Investigating the function $f(x,y)=2xy-|x|-y$.

Question

Find the local and global min/max points of $f(x,y)=2xy-|x|-y$.
Does the function has a global max/min point in the plane?

For the first question, I splitted it to when $x\ge0$ and when $x<0$.
for $x\ge0$: $f(x,y)=2xy-x-y. $
$f_x=2y-1$, $f_y=2y-1 \Longrightarrow(\frac{1}{2},\frac{1}{2})$.
$f_{xx}*f_{yy}-f_{xy}^2=-4$ which means this is a "Saddle" point.

When $x < 0$: $f(x,y)=2xy+x-y$
$f_x=2y+1$
$f_y=2x-1$
So I got $(\frac{1}{2}, -\frac{1}{2})$ which is outside the area $x<0$.

Question: Do I need to check $x=0$ seperately? I think not because when $x=0$ the absolute value doesn't matter anymore. (but not sure).

For the second question, I know that the plane isn't a bounded area. But I'm not sure why is it the plane they're asking about and not the space, so if it's the plane, that means I could just set $f(x,y)=0$? I'm not sure how to solve this question.

Any help or feedback is really appreciated, thanks in advance!

Answer 1

A local extremum can occur at three types of points:

1. boundary points,
2. internal points where the function is not differentiable,
3. internal points where the gradient vanishes.

If $f$ is defined on all of $\mathbb{R}^2$ then there are no boundary points.

The function is not differentiable when $x=0$ because of the $|x|$ term. When $x=0$ you have $f(x,y) = f(0,y) = -y$ which has no local extremum and also takes all values in $\mathbb{R}$. Therefore $f$ has no local extremum on the $y$ axis ($x=0$) and neither a global minimum nor a global maximum.

For $x>0$ you have $f(x,y)=2xy-x-y$ so $\nabla f = (2y-1, 2x-1),$ which vanishes at $(x,y)=(\frac12, \frac12).$ Setting $x=\frac12+h$ and $y=\frac12+k$ gives $$ f = 2(\frac12+h)(\frac12+k)-(\frac12+h)-(\frac12+k) = -\frac12+2hk $$ which takes values both above and below $f(\frac12,\frac12)=-\frac12$ (even for small $h,k$) so this is a saddle point.

For $x<0$ you have $f(x,y)=2xy+x-y$ so $\nabla f = (2y+1, 2x-1),$ which vanishes at $(x,y)=(\frac12, -\frac12).$ This however breaks against the condition $x<0$ and we can conclude that there is no local extremum for $x<0$.

Thus, $f$ has one saddle point at $(x,y)=(\frac12,\frac12)$ and neither any local nor global extremum of any type.