I saw this question here Maximum of $x^3+y^3+z^3$ with $x+y+z=3$ and wanted to try this approach (maximizing $x^3+y^3+z^3$ subject to constraints $x+y+z=3$ and $0 \leq x, y, z \leq 2$).
I tried this:
Let $f(x,y,z) = x^3+y^3+z^3$ and $g(x,y,z) = x+y+z$. Then the Lagrange multiplier equations are:
$$ \begin{align} x+y+z &= 3 \\ 3x^2 &= \lambda \\ 3y^2 &= \lambda\\ 3z^2 &= \lambda \\ \end{align}$$
This implies $\lambda = 3$ with $x=y=z=1$ but clearly this doesn't maximize $f$ as much as setting $x,y,z=0,2,1$. What's the deal?