0
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Can you post the original problem please? $\endgroup$ Commented Nov 4, 2018 at 17:46

1 Answer 1

3
$\begingroup$

This is from a Lagrange multipliers situation?

I like to do these with no division to avoid considering division by zero. This means multiplying by the right things so that pieces of one equation appear in the other.

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.

$\endgroup$
1
  • $\begingroup$ 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! $\endgroup$
    – Jaigus
    Commented Nov 4, 2018 at 18:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .