Begin by determining the boundary lines of the inequalities and graph them.
4.5 square units
Practice makes perfect
To find the area of the bounded region, we will first graph the bounded region.
y ≥ |x|-3 & (I) y ≤ -|x|+3 & (II) x ≥ |y| & (III)
For this case, we should graph each inequality separately and then combine them on the same coordinate system. Let's start!
Inequality I
To graph the inequality, first thing we need to do is to determine its boundary line. It can be determined replacing the inequality symbol with the equals sign.
&Inequality I &&Boundary Line I
& y ≥ |x|-3 && y = |x|-3
The graph of the boundary line is the graph of parent function y=|x| translated 3 units down. Because the inequality is non-strict, the boundary line will be solid. With this, we can graph it.
Now we will decide which region we should shade by testing the point (0,0).
The point satisfied the inequality, so we can shade the region that contains the point.
The first inequality was completed.
Inequality II
We will proceed in the same way to graph the second inequality.
&Inequality II &&Boundary Line II
& y ≤ -|x|+3 && y = -|x|+3
The graph of the second boundary line is the graph of the parent function y=|x| reflected in the x-axis and translated up 3 units. Let's graph it!
We will again test the points (0,0) to decide the region that will be shaded.
Since the point satisfied the inequality, we will shade below the boundary line.
Inequality III
We will again begin by determining it boundary line.
&Inequality II &&Boundary Line II
& x ≥ |y| && x = |y|
Notice that the equation of the boundary line does not represent a function. Therefore, we will consider two cases, positive case and negative case, to graph it.
x=|y| ⇒
y ≥ 0: x=y
y < 0: x=- y
For the positive y-values, the boundary line will be x=y. For the negative y-values, it will be x=- y. With this, we can graph Boundary Line III.
To complete the graph, we will test the point (2,0).
Now that we have graphed all three inequalities, we are ready to graph the bounded region.
The overlapping section of the graph above gives us the bounded region. Notice that it is a square. Let's say a is its side length!
The area of a square can be found by squaring its side length. In this case, we should find a^2. To do so, we will apply the Pythagorean theorem. Notice that the length of the diagonal is 3 units.
