How To Solve A Minimization Problem In Linear Programming When Objective Function Are All Non Negatives? This is the Optimization Problem:
Minimize $C = x_4 + x_5$
Subject to constraints:
$x_1 + x_2 + x_4 = 2$
$2x_1 - x_3 + x_5 = 1$
I converted it to maximization:
so my objective function becomes
Maximize $P = -x_4 - x_5 $
=> $x_4 + x_5 + P = 0$
And my tableau becomes:
1  1  0  4  0 | 2    ~ constraint
2  0 -1  0  1 | 1    ~ constraint
0  0  0  1  1 | 0    ~ objective function
Basically, I'm not quite sure how to choose my pivot here, or if I set this up even correctly
 A: I suppose that all variables are positive.  Otherwise, the question wouldn't make sense.
Observe that $(x_4,x_5) = (2,1)$ is a basic feasible solution.
Your tableau should be
\begin{array}{l|rrrrr|rr}
 & x_1 & x_2 & x_3 & x_4 & x_5 & \text{RHS} & \text{ratio}\\ \hline
x_4 & 1 & 1 & 0 & \color{red}{1} & 0 & 2 \\
x_5 & 2 & 0 & -1 & 0 & 1 & 1 \\ \hline
z & 0 & 0 & 0 & 1 & 1 & 0 \\
\end{array}
Note that the $x_4,x_4$-entry should be $1$ instead of $4$ as the coefficient of $x_4$ in the first constraint is one.
Before starting the simplex method, in the tableau, you have to make the columns representing the current basis ($x_4,x_5$-columns in this case) elementary, subtract the $x_4,x_5$-rows from the $z$-row.
\begin{array}{l|rrrrr|rr}
 & x_1 & x_2 & x_3 & x_4 & x_5 & \text{RHS} & \text{ratio}\\ \hline
x_4 & 1 & 1 & 0 & \color{red}{1} & 0 & 2 & 2/1 = 1 \\
x_5 & 2^* & 0 & -1 & 0 & 1 & 1 & 1/2 = 1/2 \\ \hdashline
z & -3^* & -1 & 1 & 0 & 0 & -3 \\ \hline
x_4 & 0 & 1^* & 1/2 & 1 & -1/2 & 3/2 & 3/2 \\
x_1 & 1 & 0 & -1/2 & 0 & 1/2 & 1/2 & - \\ \hdashline
z   & 0 & -1^* & 5/2 & 0 & 3/2 & -3/2 \\ \hline
x_2 & 0 & 1 & 1/2 & 1 & -1/2 & 3/2 & - \\
x_1 & 1 & 0 & -1/2 & 0 & 1/2 & 1/2 & - \\ \hdashline
z   & 0 & 0 & 3 & 1 & 1 & 0
\end{array}
The basic solution $(x_1,x_2) = (1/2,3/2)$ gives the minimum value $0$.
