I have to find the absolute maximum of $f(x, y, z) = 2xy + 3yz$ over the set $V = \{ (x, y, z) \in \mathbb{R}^3:\ x+y+z \leq 1, x \geq 0, y \geq 0, z \geq 0\}$.
I proved the set is both bounded and closed, hence Weierstrass guarantees there exist max and min (absolute) over V.
For the minimum I haven't got much problems, I found every point of the form $(0, y, 0)$ for $y \geq 0$ is a minimum.
I am having problems for the max though. Using the lagrangian:
$$L = 2xy + 3yz - \lambda(x+y+z-1) $$
I obtain the system
$$\begin{cases} 2y = \lambda \\ 2x + 3z = \lambda \\ 3y = \lambda \\ x+y+z = 1 \end{cases}$$
Which immediately states that $y = 0$.
Yet the solutions then come in the form $\left(-\frac{3}{2}z, 0, z\right)$, but it's not good since $z$ must be positive, but then $x$ is negative...
I checked with Wolfram and it says the absolute max lies at $(0, 1/2, 1/2)$ whence $f = \frac{3}{4}$.
How do I get it??
I thought that maybe I should/could use KKT, but i actually don't know how.
The problem seems really simple, yet the solution I naively found doesn't work.
You are not given a constraint but a region/set. We go with the ordinary way to find the extrema of a polynomial function over the given set: $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=0. $$ gives $y=0$ which is a part of the boundary of the given set and $\color{red}{f=0}$ there.
On $x=0$, $f=3yz$ and on the triangle $y,z\geq0,y+z\leq1$ it has the maximum $\color{red}{f=\frac34}$ at $(y,z)=(\frac12,\frac12).$
On $z=0$, $f=2xy$ and on the triangle $x,y\geq0,x+y\leq1$ it has the maximum $\color{red}{f=\frac12}$ at $(x,y)=(\frac12,\frac12).$
On $x+y+z=1$, $f=3y-xy-3y^2$ and on the triangle $x,y\geq0,x+y\leq1$ it has the maximum $\color{red}{f=\frac34}$ at $(x,y)=(0,\frac12)$.
Conclusion: $\max=\frac34$ at $(0,\frac12,\frac12).$