# How can I find the two critical points of this system of equations?

I'm currently trying to use Lagrange Multipliers to find the 2 critical points of the function $$f(x,y,z) = \frac{1}{2}x^{2}+yz+\frac{1}{3} y^{3} - z^{2}$$ subject to $$h(x,y,z) = x+y+z-2 = 0$$

I have formed the Lagrangian for this, which is given by $$L= \frac{1}{2}x^{2}+yz+\frac{1}{3} y^{3} - z^{2} - \lambda (x+y+z-2)$$ and I have worked out that $$L_{x}= x-\lambda = 0 \\ L_{y}= z+y^{2} - \lambda = 0 \\ L_{z}= y-2z-\lambda = 0 \\ L_{\lambda}=-x-y-z+2 =0$$

How would I find the values of $x,y,z, \lambda$ that constitute the critical points of this system?

EDIT: I have deduced that $$x=\lambda \\ y=\frac{4-\lambda}{3} \\ z=\frac{2-2\lambda}{3}$$ Thus, since $x+y+z=2$, we must solve the equation $$\lambda + \frac{4-\lambda}{3} + \frac{2-2\lambda}{3} = 2$$ for $\lambda$. However, any $\lambda \in \mathbb{R}$ is a solution.

What am I doing wrong?

• Doing certain correct manipulations you obtained $0=0$. This does not mean that you have done something wrong; it just means that you were not sufficiently aiming at the target. – Christian Blatter Dec 9 '15 at 19:12

You got $x$, $y$, and $z$ from below three equations \begin{align} L_{x}&= x-\lambda = 0 \\ L_{z}&= y-2z-\lambda = 0 \\ L_{\lambda}&=-x-y-z+2 =0 \end{align} and put those back to one of those ($L_{\lambda}$), which of course always holds.
You should have put those back to this one $$L_{y}= z+y^{2} - \lambda = 0$$