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I'm asked to characterize all the critical points of this function

$f(x,y)=x^2+4y^2-4xy+2x-4y+3$

So as always I find the partial derivatives and solve the system

$f_x(x,y)=2x-4y+2=0$

$f_y(x,y)=-4x+8y-4=0$

This system has infinite solutions(i.e infinite critical points) in the line $y=1/2x+1/2$ and the Hessian $H$ has determinant $0$ therefore is useless.

Having introduced my problem now I show the solution given by the lecturer which I don't fully understand and I would appreciate if someone explained to me $WHY$ is this correct.

Here I go.

The function $f(x,y)=x^2+4y^2-4xy+2x-4y+3$ can be expressed as (Factorizing the 3 first terms)

$f(x,y)=(x-2y)^2+2(x-2y)+3$

which if I take $h(x,y)=x-2y$ and using $g(h)=h^2+2h+3$ I can express $f(x,y)=g(h(x,y))$. And here it comes what I don't understand.

Because $g(h)$ has a minimum at $h=1$ the points which $h=1$, which in this case are the line $x-2y=1$ or $y=1/2x+1/2$ are a minimum.

Why does this last step is ok?

For using this kind of argument is not necessary to $g(h)$ strictly increasing?

Does this (finding the minimun of the outer function (in this case $g$) and then solving for the point of the inside function ($h$)) always work?

Thank in advance.

mmendina
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  • Based on your formula for $g(h)$, the minimum of $g$ seems to be at $h = -1$... so the graph of $f$ would be a parabolic trough, opening upward, whose "bottom" is the horizontal line $x-2y=-1, z=2$. – Ned May 06 '20 at 13:31
  • And why that allows me to solve the problem that way? – mmendina May 06 '20 at 13:34
  • Each point $(2c - 1, c, 2)$ on the surface $z=f(x,y)$ satisfies $f(2c-1,c) <= f(x,y)$ for all points $(x,y)$, however it is NOT true that $f(2c-1,c) < f(x,y)$ in any neighborhood of $(2c-1,c)$, so I think that means you would say that $(2c-1,c)$ is NOT a local minimum (I don't remember the official definition of "local minimum" for situations like this where the point in question is "tied" with nearby points for the smallest value ). – Ned May 06 '20 at 14:16

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