# Absolute extremes of: $f(x,y)=x^3+xy+y$ in the enclosed triangle region limited by the lines $x=-1$, $y=3$, $y=x+2$.

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.

• Find the functions's extrema using the gradient, and determine whether any of those points are inside the region. Next, evaluate the function on the boundaries and find any other possible extrema there. – Dylan Nov 18 '17 at 21:52
• when you say evaluate the function on the boundaries, you mean the lines? – Omari Celestine Nov 18 '17 at 21:56
• The line segments that make up the triangle – Dylan Nov 18 '17 at 22:28

$$\frac{\partial f}{\partial x} = 3x^2 + y \quad \text{ and } \quad \frac{\partial f}{\partial y} = x+1.$$ 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 $1.$