Question

Let \(R\) be the region consisting of the points \((x, y)\) of the Cartesian plane satisfying both \(|x|-|y| \leq 1\) and \(|y| \leq 1\). Sketch the region \(R\) and find its area.

Short answer

The region \(R\) is a diamond shape with vertices at (-1,0), (1,0), (0,-1), and (0,1) on the Cartesian Plane with an area of 4 square units.

Step-by-step solution

Drawing the Region

The inequalities \(|x|-|y| \leq 1\) and \(|y| \leq 1\) define the region \(R\). Start sketching the region \(R\) by understanding each inequality separately. The inequality \(|y| \leq 1\) shows that the value of 'y' is between -1 and 1, and it does not depend on 'x'. This gives us a rectangular strip in the Cartesian plane extending from -1 to 1 along the y-axis.

Interpreting \(|x|-|y| \leq 1\)

To understand \(|x|-|y| \leq 1\), we break it down to two scenarios: 1) \(x \geq 0\) and 2) \(x < 0\). For the first scenario, \(|x| = x\) and \(|y|= y\) if \(y \geq 0\) or \(|y| = -y\) if \(y < 0\). Therefore, in the first quadrant, the equation becomes \(x - y \leq 1\) which gives a line having an intercept of 1 and slope -1. Similarly, in the fourth quadrant, the line equation is \(x+y \leq 1\) which is a line with intercept 1 and slope 1. In the case of the second scenario, the signs of x and y in the inequality are flipped and it gives line equations \(y-x \leq 1\) in second quadrant and \(-x-y \leq 1\) in the third quadrant.

Equating scenarios and finding the region

By combining the lines from the previous step and the strip from the first step, the four lines confine the area into a diamond shape on the Cartesian plane. This central diamond shape is the region R.

Calculate the Area

The diamond enclosed by the lines will have its opposite vertices on (-1,0), (1,0), (0,-1), and (0,1). All sides are of equal length (i.e., 2 units) and it is a square. The area A of a square with side length 's' is given by \(A = s^{2}\). Substituting s = 2, the area will be 4 square units.