I solved this exercise and I wanted to ask you in what way using KKT conditions, which here I did not use / wrote down, would have made (IF) the solution better / shorter or the procedure more rigorous.
I have to minimize $z = (x-3)^2 + (y-4)^2$ subject to $x + 2y \leq 8$, $4x + y \leq 10$ and $x \geq 0$, $y\geq 0$.
Solution
$$L = (x-3)^2 + (y-4)^2 - \lambda (x+2y-8) - \mu (4x + y - 10)$$
Thence
$$ \begin{cases} 2x- 6 - \lambda - 4 \mu = 0 \\\\ 2y - 8 - 2\lambda - \mu = 0 \\\\ \lambda(x+2y -8) = 0 \\\\ \mu (4x + y - 10) = 0 \\\\ x + 2y \leq 8 \\\\ 4x + y \leq 10 \end{cases} $$
Now: if $\lambda = \mu = 0$ I get $y = 4$, $x = 3$ which is not admissible.
If $\lambda \neq 0 $ and $\mu = 0$ I got $x = \frac{4}{5}$ and $y = \frac{18}{5}$ which is admissible. In particular here $\lambda < 0$
If $\lambda = 0$ and $mu \neq 0$ I got $x = \frac{27}{17}$ and $y = \frac{62}{17}$ which does not satisfy.
If $\lambda$ and $\mu$ are both $\neq 0$ I got $\lambda < 0$, $\mu < 0$ and $x = \frac{12}{7}$ and $y = \frac{22}{7}$ which is ok.
And this is indeed the absolute minimum.
I have been told from our professor that the multipliers must be positive. So why those are negative and still the result hold?
As you can see, I did not wrote down KKT conditions entirely because for what I understood from the class, the Lagrangian should hae taken into account also terms like $\rho x$ and $\sigma y$ after having turned $x \to -x$ and $y \to -y$, and also writing in the system $\rho x = 0$ and $\sigma y = 0$.
So why / how my result is correct even with less multipliers and incomplete KKT conditions?
If for some reason I understood wrong something, please point it out.
It's because your Lagrange equation is non-standard. If the constraints are
$$ \begin{aligned} 0&\geq g(x,y) \\ &=x+2y-8 \\ \\ 0&\geq h(x,y) \\ &=4x+y-10 \end{aligned} $$
The Lagrange equation in standard form is given by $z+\lambda g(x,y) + \mu h(x,y)$. You use negative sign, which is why your solution has negative multiplier.