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I think I know this one but the solution given is different than what I get so maybe I've missed something important.

From $f(x,y)=x^2y^2+x^2y+2y^2-4y$ we have $f'_x=2xy^2+2xy$ and $f'_y=2x^2y+x^2+4y-4$. Setting both equations equal to $0$ we have $(1)$ $2xy^2+2xy=2xy(y+1)=0$ and $(2)$ $2x^2y+x^2+4y-4=0$.

The points that satisfy $(1)$ is $(0,y)$, $(x,0)$ and $(x,-1)$ where $x,y\in\mathbb{R}$.

Putting $(0,y)$ in $(2)$ gives $4y-4=0$ and it follows that $(0,1)$ is one stationary point.

Putting $(x,0)$ in $(2)$ gives $x^2-4=0\iff x=2\text{ or } x=-2$ so the stationary points are $(2,0)$ and $(-2,0)$.

Similarly with $(x,-1)$ we get $-x^2-8=0$ which does not have any real solutions.

So the stationary points I got are $(0,1)$, $(2,0)$ and $(-2,0)$ but the solution that was given to me is only $(0,1)$. I know the other two points satisfy $(1)$ and $(2)$ but I'm trying to make sure I haven't missed something.

per persson
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  • I'm guessing extreme points are those points that give max and min values while stationary points include extreme points but can also be saddle points. So can I use the second partial derivative test to conclude which points aren't extreme points? – per persson Feb 03 '23 at 12:38
  • https://math.stackexchange.com/questions/1280588/extrema-of-functions-of-two-variables-necessary-and-sufficient-conditions – zkutch Feb 03 '23 at 12:38
  • Along the line $y=0$ is the function constant, therefore the two stationary points are not extremal. – user376343 Feb 03 '23 at 12:48
  • Check this graph – WindSoul Feb 03 '23 at 17:46

1 Answers1

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An easy calculation shows that $\bigtriangleup=(2y^{2}+2y)(2x^{2}+4)-(4xy+2x)^{2}$. We have $$\bigtriangleup(-2,0)=\bigtriangleup(2,0)=-16< 0$$ The proof is complete.

Aliakbar
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