# Level curves

## Homework Statement

I have to find some level curves for: $$f(x,y)=1-|x|-|y|$$

## The Attempt at a Solution

So, if we call $$S$$ at the surface given by the equation $$z=f(x,y)$$, then

$$z=1\Rightarrow{-|x|-|y|=0}\Rightarrow{x=y=0} \therefore P(0,0,1)\in{S}$$

Now, that particular case its simple, cause it gives just a point, but if I go downwards I get:

$$z=0\Rightarrow{-|x|-|y|=-1\Rightarrow{|x|+|y|=1\Rightarrow{|y|=1-|x|}}}$$

I'm not sure how to represent this. How does this look on the xy plane?

I know that:

$$|y|=\begin{Bmatrix} 1-|x| & \mbox{ si }& y\geq{0}\\-1+|x| & \mbox{si}& y<0\end{matrix}$$

And

$$|x|=\begin{Bmatrix} x & \mbox{ si }& x\geq{0}\\-x & \mbox{si}& x<0\end{matrix}$$

But it don't helps me to visualize the "curve". I know actually that it looks like a parallelogram, but thats because I've used mathematica to compute the surface :P I don't know how to deduce it analytically.

vela
Staff Emeritus
Homework Helper
Just analyze it by quadrant. For example, in the second quadrant, x<0 and y>0, so |x|=-x and |y|=y. Therefore, you get |y| = 1-|x| ⇒ y = 1+x.

Office_Shredder
Staff Emeritus
Gold Member
2021 Award
|x|+|y|=1. You can break it up into four cases:
|x|=x, |y|=y
|x|=x, |y|=-y
|x|=-x, |y|=y
|x|=-x, |y|=-y

For each quadrant, you will have the equation of a line. Graph those lines in the correct quadrants

Thanks!

May I ask here about another doubt I got? its a simple question, I don't know if I should make another topic just for it. I wanna know about the range of the function $$f(x,y)=x^2+y^2-2xy$$ I think that its all the real numbers, but I'm not sure.

vela
Staff Emeritus
Homework Helper
You should start a new thread for your question. (And it's not all real numbers.)

HallsofIvy
May I ask here about another doubt I got? its a simple question, I don't know if I should make another topic just for it. I wanna know about the range of the function $$f(x,y)=x^2+y^2-2xy$$ I think that its all the real numbers, but I'm not sure.
It might help to note that $$f(x,y)= x^2- 2xy+ y^2= (x- y)^2$$.