# Determine the optimal point satisfying the given condition

We are given the following linear problem: $$\max\left\{5x_1-2x_2\mid3x_1+x_2\leq7,4x_1-2x_2\leq3,x_1\geq0,x_2\geq0\right\}$$. If there exists an optimal point where at least one of its coordinates is not zero, then $$x_1$$ and $$x_2$$ shall differ by at least $$1$$ whereby the second coordinate mustn't be lower than the first coordinate.

I don't know how this should all be possible.

First I tried to solve the problem without respecting the conditions, just by using the Simplex algorithm:

Thus optimal point is $$x= \begin{pmatrix} x_1\\ x_2 \end{pmatrix}= \begin{pmatrix} \frac{17}{10}\\ \frac{19}{10} \end{pmatrix}$$ , clearly not satisfying the conditions above...

How is it possible to determine an optimal point satisfying these conditions? I don't know but I believe I somehow need to apply these conditions while doing the Simplex algorithm? Maybe in the middle table, increase $$\frac{19}{4}$$ by $$1$$ and continue with that? Or a complete different attempt? :o

The optimal solution is at $$x = [1.7,1.9]^T$$, so your answer is correct. Also, the second coordinate (I believe this is $$x_2$$) is greater than the first one.
• "Differ by $1$" is clearly $1$ and not $0.1$. I'm not sure if that is really correct your answer because it seems too easy and then I also wonder why there is a condition at all in the given example. Dec 3, 2018 at 18:57