What is the area of the set of points $P(x ; y)$ in the Cartesian coordinate system for which $|x+y|+|x-y| \leq 4 ?$
Solution. Let's examine when the expressions inside the absolute values change sign:

$x-y$ changes sign if $x-y=0$, that is, if $x=y$.

$x+y$ changes sign if $x+y=0$, that is, if $y=-x$.

Let's plot these two lines. They divide the coordinate plane into 4 regions. On the 1st diagram, we shaded one such region, let this be the region $A$. The points in this region must satisfy the inequality $(x+y)+(x-y) \leq 4$, which holds if $2 x \leq 4, 0 \leq x \leq 2$.

![](https://cdn.mathpix.com/cropped/2024_05_02_15433c7eb1bba81992efg-1.jpg?height=482&width=501&top_left_y=434&top_left_x=801)

1. diagram

Moving further in the positive direction, the next region is $B$. Here,

$$
(x+y)-(x-y) \leq 4, \quad \text { hence } \quad 0 \leq y \leq 2
$$

In the region $C$, $-(x+y)-(x-y) \leq 4$, that is, $0 \geq x \geq-2$, and finally, in the region $D$, $-(x+y)+(x-y) \leq 4$, that is, $0 \geq y \geq-2$.

By connecting the points, we get a square with a side length of 4 (as shown in the 2nd diagram), which has an area of 16 square units.

![](https://cdn.mathpix.com/cropped/2024_05_02_15433c7eb1bba81992efg-1.jpg?height=485&width=499&top_left_y=1378&top_left_x=802)

2. diagram