Answer
All points that lie on the line $x=2y$ are critical points
Work Step by Step
Let us first find the critical points of the function where $f_x$ and $f_y$ are equal to 0, assuming first and second partial derivatives are continuous.
$f_x=2x-4y$ and $f_y=8y-4x$
There are an infinite number of critical points because the equations are integer multiples of each other. All points that lie on the line $x=2y$ is a critical point.
$f_{xx}=2$ and $f_{yy}=8$ and $f_{xy}=-4$.
$D(x,y)=(2)(8)-(-4)^2=0$ for each point on the line.
To show that the critical points are local and absolute minima, let us rewrite the function in a different format.
$f(x,y)=(x+2y)^2+2$. The expression (x+2y) is squared, which means it cannot be negative. Therefore, the function has an absolute minimum, which is given by the critical points.
