Extrema of $x+y+z$ subject to $x^2 - y^2 = 1$ and $2x + z = 1$ using Lagrange Multipliers

Question

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?

Answer 1

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.

Answer 2

Consider the constraints: you can simplify them in $$ y=\pm\sqrt{y^2+1}, \qquad z=\mp 2\sqrt{y^2+1}+1$$ for all $y\in\mathbb{R}$. In this way you can convert the constrained optimization in two unconstrained optimizations: substitute the two solutions in the objective function and you will find $$- \sqrt{y^2+1}+y+1, \qquad \sqrt{y^2+1}+y+1.$$

These two real functions are monotone increasing so they don't possess any stationary point.

Answer 3

Note: There is no extremas with the current situation...So I slightly changed one of the

"constraints" to:

$x^2 - 2y^2 = 1$, and to show you how to solve if using one parameter $LM$ method.

$x + y + z = x + y + (1 - 2x) = y - x + 1$, subject to: $x^2 - 2y^2 = 1$.

Define $f(x,y) = -x + y + 1$, and $g(x,y) = x^2 - 2y^2$. Then:

$\nabla f = \lambda\cdot \nabla g \to (-1,1) = (2x\lambda,-4y\lambda)$. Thus:

$-1 = 2x\lambda$

$1 = -4y\lambda$.

Add the above equations: $2\lambda(x - 2y) = 0 \to \lambda = 0$ or $x = 2y$. But $\lambda$ cannot be $0$, so $x = 2y$, and with $x^2 - 2y^2 = 1$ we get:

$4y^2 - 2y^2 = 1 \to 2y^2 = 1 \to y = \pm \dfrac{1}{\sqrt{2}}$. So $x = 2y = \pm \sqrt{2}$.

$f_{min} = f(\sqrt{2},\frac{1}{\sqrt{2}}) = \dfrac{\sqrt{2} - 1}{\sqrt{2}}$, and

$f_{max} = f(-\sqrt{2},-\frac{1}{\sqrt{2}}) = \dfrac{\sqrt{2} + 1}{\sqrt{2}}$