# Maximize $x_1$ and $x_2$

I have the following question to tackle:

Maximize $x_1$ and $x_2$ for:

$$x_1, x_2 \geq 0$$

$$-x_1 + x_2 \leq 5$$

$$x_1 + 4x_2 \leq 45$$

$$2x_1 + x_2 \leq 27$$

$$3x_1 - 4x_2 \leq 24$$

So I just wrote out these constraints as the following functions:

$$y \leq -x+5$$

$$y \leq \dfrac{1}{4}x+ \dfrac{45}{4}$$

$$y \leq -2x + 27$$

$$y \geq \dfrac {3}{4}x - 6$$

But then I realized; I have no idea what to do here. What do they mean with 'maximize $x_1$ and $x_2$?! From what I can see with my graphing calculator, $y\leq -x+5$ seems to be the only constraint which actually matter, and it has 2 vertices (namely the y-intercept and the x-intercept). Do they perhaps mean that you should maximize $x_1$ and $x_2$ seperately? (i.e. the maximum of $x_1$ is the x-intercept and the maximum of $x_2$ is the y-intercept?) I haven't got the faintest idea.

Also, I don't get how we can have the constraint $y \geq \dfrac{3}{4}x -6$. It seems to contradict the 3 above constraints!

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"Do they perhaps mean that you should maximize x1 and x2 seperately? ... I haven't got the faintest idea." Neither do we. "Maximize $x_1$ and $x_2$" does not make sense. – leonbloy Apr 1 '13 at 21:04

## 1 Answer

This is a linear programming problem. I suggest you reading about simplex method. Maximizing $x_1 + x_2$ is the same as minimizing $-x_1-x_2$.

I will present you a geometric solution. Draw all the conditions in the plane and if you do it right you should get an area bounded by 6 lines: $x=0$, $y=0$, $y=x+5$, $y=\frac{45}{4} - \frac{x}{4}$, $y=27-2x$, and $y=-6+\frac{3x}{4}$. The solution is then one of the vertices of this polygon. Its vertices are: $(0,0),(0,5),(5,10),(9,9),(12,3)$ and $(8,0)$. Choose the right one.

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I think substituting $+$ for and is a reasonable guess. – Ross Millikan Apr 1 '13 at 21:27
Choose the right one. Really, you think that helps? – Ylyk Coitus Apr 1 '13 at 22:16
@RossMillikan What do you mean? – Ylyk Coitus Apr 1 '13 at 22:19
@YlykCoitus: This answer assumes we are being to maximize $x_1+x_2$ instead of $x_1$ "and" $x_2$. Sometimes people read a plus sign as "and". I was just saying that seemed a good way to translate the problem to one that could be understood and solved. – Ross Millikan Apr 1 '13 at 22:24
@RossMillikan Oh, that's what you meant. But I still don't get how to do this. Isn't this answer wrong? The constraint $-x_1 + x_2 \leq 5$ makes the other ones redundant, right? Also, I still don't know what he means by 'choose the right one'. Right doesn't sound too mathematical for me. – Ylyk Coitus Apr 1 '13 at 22:28