I have a minimization problem here:
minimize the cost function $C= 12x + 40y +30z$ subject to
\begin{align} x + 2y +2z \geq \, & 2\\ -x -y -3z \geq\, & -1\\ -x +2y + z \geq\, & -2 \\ \ \\ x \geq 0, y \geq 0, z &\geq 0 \end{align}
So I made the matrix out of that and transposed it. Then i formulated the new problem as a maximization problem and switched $C$ to $P$ and used new slack variables
after finding the pivot columns and doing the row operations, I finally got the optimal solution which is $P=40$ at $(28,16,0)$ now according to my teachers answers this should be $P=40$ at $(0,1,0)$ why is that I don't understand ?
I think this problem can be solved by inspection. Split the inequality into 3 inequalities, such that:
$$ x + 2y +2z \ge 2 - x - y -3z \;\rightarrow\; 2x + 3y + 5z \ge 2$$ $$2 - x- y -3z \ge -1 -x +2y + z \;\rightarrow\; -3y -4z \ge -3$$ $$ -1-x+2y+z \ge -2 \;\rightarrow\; -x + 2y + z \ge -1$$ $$x,y,z \ge 0$$
If you want to minimize $f(x,y,z) = 12x + 40y + 30z$, clearly one should try to focus on the value of $x$, since has the smallest multiplier. From the previous inequalities is easy to see that $x=1$ so $f(x,y,z) = 12$.
$$2\cdot1 + 3\cdot0 +5\cdot0 \ge 2$$ $$-3\cdot 0 -4\cdot 0 \ge -3$$ $$-1 + 2\cdot 0 + \cdot 0 \ge -1$$