0:

R2+R4->R4



We write the system of equations for that matrix:



Solve the bottom equation for P




We want P to be large as possible.  All the variables 
are non-negative.  We can keep the whole 30 for P if
we choose y=0, z=0, v=0.  So we substitute those:



P has maximum value 30 when x=0, y=0, and z=0.

[If this were a realistic problem, the solution would tell us
not to make any products, for our profit would be 30 gazillions
of dollars by making no products at all!]

The inactive variables are the ones we set = 0, y, z and v.
You can recognize them in the final tableau by the variables 
that DO NOT have 0 at the bottom of their columns. 

Edwin