I have a homework question on solving a game matrix geometrically.
$m =$ $\begin{bmatrix}1 & 11\\7 & 2\end{bmatrix}$ (after adding the constant $k$ to ensure it's a positive matrix)
The first of the linear programming problem I got from this was:
$Minimize$ $$C = x + y$$
$Subject$ $to$ $$x + 7y >= 1$$ $$11x + 2y >= 1$$ $$x + y >= 0$$
I then rearranged the equations, solved for the intersections algebraically, and got the following solution "corners":
$$x = 1/15, y = 2/15, x + y = 3/15 $$ $$x = -1/6, y = 1/6, x + y = 0 $$ $$x = 1/9, y = -1/9, x + y = 0 $$
Since the goal is to minimize $x + y$, the bottom 2 solutions seem to both be equally correct (since they're both $0$). When I checked the answer using the following calculator though, it's saying that the answer is the first solution I got ($x = 1/15, y = 2/15$):
I haven't supplied any constraints saying that either $x$ or $y$ can't be negative. Why is the first solution the correct one?
The conditions $x\ge0,y\ge0$ are stronger than the condition $x+y\ge 0$.
If your maximum over all $(x,y)$ such that $x+y\ge0$ happens to have both $x$ and $y$ positive, then it is also feasible when you maximize over all $(x,y)$ such that $x\ge0,y\ge0$.
It also is better than any $(x,y)$ where $x\ge 0,y\ge 0$, since all of these points are included in the set of all $(x,y)$ such that $x+y\ge 0$.
So it's the maximum over the smaller set as well.