Problem 24 - 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. $$\left\\{\begin{aligned}x &>2 \\\y &<12 \\\2 x-4 y &>8\end{aligned}\right.$$

Step-by-Step Solution

Answer

Vertices: (2, 12), (2, -1), (28, 12). The solution set is unbounded.

Step 1: Set Up Each Inequality

We will rewrite each inequality in a form that makes graphing easier.1. The first inequality is \(x > 2\). This is a vertical line at \(x = 2\), and the region to the right of this line (not including the line) is the solution.2. The second inequality is \(y < 12\). This is a horizontal line at \(y = 12\), and the region below this line (not including the line) is the solution.3. For the third inequality, \(2x - 4y > 8\), first solve for \(y\): \[ 2x - 4y = 8 \Rightarrow 4y = 2x - 8 \Rightarrow y = \frac{1}{2}x - 2 \] This inequality represents a line with a slope of \(\frac{1}{2}\) and a y-intercept of \(-2\). The region above this line (not including the line) is the solution.

Step 2: Graph the Boundary Lines

Draw the lines from each inequality on the coordinate plane.1. Draw a dashed vertical line at \(x = 2\).2. Draw a dashed horizontal line at \(y = 12\).3. Draw a dashed line for \(y = \frac{1}{2}x - 2\), using the y-intercept \(-2\) and slope \(\frac{1}{2}\). This means starting at point (0, -2) and going up 1 unit and to the right 2 units for each point.The dashed lines indicate that points on the line are not included in the solutions (strict inequalities).

Step 3: Shade Each Inequality Region

Shade the appropriate region for each inequality on the graph.1. For \(x > 2\), shade the area to the right of the line \(x = 2\).2. For \(y < 12\), shade the area below the line \(y = 12\).3. For \(y > \frac{1}{2}x - 2\), shade the area above the line \(y = \frac{1}{2}x - 2\).The solution to the system is the area where all shaded regions overlap.

Step 4: Identify Vertices of the Solution Region

Find the intersection points of the boundary lines to discover the vertices of the overlapping region.1. Intersection of \(x = 2\) and \(y = 12\): The vertex is (2, 12).2. Intersection of \(x = 2\) and \(y = \frac{1}{2}x - 2\): Substitute \(x = 2\) into \(y = \frac{1}{2}x - 2\): \[ y = \frac{1}{2}(2) - 2 = 1 - 2 = -1 \] The vertex is (2, -1).3. Since \(y < 12\) does not intersect any new boundary other than with the vertical lines, check the line \(y = \frac{1}{2}x - 2\) with \(y = 12\): \[ 12 = \frac{1}{2}x - 2 \Rightarrow 14 = \frac{1}{2}x \Rightarrow x = 28 \] The vertex is (28, 12).These points form the vertices of the solution region.

Step 5: Analyze the Solution Set

Determine whether the solution region is bounded or unbounded.The solution region is contained within the lines \(x = 2\), \(y = 12\), and \(y = \frac{1}{2}x - 2\). The region extends out indefinitely in one direction due to \(x > 2\) and \(y > \frac{1}{2}x - 2\) constraints not intersecting with additional bounding lines in all directions, making the solution region unbounded.

Key Concepts

Graphing Inequalities

Graphing inequalities involves plotting lines or curves on a coordinate plane, which represent the boundaries of the solutions to inequalities.
These lines are usually dashed, indicating that points directly on the line are not included in the solutions for strict inequalities.
Understanding how to graph each inequality correctly is crucial for identifying the correct solution region.
When dealing with linear inequalities, consider these steps:

• Convert the inequality into an equation to find the corresponding boundary line.
• For inequalities like \(x > 2\), plot a vertical dashed line at \(x = 2\); this shows that points right of the line are potential solutions.
• For \(y < 12\), draw a horizontal dashed line at \(y = 12\), shading below this line to indicate possible solutions.
• Lastly, for an inequality such as \(2x - 4y > 8\), it might be easier to re-arrange it into a slope-intercept form (\(y = \frac{1}{2}x - 2\)) before graphing. Draw this line using its slope and y-intercept, shading the relevant area accordingly.

Mastering graphing these inequalities is an essential step in finding where all conditions of the system are satisfied simultaneously. This involves understanding how to shade in the correct areas and the role of boundary lines in this process.

Solution Region

The solution region of a system of inequalities is the area where the shaded regions from each inequality overlap.
Identifying this region helps determine which points satisfy all inequalities simultaneously.
Any point within this overlapping region will make every inequality in the system true.To find the solution region:

• First, graph each inequality on the same coordinate plane, shading the region that satisfies each respective inequality.
• Examine where all the shaded regions intersect; this intersection area is your solution region.
• Make sure to carefully assess the nature of the lines involved — dashed for strict inequalities and solid for inclusive inequalities (≥ or ≤).

In our example problem, the solution region emerges where all three shaded areas: right of \(x = 2\), below \(y = 12\), and above \(y = \frac{1}{2}x - 2\), integrate. This systematic approach simplifies the identification of solution regions in any given set of inequalities.

Vertices of Polygons

The vertices of the polygon formed by the solution region are found by identifying where the boundary lines intersect.
These vertices are particularly important as they define the corners of the polygon, encapsulating the solution region.
Calculating these intersection points provides a clearer picture of the bounds of the solution region.To find vertices:

• Determine the intersection of each pair of boundary lines mathematically. This often involves solving two equations simultaneously.
• In the case of our example, the vertices were found by checking the intersection points: the lines \(x = 2\), \(y = 12\), and \(y = \frac{1}{2}x - 2\) formed specific points like (2, 12), (2, -1), and (28, 12).
• Each of these vertices gives a corner for the polygon that bounds the solution region.

These vertices help ascertain whether the solution region is closed (bounded) or extends indefinitely (unbounded). In the current example, the solution region is unbounded, as one particular direction allowed for limitless extension due to the unrestricted nature of one of the inequalities. This understanding ensures a comprehensive grasp of the solution set's limits and possibilities.