Finding the Critical Point of a Multivariable Function

[danny_manny]

Homework Statement

find the critical point of,

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

Homework Equations

none

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

 

[Joffan]

OK here's the function, as I read it from your post:

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

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.

 

[danny_manny]

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

 

[danny_manny]

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

Dan.

 

[Joffan]

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

 

[danny_manny]

lost at that point.

 

[danny_manny]

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

 

[ehild]

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

ehild

 

[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.

 

[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

 

[ehild]

How did you get point (1,1)?

ehild

 

[SammyS]

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

thanks for your assistance so far.
Dan

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.​

 

[danny_manny]

How did you get point (1,1)?

ehild

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

 

[ehild]

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