About using the lagrange multiplier. Today I have seen a question like the following;
$x+y+z=5$ and $xy+yz+xz=3$ and $x,y,z \in \mathbb{R}^+$
What is the maximum value $x$ can get?
Now it is pretty obvious that question is solvable using many simple elementary methods like say $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=25$
$25=x^2+y^2+z^2+6 \rightarrow x^2+y^2+z^2=19$
Applying Cauchy-Schwarz $(y^2+z^2)(1+1)\geq (z+y)^2$
$2(19-x^2)\ge(5-x)^2$
$x_{max}=\dfrac{13}{3}$
However I would also like to solve it by the lagrange multiplier, here are my efforts
$f(x,y,z,k)=x+5-(y+z)+k(x+y+z-5)$
$f_k'=0$
$f_x'=k+1$
$f_y'=k-1$
$f_z'=k-1$
$k=1,-1$ and $x$ goes nowhere near $\dfrac{13}{3}$

So my question is where am I wrong and how can I fix it?
Besides are there maxima minima questions that can't be solved with lagrange,
is this one of them?


We will use the quadratic equation approach to determine the min and max values of $x$. We have: $y+z = 5-x$, and $yz + x(y+z) = 3\implies 3- x(5-x)= yz\implies x^2-5x+3= yz\le \dfrac{(y+z)^2}{4}= \dfrac{(5-x)^2}{4}\implies 4x^2-20x+12 \le 25-10x+x^2\implies 3x^2-10x-13 \le 0 \implies (3x -13)(x + 1)\le 0\implies -1 \le x \le \dfrac{13}{3}\implies x_{\text{min}} = -1, x _{\text{max}} = \dfrac{13}{3}$ .
