# Critical point multivariable

1. Jan 23, 2012

### danny_manny

1. The problem statement, all variables and given/known data
find the critical point of,

x^(2)+y^(2)+2x^(-1)y^(-1)

2. Relevant equations
none

3. The attempt at a solution
Ok so first we differentiate the function such that fx = 0 and fy=0

doing this yields,

fx = 2x-2x^(-2)y^(-1)
fy = 2y-2x^(-1)y^(-2)

both set to equal this is where i lose it,
so i can i get 1/x^(3) = y but subbing this into fy helps me little.

any help would be appreciated thanks,
dan

2. Jan 23, 2012

### Joffan

$f = x^{2} + y^{2} + \frac{2}{xy}$

You get two lines describing the stationary points for each partial derivative
$\frac{\partial f}{\partial y} = 0 ; 2y - \frac{2}{xy^{2}} = 0$ therefore $x = y^{-3}$

and similarly $y = x^{-3}$ from $\frac{\partial f}{\partial x}$

So the place(s) where those lines cross is/are stationary.

Last edited: Jan 23, 2012
3. Jan 23, 2012

### danny_manny

book gives the answer as (-1,-1) and (1,1)

4. Jan 23, 2012

### danny_manny

I'm totally lost on how the book arrives at the aforementioned answers.

Dan.

5. Jan 23, 2012

### Joffan

How can you find out where $x=y^{-3}$ crosses $y=x^{-3}$ ? Or are you lost earlier than this?

6. Jan 23, 2012

### danny_manny

lost at that point.

7. Jan 23, 2012

### danny_manny

Subbing either of those values into fx or fy doesn't seem to help me :(.
Dan.

8. Jan 23, 2012

### ehild

Your equation are identical with xy3 and x3y=1. Divide them and solve for x/y. What do you get?

ehild

9. Jan 23, 2012

### Joffan

Substituting the two equations into each other, as you said, you get x=(x-3)-3 (and/or y=(y-3)-3). From there it shouldn't be too hard.

10. Jan 24, 2012

### danny_manny

ok i get how the book gets point (1,1) but how does it get (-1,-1).

thanks for your assistance so far.
Dan

11. Jan 24, 2012

### ehild

How did you get point (1,1)?

ehild

12. Jan 24, 2012

### SammyS

Staff Emeritus
You have x3y = 1 and xy3 = 1.

You could graph these.

Or use algebra to solve them simultaneously.
Solving the way Joffan suggested is one way. Maybe the exponents involved confused you.

Alternatively, you can notice that since x3y and xy3 each equal 1, they equal each other.

So, x3y = xy3 → x2 = y2 → x = ±y.

Substituting that back into either equation gives ±x4 = 1. (or ±y4 = 1)

Only the + sign works for real numbers, so you have x = y.

Finally you get x4 = 1.

Solving this for x should give the two answers you're looking for. ​

13. Jan 25, 2012

### danny_manny

subbing y=1/x^(3) into x=1/y^(3), which leads to 1=x^(8).
so x = 1 and y = 1

14. Jan 25, 2012

### ehild

But x=-1 is also solution, isn't it? (-1)^8=1.

ehild