Math Help - Level curves

1. Level curves

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

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.

2. Write $y=\pm(1-|x|)$. Can you graph $y=1-|x|$?

3. A level curve for f(x)= 1- |x|- |y| is given by 1- |x|- |y|= C, where C is a constant, or |x|+ |y|= 1- C.

Now, it should be obvious that, since the left side of that is never negative, C cannot be larger than 1.

If C= 1, the equation becomes |x|+ |y|= 0 which, since neither |x| nor |y| can be negative, is only true when x= y= 0. The level curve is the single point (0, 0).

For C< 1, as always with absolute values, the simplest thing to do is to break the problem into cases.
1. If x and y are both positive, |x|+ |y|= x+ y= 1- C. That is a straight line but remember to only draw it in the first quadrant.

2. If x< 0 and y> 0 then |x|+ |y|= -x+ y= 1- C. Again, a portion of a straight line but now in the second quadrant.

3. If x< 0 and y< 0 then |x|+ |y|= -x- y= 1- C. A portion of a straight line in the third quadrant.

4. If x> 0 and y< 0 then |x|+ |y|= x- y= 1- C. A portion of a straight line in the fourth quadrant.

If you draw those four segments for one value of C, say C= 0, it should be easy to see what the other level curves are.