Problem 39 - Precalculus: Mathematical for Calculus

Question

Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{aligned}x^{2}+y^{2} &<9 \\\x+y &>0 \\\x & \leq 0\end{aligned}\right.$$

Step-by-Step Solution

Answer

Vertices: \((-\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}})\), Bounded solution.

Step 1: Understanding the Inequalities

The given system of inequalities consists of three parts:1. \( x^2 + y^2 < 9 \): This represents the interior of a circle centered at the origin (0,0) with a radius of 3.2. \( x + y > 0 \): This inequality represents the region above the line \( y = -x \).3. \( x \leq 0 \): This restricts the solution to the left side of the y-axis (including the y-axis).

Step 2: Graphing the Circle

Draw a circle centered at (0,0) with radius 3 on the coordinate plane. The inequality \( x^2 + y^2 < 9 \) indicates that solutions are inside this circle. Dashed lines are used to show that points on the circle are not included in the solution.

Step 3: Graphing the Line

Graph the line \( y = -x \) which passes through the origin. Use a dashed line to indicate that the points on the line itself are not part of the solution. The solution includes the region where \( y > -x \), which is above this line.

Step 4: Shading the Intersection

The valid solution region is where all inequalities overlap. Start by shading the area inside the circle, then restrict it to the left of the y-axis (including it), and above the line \( y = -x \). This overlapping region is the solution set.

Step 5: Finding Vertex Coordinates

To find vertices, look at intersections of boundary lines and the circle:- Intersection of \( y = -x \) and \( x^2 + y^2 = 9 \) gives the points where both are equal.Solving, we get:- \( 2x^2 = 9 \rightarrow x^2 = 4.5 \rightarrow x = -rac{3}{ ext{sqrt}(2)} \) and \( y = rac{3}{ ext{sqrt}(2)} \).Only consider \( x \leq 0 \), vertex: \( (-rac{3}{ ext{sqrt}(2)}, rac{3}{ ext{sqrt}(2)}) \).

Step 6: Determining Boundedness

The solution is entirely within the circle and no part of the solution extends to infinity; therefore, the solution set is bounded by the constraints and the circle.

Key Concepts

Graphing Inequalities

When graphing inequalities, we represent regions rather than just lines or curves. This is the graphical depiction of where the inequality holds true. For instance, let's consider the inequality \( x^2 + y^2 < 9 \). This defines all points inside a circle centered at \((0,0)\) with a radius of 3, excluding the circle itself. This boundary is often shown with a dashed line to indicate that points on the circle aren't part of the solution.
Another inequality, \( x + y > 0 \), represents an area above the line \( y = -x \). This line splits the plane into two halves. The final inequality, \( x \leq 0 \), confines the solution set to the left half, including the y-axis.
To graph these manually or using software, always remember:

• Use dashed lines for strict inequalities (e.g., \(<\) or \(>\)).
• Shade the region where the inequalities overlap, highlighting the solution set.

Bounded Solution Set

In the context of graphing, a bounded solution set is one that is enclosed from all sides within a finite region. For our given system of inequalities, the solution lies entirely within a circle having a boundary at \( x^2 + y^2 = 9 \). This circle effectively caps any further extension of the solution.
The combination of constraints \( x + y > 0 \) and \( x \leq 0 \) ensures that the region does not extend infinitely in any direction. Instead, it neatly fits within the circle, creating a clearly defined solution space.
To determine if a solution is bounded, consider:

• Whether the graphical representation of solutions is enclosed within curves or lines, such as circles or other defined boundaries.
• Absence of open paths leading to infinity.

This certainty allows us to conclude that our solution is limited and enclosed, making it bounded.

Vertices of Intersections

Vertices of intersections refer to points where the boundaries of different inequalities meet. These points often are valuable solutions because they outline the shape of the feasible region.
In our system of inequalities, intersecting entities include the circle \( x^2 + y^2 = 9 \) and the line \( x + y = 0 \). Solving these equations simultaneously gives potential vertices.
By substituting \( y = -x \) into the circle's equation, we derive equations like \( 2x^2 = 9 \), solving which gives the vertices in terms of coordinates. Remember to apply all system constraints to validate the coordinates:

• Ensure \( x \leq 0 \) to discard any invalid points.
• Integrate all inequalities to confirm the vertex is within the solution set.

Finding these coordinates is crucial for understanding the boundaries of the solution space.