Wolfram Alpha tells me that $xy^2$ only has the stationary points (0,0).... but why?
We get the gradient $$(y^2 , 2xy)$$ and this is surely 0 as long as $y=0$ meaning $x$ can be whatever it wants to be and it'll still be a stationary point?
Asked
Active
Viewed 72 times
2
-
@Amzoti: Just as it says in the question, the solution is that $y$ must be zero, while $x$ can be arbitrary. Surely WA's answer is wrong (or incomplete, rather) here. – Hans Lundmark Nov 02 '14 at 15:02
-
Yes, WolframAlpha is wrong. Someone tell @in_wolframAlpha_we_trust the bad news... – Nov 04 '14 at 11:10