Find the global max and min of $f(x,y)=18-6x+10y$ on a closed triangular region with vertices $(0,0),(10,0),(10,11)$

Question

So I found $f_x=-6$ and $f_y=10$, which then $-6=0$ and $10=0$. This is were I'm confused, because I'm not sure what my critical points are. and When its along the $x=0$ it is should be $0\leq y \leq 10$:$f(x,y)=18+10y$, meaning $f'(x,y)=10$. This is another area where I'm confused because again we'd have to set $10=0$ but that doesn't help to find $y$ to give me a point.

Answer 1

$g(x,y) = -6x + 10y$ is a linear function on $(x\quad y)$, and the triangular region restriction can be described as linear restrictions $a^T (x\quad y) \leq b$. The optimum is reached on the vertices of this region.

$g(0,0) = 0,\quad g(10,0) = -60,\quad g(10,11) = 50$.

So the global max of $f$ is $68$ and the minimum $-42$.