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Question: $$\text{Maximize } z=2x_1-6x_2$$

Subject to,

$$-x_1-x_2-x_3\le-2$$ $$2x_1-x_2+x_3\le1$$ $$x_1,x_2,x_3\ge0$$

My work:

$-2x_1+6x_2+z=0$

enter image description here

Then,

${R2\over2}\Rightarrow R2_{new}$,

$R2_{new} + R1\Rightarrow R1_{new}$,

$2\cdot R2_{new}+ R3\Rightarrow R3_{new}$.

The new matrix:

enter image description here

The last row has no negative value, and it looks like the job is done, but the answer is wrong. I think the mistake should be made on calculating, but I can't figure it out... Help!

Neuer
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  • I only have the final solution, not step by step. $(x_1,x_2,x_3)=(0,0.5,1.5)$ and z=-3 – Neuer Jan 25 '15 at 03:41
  • Actually the problem is "Maximize $2x_1-6x_2$" and I just add z myself. All the others are correct. – Neuer Jan 25 '15 at 03:54
  • The vector of right sides can't contain negative numbers! You have to multiply first inequality by -1, after that you can enter it into the table, but you will need an extra variable, which leads to simplex method with two phases. You can check, that $(0,0,0)$ is not a feasible solution, so you can't solve this problem just by using one phase. – Stefan Gyürki Jan 25 '15 at 11:39

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