Using Lagrangian multiplier method with multiple constraints So I am trying to find the minimum and maximum of the function $f(x,y,z)= x^2 + y^2 - z^2$ on the curve defined by $y^2 + z^2 =1$ and $x=y.$
My work thus far is the following:
$\text{Proof.}$ Let $g=y^2 + z^2 -1=0, h=x-y=0,$ and taking partials, $$f_x = 2x, f_y = 2y, f_z = -2z$$ $$g_x = 0, g_y = 2y, g_z = 2z,$$ $$h_x = 1, h_y = -1, h_z = 0$$ and by definition of the Lagrange multiplier with multiple constraints, we have $\nabla f = \lambda \nabla g + \mu\nabla h$ corresponding to each parameter. Thus, $$\nabla f_x = \lambda g_x + \mu\nabla h_x \implies x = \frac{\mu}{2}$$ $$\nabla f_y = \lambda g_y + \mu\nabla h_y \implies y = \lambda y - \frac{\mu}{2}$$ $$\nabla f_z = \lambda g_z + \mu\nabla h_z \implies z = -\lambda z$$ Now from here I am having issues because I cannot find a solution for $y,z$ Can someone please help me? Thank you.
 A: Assuming your setup is correct up to this point, basically there is a bunch of case work to do. If you're allowed to divide by whatever you want then you can do this:
$$y=\lambda y - \mu/2 \Rightarrow y=\frac{\mu/2}{\lambda-1} \\
z = -\lambda z \Rightarrow \lambda = -1 \\
\Rightarrow y = \frac{-\mu}{4} \\
\Rightarrow x = \frac{-\mu}{4}$$
However we also know $x=\frac{\mu}{2}$, so $\mu=0$, and so you get $z=\pm 1$ so you have the points $(0,0,1)$ and $(0,0,-1)$. You probably don't actually care but the values of the Lagrange multipliers are $\lambda=-1$ and $\mu=0$.
Now what assumptions did you make along the way? You assumed $\lambda \neq 1$ and $z \neq 0$. What happens if you violate one or both of those assumptions?
Incidentally I would say that an easier approach to this particular problem would be to just substitute $x=y$ and do a one-constraint Lagrange problem. Thus you look at $f(y,z)=2y^2-z^2$ subject to $y^2+z^2-1=0$, so $4y=\lambda 2y$ and $-2z=\lambda 2z$, from which you readily conclude that at least one of $y$ or $z$ must be zero (since $\lambda$ cannot be both $2$ and $-1$ at the same time).
A: The symmetries in the geometrical arrangement will help somewhat with locating solutions.  The constraint surfaces are a circular cylinder with its symmetry axis along the $ \ x-$axis and a plane cutting obliquely through the cylinder.  The intersection curve is then an ellipse symmetrical about the $ \ xy-$plane, so we would expect the points at which the extremal values of the function occur to have coordinates $ \ (\pm x \ , \ \pm x \ , \ \pm z) \ = \ (\pm x \ , \ \pm x \ , \ \pm \sqrt{1-x^2}) \ \ . $

For the Lagrange equations, it is sometimes best to bring all or most of the terms to one side in order to factor them; this reduces the risk of overlooking solutions.  Here, we would have
$$ 2x \ = \ \mu \ \ , \ \ 2y \ = \ \lambda·2y - \mu \ \ \rightarrow \ \ 2y · (1 - \lambda) \ = \ -\mu \ \ , $$ $$ -2z \ = \ \lambda·2z \ \ \rightarrow \ \ -2z · (1 + \lambda) \ = \ 0 \ \ .  $$
It is now clearer from the third equation that we have the two cases:
$ \mathbf{z = 0 \ , \ \lambda \neq -1 \ } \ , $ from which it follows immediately that $ \ y^2 \ = \ 1 \ \Rightarrow \ y = x = \pm 1 \ \ , $ for which we have the function value $$ f(\pm 1 \ , \ \pm 1 \ , 0 ) \ \ = \ \ 1^2 \ + \ 1^2 \ - \ 0^2 \ = \ 2 \ \  ;  $$
and
$ \mathbf{z \neq 0 \ , \ \lambda = -1 \ } \ , $ for which we obtain $ \ 2y · (1 - [-1]) \ = \ 4y \ =  \ -\mu \ = \ -2x \ \ ; $  since the planar constraint requires $ \ y = x \ , $ we must have $ \ x = y = 0 \ \Rightarrow \ z^2 = 1 \ \ , $ leading at once to $ \ f(0 \ , \ 0 \ , \pm 1 ) \ \ = \ \ 0^2 \ + \ 0^2 \ - \ 1 \ = \ -1 \ \ . $
The former case thus gives us the absolute maximum $ \ 2 \ $ and the latter case the absolute minimum $ \ -1 \ $ on the intersection ellipse.
