# linear programming – Solution from simplex method not satisfying the given constraint function and thus not feasible?

–Attempted solution below–

I am given the following minimization linear programming problem:

$$minimize: 5x_{1} – 6x_{2} + 2x_{3}$$

$$subject to:$$

$$x_{1} + 3x_{2} + 7x_{3} = 8$$

$$-5x_{1} + 6x_{2} -3x_{3} = -8$$

$$x_{1}, x_{2}, x_{3} geq 0$$

I do not require slack variables since the constraint functions are equalities, so I applied the Simplex method as is which began with the following tableau:

$$begin{array}{|c|c|c|c|c|c|c|c|} hline x_{1} & x_{2} & x_{3} & textrm{b} \ hline 1 & 3 & 7 & 8 \ hline -5& 6 & -3 & -8 \ hline 5 & -6 & 2 & 0 \hlineend{array}$$

which made me choose the 3 in the $$x_{2}$$ column as the pivot element, which I divided that row by 3 to make the pivot element into a 1:

$$begin{array}{|c|c|c|c|c|c|c|c|} hline x_{1} & x_{2} & x_{3} & textrm{b} \ hline 1/3 & 1 & 7/3 & 8/3 \ hline -5& 6 & -3 & -8 \ hline 5 & -6 & 2 & 0 \hlineend{array}$$

then I subtracted row 2 by 6 times row 1, and added row 3 by 6 times row 1:

$$begin{array}{|c|c|c|c|c|c|c|c|} hline x_{1} & x_{2} & x_{3} & textrm{b} \ hline 1/3 & 1 & 7/3 & 8/3 \ hline -7 & 0 & -17 & -24 \ hline 7 & 0 & 16 & 16 \hlineend{array}$$

Which is where I stop since the bottom row no longer has any negative values remaining, so the resulting solution is $$x = (0, frac{8}{3}, 0)$$, but if I plug that into the second constraint function it does not equal $$-8$$, so I am confused why it does not work, I don’t think I mischose the pivot element nor did the row operations incorrectly.

Do I need to add slack variables despite the constraint function being equalities? Any hints will be appreciated!

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