The function $u(x,y,z) = xyz$ is to be maximized, under constraints: $ 0 \le x \le 1, y \ge 2, z \ge 0 $ and $ 4 - x - y - z \ge 0 $
Now I'm not quite sure how to translate the x-constraint into proper inequalaties, but my first shot would be $ h_1 = x \ge 0$ and $h_2 = 1-x \ge 0 $.
So I set up the Lagrangian like: $$ L = xyz + \lambda_1 x + \lambda_2(1-x) + \lambda_3(y-2) + \lambda_4 z + \lambda_5(4-x-y-z) $$
... resulting in FOC like:
[... irrelevant FOC omitted ...]
$$ \frac{\delta L}{\delta z} = xy + \lambda_4 - \lambda_5 = 0 $$ $$ \frac{\delta L}{\delta \lambda_4} = z \ge 0, \lambda_4 \ge 0, z \lambda_4 = 0$$
Now, as $u=xyz$, neither $x,y,z$ can be $0$ in maximum; thus, $\lambda_4 = 0$. Also, as $u$ is isotone in all $x,y,z$, the fifth constraint has to be effective in maximum, so $\lambda_5 = 0 $ either. So plugging into the first FOC from above, $xy = 0$ - so either $x$ or $y$ has to be $0$, which cannot be a maximum though!
So am I making a mistake in the way I set up the Lagrangian here? In the derivatives? Or can't I apply this method here at all? If so, why not?
Thanks for any hint!
Since the fourth constraint: $z\geq0$ will not bind at the maximum, you are correct that $\lambda_4=0$ (by the complimentary slackness condition). However, since the 5th constraint will bind, $\lambda_5\geq0$.
You are left with $xy=\lambda_5$ rather than $xy=0$.