Linear Algebra 101 - Optimizing inequalities I am considering the region contained in $\mathbb{R}^2$ consisting of all the points that satisfy all the following inequality:
$-4 \leq y < 4 \\
-9 \leq 2x + y \leq 9 \\
-9 \leq x + 2y \leq 9 \\
-4 \leq x < 4 \\
-9 \leq 2x - y \leq 9  \\
-9 \leq x - 2y \leq 9 $
Determine the maximum value of $f(x,y)= 5x +10y + \pi^2$ on this region. At which points in the region does it occur?
 A: Since $\pi^2$ is a constant, your problem is really maximizing $g(x,y)=5x+10y$, subject to the linear inequalities you're listing.
The solution is not strictly inside the region. It will actually be one of the "corner points" that define the polytope induced by the inequalities. Look up the fundamental theorem of linear programming.
Now, since you have positive coefficients in your objective function, you only need to investigate the corner points that occur in the first quadrant ($x \ge 0, y \ge 0$); any other feasible solution that lists either $x$ or $y$ as negative is dominated by one with positive values, since it increases your goal.
In the first quadrant there are three such points:
$A$, at the intersection of $y=4$ with $x+2y = 9$. It therefore has coordinates $(1,4)$.
$B$, at the intersection of $x+2y=9$ with $2x+y=9$. Solving simultaneously you get the coordinates $(3,3)$.
$C$, at the intersection of $2x+y=9$ with $x=4$. It therefore has coordinates $(4,1)$.
The value of your objective function are, for each of these points,
$$
f(1,4) = 5(1) + 10(4) + \pi^2 = 45 + \pi^2 \\
f(3,3) = 5(3) + 10(3) + \pi^2 = 45 + \pi^2 \\
f(4,1) = 5(4) + 10(1) + \pi^2 = 30 + \pi^2
$$
Since you get equal (maximum) values for points $A$ and $B$ you have alternate optimal solutions. The entire segment between $(4,1)$ and $(3,3)$ consists of optimal points.
