How to sketch level curves of the form $f(x,y)= |x|+|y|$?

Question

I'm struggling to sketch the level curves of the equation $f(x,y)= |x|+|y|$ I know for finding the level curves you have to set $f(x,y) = C$ (with c a constant). But then I have the equation $|x|+|y|=C$.

So lets say $C=0$, then $|x|+|y|=0$, but how can I sketch this. Because both x and y are greater then zero, and they have to add up to 0, therefore x and y needs to be zero, but then I can't draw anything.

So lets say $C=1$,then $|x|+|y|=1$, but how can I sketch this. Can I say that $x + y= \pm1$, therefor $y=1-x$ and $y = 1+x$ (but $x$ can't be negative, so I don't know how do this problem.)

If you know the answer, it would be very much appreciated to help me

Thank you in advance.

Answer 1

If $|x|+|y|=1$, consider four possibilities:

• $x,y\geqslant0$: then you just have $x+y=1$;
• $y\leqslant0\leqslant x$: then you just have $x-y=1$;
• $x,y\leqslant0$: then you just have $-x-y=1$;
• $x\leqslant0\leqslant y$: then you just have $-x+y=1$.

So, your curve is a square, expressed as the union of $4$ line segments. To be more precise: it's the square whose vertices are $(\pm1,0)$ and $(0,\pm1)$.

Answer 2

I just add a comment to the already proposed solutions. Since $f(x,y) = f(-x, y) = f(x, -y) = f(-x, -y)$, the level sets of $f$ are symmetric w.r.t. the reflections about the axes $(x,y) \mapsto (-x,y)$ and $(x,y) \mapsto (x,-y)$, and w.r.t. the symmetry about the origin $(x,y) \mapsto (-x, -y)$.

Given that, it is enough to draw the level set for example in the quadrant $x\geq 0$, $y\geq 0$, and then you get the whole picture using the above symmetries.