Finding extremum value of multivariable function How can I find the max and min values of $$f(x,y)=2x^2y-3xy^2+36xy$$ on and in the triangular region with vertices $(-14,-2), (0,-2), (0,12)$?
I have graphed the triangular region and calculated
$$f_x(x,y)=4xy-3y^2+36y$$
$$f_y(x,y)=2x^2-6xy+36x$$
I do not know how to continue.
Note: this is a homework question, with explicit permission to "discuss ideas"
Edit: Forgot to add that I have found the critical points to be $(0,0),(0,12),(-6,4),(-18,0)$
 A: As PJK commented, the first thing to do is to write
$$f_x(x,y)=4xy-3y^2+36y=y(4x-3y+36)=0 \tag 1$$
$$f_y(x,y)=2x^2-6xy+36x=x(2x-6y+36)=0 \tag 2$$
Now, from$(2)$, $y=6+\frac x 3$. Plug it in $(2)$ to get $$x^2+24 x+108=(x+6)(x+18)=0 \tag 3$$ So , the solutions are $(x_1=-18,y_1=0)$,  $(x_2=-6,y_2=4)$. But $(x_3=0,y_3=0)$ is also a solution as well as $(x_4=0,y_4=12)$.
Now compare with the triangle to see the solutions to be retained and, for these, compute $f(x,y)$.
A: As you have already noted, $(0, 0), (18, 0), (0, 12)$ and $(-6, 4)$ are the critical points of which only $(-6, 4)$ lies inside the triangle with $\color{blue}{f(-6, 4)=-288}$. So this point $(-6, 4)$ is one candidate for global max or min. 
Now for the boundary of the triangular region, we look for the possible max/min at each of the three straight lines bounding the triangle. For example, lets consider the side joining $(-14, -2)$ and $(0, -2)$. The equation of this line is $y=-2$ and so the function $f(x, y)$ reduces to $$f(x, -2)=-4x^2-12x-72x=-4x^2-84x$$ with  $-14\leq x\leq 0$ on this line. This is a single variable function with critical point $x=-\frac{21}{2}$ which lies within $-14\leq x\leq 0$. So the point $(-\frac{21}{2}, -2)$ is one candidate for global max/min on this edge of the triangle with $\color{blue}{f(-\frac{21}{2}, -2)=441}$. Also on the two boundary points $(-14, -2)$ and $(0, 2)$ of this line, we have $\color{blue}{f(-14, -2)=392}$ and $\color{blue}{f(0, 2)=0}$.
Proceed similarly for the other two sides and then find the global max or min by comparing all the possible candidates (colored in blue).
