If I have the matrix:
x1 x2 s1 s2 s3 z
1 4 1 0 0 0 | 12
2 5 0 1 0 0 | 2
1 3 0 0 0 1 | 4
----------------
-2 -1 0 0 0 1 | 0
Then, where x1 is the row of 1, 2, 1....x2 is the row of 4, 5, 4....etc where z = 0,0, 0 and
I'm trying to find the maximum _____ when x1 = , x2 = _, and s1 = 11, s2=0, s3=3
I tried to pivot it based upon the most negative but lost in the algebra
You have two options to solve this problem.
First Option
It seems you know the (optimal) values of the slack variables. You can use this information and 2 of the 3 constraints to calculate the optimal solution. I take the constraints 2 and 3.
$2x_1+ 5x_2 +s_2= 2$
$1x_1 +3x_2 +s_3 = 4$
Inserting the values for $s_1$ and $s_2$
$2x_1+ 5x_2 + 0= 2 \quad \Rightarrow x_1=-2.5x_2+1$
$1x_1 +3x_2 +3 = 4\quad \Rightarrow x_1+3x_2=1$
Insterting the term for $x_1$ into the second equation.
$-2.5x_2+1+3x_2=1 \Rightarrow 0.5x_2=0 \Rightarrow x_2=0.$
Thus $x_1=-2.5\cdot 0 +1=1$
Second option
If you do not know the values of the slack variables then you have to apply the simplex algorithm. The initial tableau is
$\begin{array}{|c|c|c|c|c|} \hline x_1&x_2& s_1& s_2&s_3&RHS \\ \hline 1&4&1&0&0&12 \\ \hline \color{red}2&5&0&1&0&2 \\ \hline 1&3&0&0&1&4 \\ \hline -2 & -1 &0&0&0&0 \\ \hline \end{array}$
The pivot column is the first column, because $|-2|$ is larger than $|-1|$. The pivot row is the second one, because $min\left(\frac{12}{1},\frac{2}{2},\frac{4}{1} \right)=\frac{2}{2}$
$\begin{array}{|c|c|c|c|c|} \hline x_1&x_2& s_1& s_2&s_3&RHS \\ \hline 0&3/2&1&-1/2&0&11 \\ \hline 1&5/2&0&1/2&0&1 \\ \hline 0&1/2&0&-1/2&1&3 \\ \hline 0&4&0&1&0&2 \\ \hline \end{array}$
All values of $x_i$ are non-negative. No more transformation is needed. The optimal solution is $(x_1^*,x_2^*,s_1^*,s_2^*,s_3^*)=(1,0,11,0,3)$