I am trying to find the absolute extremes of the function: $f(x,y)=x^3+xy+y$ in the enclosed triangle region limited by the lines $x=-1$, $y=3$, $y=x+2$.
So far I have been able to graph the triangle but in determining the equations of the triangle, I am not sure how to proceed.
$$ \frac{\partial f}{\partial x} = 3x^2 + y \quad \text{ and } \quad \frac{\partial f}{\partial y} = x+1. $$ Thus $\partial f/\partial y = 0$ only when $x=-1,$ thus only on the boundary. Since there is no point in the interior where both partial derivatives are $0,$ an extreme value can occur only on the boundary.
So you have $f(-1,y) = (-1)^3+(-1)y + y = -1,$ so the behavior of the function on that boundary is simple.
And $f(x,3) = x^3+3x+3,$ so you want extreme values of that function of $x$ on the interval from $-1$ to $1.$
And on the line $y=x+2$ you have $f(x,y) = f(x,x+2) = x^3 + x(x+2) + (x+2),$ so you want extreme values of that function of $x$ between $-1$ and $1.$