Prove that there are infinitely many critical points and they are all local (and absolute) minima on $x^2+4y^2-4xy+2$

I am studying for a fast approaching Calc 3 midterm exam and ran into this problem in the textbook.

Show that $f(x,y) = x^2+4y^2-4xy+2$ has an infinite number of critical points and that $D = 0$ at each one. Then show that $f$ has a local (and absolute) minimum at each critical point.

So I am able to solve the two partial derivatives to get $f_x=2x-4y$ and $f_y=8y-4x$ and that I can solve both equations for $f_x=f_y=0$ which is true for all points. I also can see that this is is true for all values: $$D = (f_{xx})(f_{yy})-(f_{xy})^2 = 0$$

However, I dont know how to go about determining that all of the critical points are local and absolute minimums. I believe that I was only taught the 2nd derivative test where you compare the value of $D$ to $f_{xx}$.

Any ideas? Thanks!

• Try writing $f$ in some other form. Does $x^2 - 4xy + 4y^2$ look in any way familiar? Oct 31 '14 at 22:20

Note that $$f(x,y)=(x-2y)^2+2\ .$$ So at any point where $x=2y$ (and there are infinitely many such points) the value of $f(x,y)$ is $2$, and for any other point the value is greater than $2$. So these points are local and absolute minimum points.