Find the maximum and the minimum of the given function $u$ in the field $\mathbb G$: $$ u=x+2y+3z,\ \ \ \ \mathbb G=\left\{x+y\leqslant 3,\ x+y\leqslant z,\ 3x+3y\geqslant z,\ x\geqslant 0,\ y\geqslant 0\right\} $$
I know how to solve these kinds of problems in the $2$-D case (with only $x$ and $y$). It is pretty straightforward there, since there is no difficulty in plotting $\mathbb G$ there. But here I am struggling to do the latter. Okay, it is easy to plot $xOy$, but it is not enough.
My attempt anyway:
First, it is clear that $u_x'=1\ne 0\Rightarrow$ there are no stationary points inside the field $\mathbb G$. Okay, now I would like to apply the method of Lagrange multipliers, but I can't do that until I explicitly define the boundaries. Could someone help me with that?
Using Lagrange multipliers is ill-advised for this problem, which is a linear programming problem and can be solved using techniques optimised for such problems. I'll just solve this problem via first principles.
For an optimal solution we need three inequalities to be tight at the putative optimum. Let's say we are maximising, and set $x+y=3$ because that allows the highest possible $z$-values from $3x+3y\ge z$. This latter equation then implies $z=9$ for our purposes. Now the constraints $x,y\ge0$, combined with the objective function $u$, give us $(x,y,z)=(0,3,9)$ as the maximum of $u$ in $G$, with value $33$.
For minimising, since $x,y$ cannot be negative, $z$ also cannot be negative by $x+y\le z$. Thus the obvious minimum is $(x,y,z)=(0,0,0)$ with value $0$.