# Find the extrema of f subject to the stated constraint

1. May 22, 2010

### number0

1. The problem statement, all variables and given/known data

Find the extrema of f subject to the stated constraint:

f(x,y) = x-y subject to x2-y2=2

2. Relevant equations

Apply the Lagrange Multiplier!

3. The attempt at a solution

This question was rather odd... I just did a problem similar to this one, and I got the answer right.

Let g(x,y) = x2-y2-2

Now let L(x,y) = f-$$\lambda$$g (where $$\lambda$$ = the Lagrange Multiplier)

Lx = 1 - 2$$\lambda$$x = 0

Ly = -1 + 2$$\lambda$$y = 0

I then solve for x and y.

I get x = $$\frac{1}{2\lambda}$$ = y

I plugged both of them into the constraint g(x,y) = x2-y2-2 = 0

Both x and y cancels out!!! and so

-2 = 0

I am sure I am doing something wrong because there is an answer! I've checked using an online calculator.

Can anyone please show me what I am doing wrong?

Last edited: May 22, 2010
2. May 22, 2010

### Dickfore

If you parametrize x and y using hyperbolic substitution:

$$\begin{array}{rcl} x & = & \sqrt{2} \; \epsilon \cosh{t}, \; \epsilon = \pm 1 \\ y & = & \sqrt{2} \sinh{t}, \; -\infty < t < \infty \\ \end{array}$$

and substitute this into the function $f(x, y)$, you will get a function of a single variable $\tilde{f}(t)$. Can you find its analytic expression? Does this function have extrema?

3. May 22, 2010

### Dickfore

you made an error in the last line. The first and second term cancel.

4. May 22, 2010

### cronxeh

Yes you right I made a mistake, the substitution is the way to go

5. May 22, 2010

### number0

Sorry guys, but I still do not know what to do if the x's and y's cancels out...

6. May 22, 2010

### Dick

x^2-y^2=2 is the same as (x+y)(x-y)=2 is the same as (x-y)=2/(x+y). Have you considered the possibility it may not have any extrema?

7. May 22, 2010

### number0

Yes... but, I said I used the online calculator and it said the values for the extrema.

http://www.wolframalpha.com/input/?i=maximize+x-y+on+x^2-y^2-2%3D0

Note*** I may be misreading the values on the online calculator

8. May 22, 2010

### Dick

Don't trust wolfram alpha for everything, I don't even know what it's trying to say. I think it's simply confused. You've correctly concluded a local extremum exists if 0=(-2). It doesn't. There aren't any local extrema.

9. May 22, 2010

Thank you!