# Strict inequalities in an LP problem

So I have to formulate an LP problem out of this scenario converting 3 variables into just 2 variables in order to use the graphical method. I guess I do that by using the demand constraint:

I think we should maximise

$$z = 280x_1 + 240x_2 + 320x_3$$

s.t.

$$x_1 + x_2 + x_3 = 1000$$

$$400 \le x_1 \le 600$$

$$200 \color{red}{<} x_2 \color{red}{<} 300$$

$$100 \color{red}{<} x_3 \color{red}{<} 400$$

$$x_2 \le 1.5x_3$$

Is that right? If so, I thought to convert as follows:

$$\max \ z = 280(1000-x_2-x_3) + 240x_2 + 320x_3$$

s.t.

$$400 \le x_2 + x_3 \le 600$$

$$200 \color{red}{<} x_2 \color{red}{<} 300$$

$$100 \color{red}{<} x_3 \color{red}{<} 400$$

$$x_2 \le 1.5x_3$$

which produces this feasible region

Is that right? If the inequalities are not strict then, I guess I just check the endpoints:

$$(200,400), (300,300), (300,200), (240,160), (200,200)$$

If they are strict, is the optimal solution $(240,160)$ ? It seems to give a $z$ value of $276,800$, but the points $(200,400), (300,300), (200,200)$ seem to give $z$ values higher than $276,800$ so perhaps points close to it may give $z$ values higher than $276,800$.

Checked WolframAlpha. Yeah. So technically there is no optimal solution?

Related question: Strict inequalities in LP . I don't think using $\epsilon$, as suggested there, is needed as not all the inequalities in this problem are strict.

From Chapter 2 here.

• I would interpret e.g. "between 200 and 300 pounds" as $200\leqslant x\leqslant 300$ unless specifically told otherwise... – Math1000 Feb 29 '16 at 2:11
• @Math1000 So why isn't the ground beef stated in a similar way? – BCLC Feb 29 '16 at 11:43