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## Homework Statement

I have to find some level curves for: [tex]f(x,y)=1-|x|-|y|[/tex]

## The Attempt at a Solution

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

[tex]z=1\Rightarrow{-|x|-|y|=0}\Rightarrow{x=y=0} \therefore P(0,0,1)\in{S}[/tex]

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

[tex]z=0\Rightarrow{-|x|-|y|=-1\Rightarrow{|x|+|y|=1\Rightarrow{|y|=1-|x|}}}[/tex]

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

I know that:

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

And

[tex]|x|=\begin{Bmatrix} x & \mbox{ si }& x\geq{0}\\-x & \mbox{si}& x<0\end{matrix}[/tex]

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.