Problem 33 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Determine graphically the solution set for each system of inequalities and indicate whether the solution set is bounded or unbounded. $$ \begin{aligned} 6 x+5 y & \leq 30 \\ 3 x+y & \geq 6 \\ x+y & \geq 4 \\ x \geq 0, y & \geq 0 \end{aligned} $$

Step-by-Step Solution

Answer

The overlapping region of the given system of inequalities forms a quadrilateral with vertices at (0,6), (2,2), (4,0), and (0,4). Since the region is enclosed by the inequalities, the solution set is bounded.

Step 1: Graph each inequality

Firstly, we will graph each inequality: 1. \(6x + 5y \leq 30\) 2. \(3x + y \geq 6\) 3. \(x + y \geq 4\) 4. \(x \geq 0\) and \(y \geq 0\) To graph each inequality, we will find the boundary lines by setting equalities and then figure out which half-plane to shade for the inequality.

Step 2: Graph the inequality \(6x + 5y \leq 30\)

Rewrite the inequality as an equality: \(6x + 5y = 30\). Now, we will find two points that satisfy the equality - typically the x and y-intercepts. For the x-intercept, set \(y = 0\), then: \(6x = 30\), so \(x = 5\). For the y-intercept, set \(x = 0\), then: \(5y = 30\), so \(y = 6\). The x-intercept is (5,0) and the y-intercept is (0,6). To determine which side of the line to shade, choose a point not on the line, for example, the origin (0,0). Since \(6(0) + 5(0) \leq 30\) is true, shade the region containing the origin (and the line itself, as it's less than or equal to).

Step 3: Graph the inequality \(3x + y \geq 6\)

Rewrite the inequality as an equality: \(3x + y=6\). Find the x and y-intercepts similarly. For the x-intercept, set\(y=0\), then: \(3x=6\), so \(x=2\). For the y-intercept, set\(x=0\), then: \(y=6\). The x-intercept is (2,0) and the y-intercept is (0,6). Choose a point outside the line, let's say origin (0,0). Since \(3(0) + 0 \ngeq 6\), shade the opposite side of the line (and the line itself, as it's more than or equal to).

Step 4: Graph the inequality \(x + y \geq 4\)

Rewrite the inequality as an equality: \(x + y=4\). Find the x and y-intercepts again. For both x and y-intercepts, either \(x=0\) or \(y=0\), then: \(x=4\) or \(y=4\). The x-intercept is (4,0) and the y-intercept is (0,4). Choose a point outside the line, let's suppose origin (0,0). Since \(0 + 0 \ngeq 4\), shade the opposite side of the line (and the line itself, as it's more than or equal to).

Step 5: Graph the inequalities \(x \geq 0\) and \(y \geq 0\)

The inequalities \(x \geq 0\) and \(y \geq 0\) represent the axes themselves. Shade the region in the first quadrant (above the x-axis and right of the y-axis) as both x and y are greater than or equal to 0.

Step 6: Identify the solution region

The solution set will be in the region where all the shaded areas from steps 2-5 overlap. Graph the inequalities on the same set of axes and find the intersection of all the shaded regions.

Step 7: Determine if the solution set is bounded or unbounded

A solution set is considered bounded if it is enclosed by the inequalities, and unbounded if it extends to infinity. Observe the graph, if the region enclosed by the inequalities forms a closed figure (like a polygon), the solution set is bounded. If the region extends infinitely, the solution set is unbounded. By following these steps, you should be able to graphically determine the solution set for the given system of inequalities and establish whether the solution set is bounded or unbounded.

Key Concepts

Graphical Solution

Understanding systems of inequalities involves graphing each inequality on the same coordinate plane to find a common solution region. This is referred to as a graphical solution. The process requires converting each inequality into an equation, which forms a line on the graph. These lines act as boundaries of the solution region. Then, determine which side of the line to shade by testing a point that does not lie on the line.

• First, graph the boundary line by setting the inequality as an equation.
• Second, identify which half-plane to shade by substituting a test point into the original inequality.
• The overlapped shaded region from all inequalities' half-planes is the graphical solution.

For example, with the system of inequalities discussed, graph each of them to observe where all shaded regions intersect. This intersection represents the common solutions that satisfy all inequalities in the system.

Bounded vs Unbounded

When determining whether the solution set is bounded or unbounded, it is crucial to observe the graphical solution. A bounded solution set is enclosed within a limited region, often forming a closed shape such as a polygon on the graph. In contrast, an unbounded solution set extends indefinitely in one or more directions, meaning it is not confined by the boundaries of the graph.

• A bounded region will appear as a distinct shape with complete boundaries.
• An unbounded region will extend beyond a certain area, without forming a completely enclosed shape.

In the provided exercise, after graphing all inequalities, if the shaded region forms a closed polygon, the solution set is bounded. If it opens outwards and continues infinitely, it is unbounded.

Graphing Inequalities

Graphing inequalities on a coordinate plane involves a few precise steps to visualize the solution. Start by converting each inequality to an equation to find its boundary line. Determine the x and y-intercepts by setting one variable to zero and solving for the other. These intercepts provide points to draw the line on the graph.

• Transform the inequality into an equation to identify the boundary line.
• Calculate the x-intercept by setting y to zero and solving for x.
• Determine the y-intercept by setting x to zero and solving for y.

Once the line is drawn, shading the correct side based on the inequality sign is essential. Use a test point, usually (0,0) if not on the line, to see which side of the line satisfies the inequality. Follow these steps for each inequality, and the overlapping shaded area is your solution set. It clearly shows all the points that satisfy the system of inequalities simultaneously.