# Why is the target function minimized at (0,0) if I can get to a negative number?

I just started learning LP and I saw this Q in my textbook:

$$min : -x -y \\ S.T. : x + 2y \le 3, 2x +y \le 3, x \ge 0, y \ge 0$$

It is easy to see that the polygon created from these constraints can't have a negative coordinate. In my textbook it says the answer is $(0,0)$ since $-0 -0 = 0$ and this is the minimum value of the target function (i.e. $f(x,y) = -x -y$).

Why is that the minimum? If I choose any other values (S.T. the constraints are satisfied) I can get, for example, $(1,1)$ which will result in $-1 -1 = -2 < 0$!

Another thing that I don't understand - would there be a difference if my function was different? If the min is $(0,0)$ - wouldn't it be the case for any other $f(x,y) = a*x + b*y$ for any other $a,b$?

• Just using an online LP solver (I used comnuan.com/cmnn03/cmnn03004 for no good reason), the minimum is $x=1$, $y=1$ as you suggest. Have you checked the errata for the textbook? (And made sure you read the question right of course :D) – Luke Mathieson Nov 27 '17 at 8:26
• Checked. Twice :-( I just can't understand this. About my 2nd Q - the min will be the same for f(x,y) = a(x+y) as well, wouldn't it [a < 0]? – CIsForCookies Nov 27 '17 at 9:18

Is easy to prove that $(0,0)$ is the maximum of your suggested target function, so it may be an errata of your text.
Your second question is way more interesting, the answer is no, if you change your target function to another linear function the maximum or minimum may change, suppose you change your target function to $x+y$, the minimum now is $(0,0)$, previously it was $(1,1)$.
However, as you suggested in the comments, if you only change your function by a positive scaling factor $a$, the maximum/minimum remain the same. This is easy to prove as $$\alpha\leq \beta\iff a\alpha\leq a\beta$$ holds (I leave the details to you).