6 Consider the linear programming problem:
$$\begin{array} { l r }
\text { maximise } & P = 3 x - 5 y + 4 z ,
\text { subject to } & x + 2 y - 3 z \leqslant 12 ,
& 2 x + 5 y - 8 z \leqslant 40 ,
\text { and } & x \geqslant 0 , y \geqslant 0 , z \geqslant 0 .
\end{array}$$
- Represent the problem as an initial Simplex tableau.
- Explain why it is not possible to pivot on the \(z\) column of this tableau. Identify the entry on which to pivot for the first iteration of the Simplex algorithm. Explain how you made your choice of column and row.
- Perform one iteration of the Simplex algorithm. Write down the values of \(x , y\) and \(z\) after this iteration.
- Explain why \(P\) has no finite maximum.
The coefficient of \(z\) in the objective is changed from + 4 to - 40 .
- Describe the changes that this will cause to the initial Simplex tableau and the tableau that results after one iteration. What is the maximum value of \(P\) in this case?
Now consider this linear programming problem:
$$\begin{array} { l l }
\text { maximise } & Q = 3 x - 5 y + 7 z ,
\text { subject to } & x + 2 y - 3 z \leqslant 12 ,
& 2 x - 7 y + 10 z \leqslant 40 ,
\text { and } & x \geqslant 0 , y \geqslant 0 , z \geqslant 0 .
\end{array}$$
Do not use the Simplex algorithm for these parts. - By adding the two constraints, show that \(Q\) has a finite maximum.
- There is an optimal point with \(y = 0\). By substituting \(y = 0\) in the two constraints, calculate the values of \(x\) and \(z\) that maximise \(Q\) when \(y = 0\).