Using the two-phase simplex method I am asked to
Minimise: z= 3x2-x3+8
Subject to:
x1+x2+x3=10,
2x1+3x2+x3=15
x1,x2,x3>=0
I am not sure how to go about this question as I have only used this method with maximsie problems. My initial thought was that the problem is already in canonical form, so there is no need to add slack, surplus or artificial variables and the problem can be solved using the simplex method? However I have tried this and it doesnt work. The answers to this question are x1=5, x2=0, x3=5 and z=3. If anyone could show how they get to this stage I would really appreciate it. Ps. sorry for the poor markdown I am still new e.g. x2 means 'x small 2' not 'x squared'.
You have to use artificial variables ($a_i$), because of the equality signs.
$\begin{array}{|c|c|c|c|c|c|} \hline x_1 &x_2&x_3&a_1&a_2&RHS \\ \hline 0&3&-1&0&0&-8 \\ \hline 1&1&\color{red}{ \boxed{ 1}} &1&0&10 \\ \hline 2&3&1&0&1&15 \\ \hline \end{array}$
$\color{red}{ \boxed{ 1}}$ is the first pivot element. Both artificial variables have to be kicked out of the base.
$\begin{array}{|c|c|c|c|c|c|} \hline x_1 &x_2&x_3&a_1&a_2&RHS \\ \hline 1&4&0&1&0&2 \\ \hline 1&1&1 &1&0&10 \\ \hline \color{red}{ \boxed{ 1}}&2&0&-1&1&5 \\ \hline \end{array}$
$\begin{array}{|c|c|c|c|c|c|} \hline x_1 &x_2&x_3&a_1&a_2&RHS \\ \hline 0&2&0&2&-1&-3 \\ \hline 0&-1&1 &2&-1&5 \\ \hline 1&2&0&-1&1&5 \\ \hline \end{array}$
Et voila the optimal solution is $(x_1,x_2,x_3,z)=(5,0,5,3)$