Extrema of $x+y+z$ subject to $x^2 - y^2 = 1$ and $2x + z = 1$ using Lagrange Multipliers 
Find the extrema of $x+y+z$ subject to $x^2 - y^2 = 1$ and $2x + z = 1$ using Lagrange multipliers.

So I set it up:
$$
1 = 2x\lambda_1 + 2\lambda_2 \\
1 = -2y\lambda_1 \\
1 = \lambda_2
$$
Plug in for $\lambda_2$:
$$
1 = 2x\lambda_1 + 2 \\
1 = -2y\lambda_1 \\
$$
So we work with:
$$
1 = 2x\lambda_1 + 2 \\
1 = -2y\lambda_1 \\
1 = x^2 - y^2 \\
1 = 2x + z
$$
After some algebra I got $x = y$ as a solution but that's impossible because of the constraint $1 = x^2 - y^2$. What am I missing?

The constraints define two curves in ${\mathbb R}^3$ as follows: The constraint $x^2-y^2=1$ defines a hyperbolic cylinder $Z$ consisting of two sheets, which can be parametrized as follows:
$$(t,z)\mapsto(\pm\cosh t,\sin t, z)\qquad(-\infty<t<\infty, \ -\infty<z<\infty)\ .$$
Intersecting these two sheets with the plane $z=1-2x$ produces thw two curves
$$\gamma_\pm:\ t\mapsto(\pm\cosh t,\sinh t, 1\mp2\cosh t)\qquad(-\infty<t<\infty)\ .$$
The pullback of $f(x,y,z):=x+y+z$ to $\gamma_\pm$ computes to
$$\phi(t)=1+\sinh t\mp\cosh t\ ,$$
so that we obtain
$$\phi'(t)=\cosh t\mp\sinh t=e^{\mp t}>0\qquad(-\infty<t<\infty)\ .$$
This shows that there are no conditionally stationary points of $f$ on the two curves, and explains why Lagrange's method didn't produce any solutions.
