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Let $a \in R$. For which values of $a$ does the following LP problem have an optimal solution? $$max(2x_1+6x_2+3x_3)$$ $$-3x_2+ax_3 \geq 3$$ $$x_1+5x_2+2x_3=4$$ $$x_1, x_2, x_3 \geq 0$$

I solved it with the Big M Method and found that the LP problem has an optimal solution for $a>\frac{3}{2}$ and the value of the objective function is $8-\frac{3}{a}$. Is this correct?

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Graphically, yes it is But ..... read it.
The second condition is a plane cuts the three axis at $4,4/5,2$ respectively where the first one is the above the plane parallel to the first axis and cuts the second and third axes at $-1,3/a$ respectively. If the intersection of these plane lie out side the first quad, the feasible region is empty. So, we have $$\frac{3}{a}<2 \implies a>\frac{3}{2} $$ Now you only a triangle with corners $(0,0,2),(-6/a,0,3/a), (0,(4a-6)/(6+5a),27/(6+5a))$. The first point gives $6$ and the second gives $8-3/a$. Finally, the values of $a$ is true but the max. value is also true

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  • $\begingroup$ Shouldn't it be:$$\frac{3}{a}<2 \Rightarrow a>\frac{3}{2}$$? So then it would be: $$\frac{3}{a}<2 \Rightarrow -\frac{3}{a}>-2 \Rightarrow 8-\frac{3}{a}>6$$ $\endgroup$ – Mary Star Feb 9 '14 at 23:16
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    $\begingroup$ you are right, i will edit it now $\endgroup$ – Semsem Feb 9 '14 at 23:21
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    $\begingroup$ I did the same mistake at the beginning...I will edit my post too... $\endgroup$ – Mary Star Feb 9 '14 at 23:23
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    $\begingroup$ now the value .5 is not valid $\endgroup$ – Semsem Feb 9 '14 at 23:25
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    $\begingroup$ yes it is, glad to help out $\endgroup$ – Semsem Feb 9 '14 at 23:27

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