anonymous
  • anonymous
set dy/dx=0 -(x^2-y)/(y^2-x)=DY/DX
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TuringTest
  • TuringTest
this is not the full prblem, is it?
anonymous
  • anonymous
yes it is
TuringTest
  • TuringTest
I was clearly asking caty.861

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
no, I found the dy/dx of x^3+y^3=3xy and got -(y-x^2)/(y^2-x)
anonymous
  • anonymous
o my bad
anonymous
  • anonymous
OMG I NEED HELP WIT MY PROBLEM!!!!!
TuringTest
  • TuringTest
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
anonymous
  • anonymous
so it would be does not exist?
TuringTest
  • TuringTest
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 \(x=\sqrt y\)
anonymous
  • anonymous
wouldn't it be plus or minus \[\sqrt{y}\] ?
TuringTest
  • TuringTest
yes, my mistake
anonymous
  • anonymous
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)
TuringTest
  • TuringTest
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....

Looking for something else?

Not the answer you are looking for? Search for more explanations.