Let's consider $f:\mathbb{R}^2\to\mathbb{R}$ with $f(x,y):=2x^3-12x+3y^2+6xy$. Find the global maximum and global minimum on $D:=\{(x,y)\in\mathbb{R}^2\mid x\geq 0,y\geq0,x+y\leq 1\}$.
My approach:
First, by solving $Df(x,y)=(0,0)$ we easily find a local minimum at $(2,-2)$ with $f(2,-2)=-20$. There are no other extreme points on $\mathbb{R}^2$ (there exists a saddle point at $(-1,1)$ which we ignore).
The existence of global extreme points on $D$ is guaranteed by compactness of $D$ and continuity of $f$. It suffices if we only look at the boundary of $D$ because possible extreme points in the interior have already been covered by the condition $Df(x,y)=(0,0)$; note that $(2,-2)\notin D$.
Next,
a.) $f(0,0)=0$.
b.) If $x=0$ and $0<y\leq 1$, then $0<f(x,y)\leq 3$.
c.) If $0<x\leq 1$ and $y=0$, then $f(x,0)$ is monotonically decreasing and $-10\leq f(x,0)<0$.
d.) If $0<x<1$ and $0<y<1$ and $x+y<1$, then this condition defines the interior of $D$.
e.) If $0<x<1$ and $0<y<1$ and $x+y=1$, then we can apply the method of Lagrange multipliers which yields\begin{align*}&6x^2-12+6y=\lambda\\&6y+6x=\lambda\\&x+y=1\\&\implies x^2-x-2=0\implies x_1=-1\land x_2=2, \text{ where } x_1,x_2\notin D\end{align*}So this condition doesn't deliver any possible global extreme points.
Condition b.) delivers the global maximum, namely $f(0,1)=3$ and c.) delivers the global minimum on $D$, namely $f(1,0)=-10$.
I am not quite sure if I can prove it like this because in the sample solution there are different cases where the Lagrange multipiers are used. But I don't think that this is necessary, isn't it?