Problem 27 - Algebra and Trigonometry

Question

19–40 Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{array}{c}{x^{2}+y^{2} \leq 4} \\ {x-y>0}\end{array}\right.$$

Step-by-Step Solution

Answer

Vertices: \((\sqrt{2}, \sqrt{2})\) and \((-\sqrt{2}, -\sqrt{2})\), solution set is bounded.

Step 1: Understand the Inequalities

We have the system of inequalities \( x^2 + y^2 \leq 4 \) and \( x - y > 0 \). The first inequality describes a circle centered at the origin with radius 2. The second inequality represents the region where \( x > y \). Our task is to find the region that satisfies both conditions.

Step 2: Graph the Circle

Graph the circle represented by \( x^2 + y^2 = 4 \). This is a circle centered at (0,0) with a radius of 2. The inequality \( x^2 + y^2 \leq 4 \) indicates that we include the area inside and on the circle.

Step 3: Graph the Line

Graph the line \( x = y \), the boundary for the inequality \( x - y > 0 \). This line is diagonal through the origin with a slope of 1. The area where \( x > y \) is above this line. Shade the region above the line.

Step 4: Determine the Solution Region

Identify the intersection of the shaded regions from the circle and above the line \( x = y \). This intersection is the solution set of the system of inequalities. The solution region is a section of the circular disk split by the diagonal line \( x = y \).

Step 5: Find the Vertices

The vertices of the solution region are where the circle intersects with the line \( x = y \). Solving the system \( x^2 + x^2 = 4 \) and \( x = y \), we find that \( x = \sqrt{2} \) and \( x = -\sqrt{2} \). Therefore, the vertices are \( (\sqrt{2}, \sqrt{2}) \) and \( (-\sqrt{2}, -\sqrt{2}) \).

Step 6: Is the Solution Set Bounded?

A region is bounded if it can be enclosed within a circle of finite radius. In this case, the solution is a section of a circle, so it is bounded. Thus, the solution set is bounded.

Key Concepts

Graphing Inequalities

Graphing inequalities is a vital skill when solving systems of inequalities. It involves representing the inequality on the coordinate plane to visually understand the solution set. When graphing, start by identifying the boundary line or curve for each inequality. For a quadratic inequality like \( x^2 + y^2 \leq 4 \), you will begin by graphing the equation \( x^2 + y^2 = 4 \), which describes a circle. This circle is centered at the origin \((0,0)\) with radius 2. Since the inequality is \( \leq \), include all points on and inside the circle.

Next, address linear inequalities, such as \( x - y > 0 \). Graph the boundary line \( x = y \), which passes through the origin, with a 45-degree angle. Here, the inequality specifies \( x \) must be greater than \( y \), so we shade above this line. Ensure the graph displays the overlapping regions clearly, as this indicates the common area satisfying all the inequalities.

Bounded Solution Sets

When dealing with systems of inequalities, a bounded solution set refers to a region in the plane that is enclosed within a finite distance. In other words, this region can be encapsulated inside a circle or any other closed shape without extending indefinitely. A bounded set ensures that the solutions are limited in space.

In many exercises, such as the one given, bounded regions appear when the inequalities define a closed area, like a section of a circle. The solution for the system \( x^2 + y^2 \leq 4 \) and \( x - y > 0 \) forms a bounded region because it consists of a segment of the circle divided by the line. Since this part of the circle's disk doesn't stretch outwards infinitely, it is considered bounded.

Understanding whether a solution set is bounded helps determine the nature of the solutions you're working with, playing an important role in real-world applications, such as optimization problems where constraints must be met within enclosed areas.

Vertices of Solution Regions

Identifying the vertices of a solution region involves finding the points where boundary lines or curves intersect. These points mark the corners of the region, solidifying its shape on the graph and providing precise coordinates that define the edges of the solution space.

In the given exercise, the vertices are where the circle \( x^2 + y^2 = 4 \) meets the line \( x = y \). To calculate these vertices, solve both equations simultaneously, leading to solutions \( x = y \). Substituting into the circle's equation, we set \( x^2 + x^2 = 4 \), simplifying to \( 2x^2 = 4 \), and solving for \( x \), we find \( x = \sqrt{2} \) and \( x = -\sqrt{2} \). Thus, the vertices are \((\sqrt{2}, \sqrt{2})\) and \((-\sqrt{2}, -\sqrt{2})\).

These coordinates are critical for understanding the shape and limits of the solution region, providing essential points that can guide further analysis or application of the system in various settings.