# 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

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 $0$.