# How can I solve this particular non-linear system of equations?

Here is the original problem:

Find the extrema of $$f$$ subject to the stated constraints.

$$f(x,y) = x - y$$, subject to $$x^2-y^2=2$$

I'm solving a problem involving Lagrange's multipliers, and I've got this system of equations to solve:

$$1 = \lambda2x$$

$$-1 = -\lambda2y$$

$$x^2-y^2-2 = 0$$

I tried solving for x and y in terms of lambda and plugging them in $$g(x)$$ (the third equation), however that didn't get me anywhere as it gave me the inequality -2 = 0.

• Can you post the original problem please? Commented Nov 4, 2018 at 17:46

You have $$1=\lambda2x$$ You can multiply by $$-y$$ so that the right side contains the right side from the second equation. $$-y=x(-\lambda2y)$$ Then make use of the second equation $$-y=x(-1)$$ So $$y=x$$. Put that in y our third equation: $$x^2-x^2-2=0$$ and there is a problem. This says $$-2=0$$, so there are no solutions. In the context of your original problem, that makes sense. The curve you are restricted to is a hyperbola, and the line $$y=x$$ is one of its asymptotes. There will be no place where the function takes extremal values. Either you can keep moving along the hyperbola closer and closer to that asymptote, with $$f$$ increasing (or is it decreasing?) the whole time, or you can move away from the asymptote forever with $$f$$ always reducing (or is it growing?).
Another way to say that is, the gradient of $$f$$ is never parallel to this hyperbola.
• Thanks, this was indeed a Lagrange multipliers situation. I edited the question to include the problem in it's original form. I got to the same situation by plugging in $1/2\lambda$ for x and y in the constraint equation; but I thought there might be another way to go about it. Thanks for the insight! Commented Nov 4, 2018 at 18:13