Question

set dy/dx=0
-(x^2-y)/(y^2-x)=DY/DX

Answer 1

this is not the full prblem, is it?

Answer 2

yes it is

Answer 3

I was clearly asking caty.861

Answer 4

no, I found the dy/dx of x^3+y^3=3xy and got -(y-x^2)/(y^2-x)

Answer 5

o my bad

Answer 6

OMG I NEED HELP WIT MY PROBLEM!!!!!

Answer 7

I got \[\frac{dy}{dx}={y-x^2\over y^2-x}=0\]the thing here being that we can't solve your original function for y explicitley, so we can't solve for x with a unique solution

Answer 8

so it would be does not exist?

Answer 9

so we wind up with\[y-x^2=0\implies x=\sqrt y\]now if we could solve for y explicitly, we could continue and get an actual number answer for x, but as it is we only know that the critical points occur when \(x=\sqrt y\)

Answer 10

wouldn't it be plus or minus \[\sqrt{y}\] ?

Answer 11

yes, my mistake

Answer 12

ok. from here, I need to find the local maximum of the 1st equation x^3+y^3=3xy but I don't know how. I already go the dy/dx which is -(x^2-y)/(y^2-x)

Answer 13

if we could solve x^3+y^3=3xy for y, then we could take the derivative, set it to zero, and solve for x
that is what I did above, but as you can see since we cannot solve explicitly for y, we are left with \(x=\pm\sqrt y\), which are your critical points
we can't get a number though, so....