Problem 31 - Precalculus: Mathematics for Calculus

Question

Graph the solution set of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{aligned} x+y & \leq 4 \\ y & \geq x \end{aligned}\right.$$

Step-by-Step Solution

Answer

Vertices: (0,0), (2,2), (4,0). Solution set is bounded.

Step 1: Graph the first inequality

First, let's graph the inequality \(x + y \leq 4\). Rewrite the inequality as \(y \leq -x + 4\), which represents a line with a slope of -1 and a y-intercept of 4. Draw the line \(y = -x + 4\) on the graph. Since the inequality is \(\leq\), shade the region below the line.

Step 2: Graph the second inequality

Now, graph the inequality \(y \geq x\). This can be rewritten as \(y - x \geq 0\), where the line is \(y = x\). This line has a slope of 1 and goes through the origin (0,0). Plot the line \(y = x\) on your graph. Since the inequality is \(\geq\), shade the region above this line.

Step 3: Find the intersection points

The solution set is the overlap of the two shaded regions. To find the intersection points, solve the system of equations \(x + y = 4\) and \(y = x\). Substitute \(y = x\) into \(x + y = 4\) to get \(x + x = 4\), or \(2x = 4\). Solving this gives \(x = 2\). Substitute back to find \(y = 2\). Thus, one vertex is at (2, 2).

Step 4: Identify vertices and solution set

To find other vertices, analyze the constraints. The line \(y = x\) intersects the x-axis at (0, 0). The line \(y = -x + 4\) intersects the x-axis at (4, 0). Therefore, the vertices of the feasible region are (0, 0), (2, 2), and (4, 0).

Step 5: Determine if the solution set is bounded

A bounded solution set is one that is enclosed, forming a polygon. Since the solution can be enclosed by straight lines connecting points (0,0), (2,2), and (4,0), the solution set is a triangle, hence it is bounded.

Key Concepts

Graphing Inequalities

Graphing inequalities is an essential step in understanding systems of inequalities. To graph an inequality like \(x + y \leq 4\), first treat the inequality as an equation: \(x + y = 4\). This equation represents a straight line on a coordinate plane.
Once you have the line, to determine which side of it to shade (since it's a linear inequality), you select a test point. Often, the origin \((0,0)\) is used unless it lies on the line. Substitute this point into the original inequality; if true, shade that side. Otherwise, shade the opposite side.
For \(y \geq x\), rewriting it gives the line \(y=x\). Here, test a point like \((1,0)\). Since \(0 \geq 1\) is false, you shade the area above the line. Combined, these steps show visually where all solutions of the system are located.

Intersection Points

Intersection points are crucial in finding where two or more inequalities meet – these are where the shading overlaps on your graph.
To find these points, solve the equations associated with your inequalities simultaneously. For instance, with \(x + y = 4\) and \(y = x\), substitute to find intersection points. Substitute one into the other: \(x + x = 4\), simplifies to \(x = 2\). Substituting back gives \(y = 2\) at the point \((2, 2)\).
This point is essential as it often serves as a vertex for the solution set, marking a boundary around which the feasible solutions lie.

Bounded Solution Set

Determining the nature of the solution set, whether bounded or unbounded, helps understand the limits of possible solutions.
With our inequalities, after graphing, the solution set forms an enclosed polygon – specifically, a triangle with vertices at \((0,0)\), \((2,2)\), and \((4,0)\). Since the solution can be bounded by lines, it does not extend to infinity, making it a bounded set.
If this set were unbounded, the feasible region would stretch infinitely, like a line or a ray, not forming a closed shape.

Vertices of the Solution Set

Vertices are the corner points that enclose the feasible region of the solution set. To find them, identify where different boundary lines intersect. In this example, each inequality contributes to these lines.
The intersection points found earlier, \((2,2)\), and the intercepts \((0,0)\) and \((4,0)\) from the graph are the vertices.
These points outline the shape containing all possible solutions, a triangle in our case, making them key to visualizing and solving systems of inequalities effectively. By connecting these vertices, the boundary of the solution space is completely defined.