Question

How can I find the values of x and y when

dz/dx = 2xe^{y^2-x^2}(1-x^2-y^2) = 0

dz/dy = 2ye^{y^2-x^2}(1+x^2+y^2) = 0 ?

Answer 1

I have to evaluate one function for the other in order to obtain the x and y-values but Im not sure how to take care of e...

Answer 2

I'm not sure how solving for \(\ f_{y} = 0\) would give me y = 0.

Answer 3

what was the original question

Answer 4

Can you set them equal to each other, and then divide everything away leaving you with x=y ?

Answer 5

finding the local maximum/minimum and saddle points of the function.

The function is: \(\ f(x) = (x^2 +y^2)e^{y^2-x^2}\)

Answer 6

ahh, thats a feather of a different bird

Answer 7

Oh whoops I didn't notice the difference in signs.

At the very least, e^r is always a positive number so you can safely divide it out.

Answer 8

they are partials right ?

Answer 9

In the example problems I'm told to find the derivative w.r.t both x and y, then substitute a value of x or y in place of the other.

Therefore I've found the partials with respect to both, but i don't understand how to evaluate one to put in place of the other.

Answer 10

there is a formula here
http://tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx

Answer 11

the solution states to let \(\ f_y = 0\) implying that y = 0, but how did they come to that conclusion? :\

Answer 12

so did you set them equal to each other?

Answer 13

Well for example for \[\LARGE z_x=0=2xe^{y^2-x^2}(1-x^2-y^2)\] then either \[\LARGE x=0, \ or, \ e^{y^2-x^2}=0, \ or, \ (1-x^2-y^2)=0\]

Answer 14

If I give you something like \[\LARGE a*b=0\] then you agree that either a=0 or b=0 or both correct? Same idea here in how they figured it out.