Problem 37 - 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. $$\left\\{\begin{array}{r}{x^{2}+y^{2} \leq 8} \\ {x \geqq 2} \\ {y \geq 0}\end{array}\right.$$

Step-by-Step Solution

Answer

Vertices are (2,0), (2,2), and (√8,0); solution set is bounded.

Step 1: Understand the Inequalities

The system of inequalities consists of: 1) \(x^2 + y^2 \leq 8\), which represents a circle with center at the origin (0,0) and radius \(\sqrt{8}\). 2) \(x \geq 2\), which is a vertical line where all points to the right or on this line are included. 3) \(y \geq 0\), which represents the entire upper half of the plane, including the x-axis.

Step 2: Graph the Circle

First, recognize the circle's equation, \(x^2 + y^2 = 8\), which has a radius of \(\sqrt{8} \approx 2.83\). Graph this circle, but since it is \(\leq 8\), shade the region inside the circle. Note that it includes the boundary of the circle.

Step 3: Graph the Line and Half-Plane for x

Graph the vertical line \(x = 2\). Since \(x \geq 2\), shade the area to the right of this line as it includes all x-values greater or equal to 2.

Step 4: Graph the Horizontal Axis

Note that \(y \geq 0\) requires you to include all points where y is zero or positive. Shade above the x-axis, including it, since it covers the upper half-plane.

Step 5: Determine the Intersection and Vertices

The solution to the system is where all three regions intersect. The intersection is bounded by the circle, and the lines \(x=2\) and \(y=0\). Find the vertices by locating where these lines and the circle intersect. Solve for intersections: \(2^2 + y^2 = 8\) gives \(y = \sqrt{4} = 2, y = -2\) but \(y \geq 0\), so only \(y=2\) is valid. Thus, the vertices are \((2, 0)\), \((2, 2)\), and where the circle intersects \(x=\sqrt{8}\), or the boundary of the circle edge with \(x \leq \sqrt{8}\) and \(y=0\).

Step 6: Verify Whether the Solution Set is Bounded

A bounded solution set is one that is contained within a finite region and does not extend to infinity. Here, the intersection is contained within the circle and is limited by \(x = 2\), thus confirming it is bounded within these geometric confines.

Key Concepts

Circle Graphing

When dealing with systems of inequalities, especially those involving circles, it's crucial to understand how to graph a circle given its equation. The equation \(x^2 + y^2 \leq 8\) represents a circle on the coordinate plane. This equation tells us the center of the circle is at the origin \((0,0)\), and its radius is \(\sqrt{8}\), which approximately equals 2.83. This radius indicates the distance from the origin to any point on the circle's edge.

In the case of \(\leq\), not only the boundary circle but also all points inside it are included. To graph this, use a solid line for the circular boundary to signify that the edge is part of the solution. Then, shade the entire region within the circle. This shading highlights the inclusivity of all the interior points, clearly showing the area fulfilling the first inequality.

Intersection of Regions

The intersection of regions in a system of inequalities tells us where all conditions are true simultaneously. In our exercise, we need to find the overlap between multiple areas: the interior of a circle, a vertical band, and a horizontal stretch.

The first inequality \(x^2 + y^2 \leq 8\) covers the area within a circle. The inequality \(x \geq 2\) introduces a vertical line at \(x=2\), shading everything rightward. Then, \(y \geq 0\) includes the half-plane above the x-axis. The true solution set is where these shaded areas overlap.

• The circle section limited by \(x \geq 2\) narrows the solution to the right side of the circle.
• Similarly, fulfilling \(y \geq 0\) means concentrating on the top half.
• The intersection thus forms in a corner bound by these conditions.

This visualization helps in clearly identifying where all conditions align, often depicted as the smallest common shaded area on the graph.

Graphical Solution

Finding a graphical solution for a system of inequalities involves visualizing where constraints intersect and determining vertices. For this exercise, our graphical solution focuses on where all the provisions coincide within the circle and half-planes.

To precisely find the vertices, calculate intersection points mathematically. The critical vertices here are found by solving \(x^2 + y^2 = 8\) where other lines meet. Computation shows:\(y = 2\) when intersecting vertical \(x=2\), forming the vertex \((2,2)\). Other points include \((2,0)\), the meeting of \(x=2\) and horizontal \(y=0\). Lastly, where \(x=\sqrt{8}\), ensuring it does not exceed the circle's maximum reach.

Concluding graphical analysis involves determining if the solution set is bounded, confirming it's enclosed by the circle and the lines. This entails visually and mathematically verifying it doesn't extend infinitely in any direction.