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$f(x,y)=x^2+4y^2-4xy+2$

So, $f_x=2x-4y$ and $f_y=8y-4x$

To find the stationary points we have to equal the partial derivatives to $0$:

$2x-4y=0$

$8y-4x=0$

Because we cannot find an $x$ and $y$ via the system of equations, does that mean that there are a infinite amount of $x$ and $y$ that satisfy the equation, thus proving that there indeed are infinite stationary points?

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    $\begingroup$ Note that $f(x,y)=(x-2y)^2$+2, so $f(x,y)$ has minimum value of $2$ and this is achieved along the entire line $x=2y$. So yeah, there's an entire line of stationary points. $\endgroup$
    – Semiclassical
    Sep 10, 2021 at 23:48
  • $\begingroup$ @Semiclassical yeah, that seems as a good way to do it. Is my method acceptable as a proof or not, just so I can be sure. Thanks! $\endgroup$
    – EL02
    Sep 10, 2021 at 23:55
  • $\begingroup$ It's probably acceptable, depending on how you use it. I think the assertion that "we cannot find an $x$ and $y$ via the system of equations" could be explained a bit better, not just because it's an assertion provided without proof, but also because this same phrase could just as easily describe a situation where no stationary points exist. $\endgroup$
    – Theo Bendit
    Sep 11, 2021 at 0:24

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Under the observation by Semiclassical: $$f(x,y)=(x-2y)^2+2,$$ you can see that that $f(x,y)\ge2$ since $(x-2y)^2\ge 0$, with $f(x,y)=2$ when $x-2y=0$. Then $x-2y=0$ is a critical line of minima.

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  • $\begingroup$ the gradient vanish at the indicated line and we can see how the hessian is singular but has rank one. $\endgroup$
    – janmarqz
    Sep 11, 2021 at 1:10

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