Consider the function $f(x,y) = x^3 - 6xy + y^2 + 15x$.
Find the absolute extrema of $f$ over the triangular region in the $xy$-plane with vertices (0,0), (0,4), and (2,4).
I am using this graph to see that I am getting all of the points needed: https://www.desmos.com/calculator/tnyu81h2j5
I was able to find the interior point just fine, but I'm having trouble finding all the boundary points.
For x = 0, I got (0,0):
$f(x,y) = f(0,y) = y^2$
$f' = 2y = 0 \implies y = 0$
For y = 4, I got $(\sqrt{3}, 4)$:
$f(x, y) = f(x, 4) = x^3 -24x + 16 + 15x = x^3 - 9x +16$
$f' = 3x^2 - 9 = 0$
$x = +/- \sqrt{3} \implies -\sqrt{3}$ is not in region
But I am having trouble with y = 2x:
$f(x, y) = f(x, 2x) = x^3 - 12x^2 + 4x^2 + 15 = x^3 - 8x^2 + 15x$
$f' = 3x^2 - 16x + 15$
$x = \frac{16 \pm \sqrt{76}}{6}$
These x-values are not lining up with the remaining two critical points on the boundary. What am I doing wrong? How do I find these other points?
Regarding the line $y=2x$. You are doing the right thing when looking at what happens for the trinomial $x^3-8 x^2 +15x$.
You have to study it on the interval $x \in [0,2]$. The only root of the derivative in that interval is $x_1=\frac{16 + \sqrt{76}}{6}$ (the other one is outside the interval). $f(x,2x)$ is having a maximum at $x_1$. You can then compare the value of this maximum vs. the other local maximum that you found.