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The given function is $ f(x, y) = x^2y^2 $. I used the gradient of $f$ to find the stationary points (not sure if these are correct).

$$ \nabla f(x, y) = \mathbf{0} $$ $$ \langle 2xy^2, 2x^2y \rangle = \langle 0, 0 \rangle $$ $$ 2xy^2 = 0 \; \mathrm{and} \; 2x^2y = 0 $$

I found the stationary points to be $(x, 0)$ and $(0, y)$, since points of the two forms are needed such that $\nabla f(x, y) = \mathbf 0$.

I attempted to use the second-order partial-derivative test on those two points.

$$\begin{align} det(H_f) & = f_{xx}f_{yy} - (f_{xy})^2 \\ & = (2y^2)(2x^2) - (4xy)^2 \\ & = -12x^2y^2 \end{align}$$

$$ det(H_f(x, 0)) = -12x^2(0^2) = 0 $$ $$ det(H_f(0, y)) = -12(0^2)y^2 = 0 $$

Given that the discriminant is $0$ in separate tests with each point, the test does not say anything about them. Is there a way to get around this?

user369262
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    Hint: you can study the sign of the function...$x^2y^2\ge 0$ $\forall (x,y)\in \mathbb R^2$. – MattG88 Feb 19 '17 at 21:53
  • Try to visualize the function as a surface $z = x^2y^2$. What happens to the value of z along the axes? What happens along the lines $y = x $ and $y = -x$? – victoria Feb 20 '17 at 07:59
  • @victoria I'm guessing $z$ increases along the lines $y = x$ and $y = -x$. I believe $z = 0$ along the axes. – user369262 Feb 20 '17 at 13:51
  • No guessing -- know. If either $x = 0$ or $y = 0$, $x^2y^2 = 0$, So the 3-D graph is zero along the axes. Along the lines $y = x$ and $y = -x$ noting that $(-x)^2 = x^2$, $x^2y^2 = x^2x^2 = x^4$. Therefor the graph rises very steeply (somewhat similar to a parabola but much steeper) as you move away from zero along these lines. Imagine a stretch cloth held flat along the axes and pulled up at all four corners. – victoria Feb 21 '17 at 07:10
  • So there are minima at the points of the form $(x, 0)$ and $(0, y)$? – user369262 Feb 22 '17 at 17:16

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