Problem 17 - Precalculus: Functions and Graphs

Question

Sketch the graph of the system of Inequalities. $$\left\\{\begin{array}{l}x+2 y \leq 8 \\\0 \leq x \leq 4 \\\0 \leq y \leq 3\end{array}\right.$$

Step-by-Step Solution

Answer

The solution is the polygonal region bounded by \(x = 0\), \(x = 4\), \(y = 0\), \(y = 3\), and below \(x + 2y = 8\).

Step 1: Graph the First Inequality

Start by graphing the line represented by the equation \(x + 2y = 8\). This is the boundary line for the inequality \(x + 2y \leq 8\). To graph, find the intercepts: for the x-intercept, set \(y = 0\), giving \(x = 8\); for the y-intercept, set \(x = 0\), giving \(y = 4\). Plot these intercepts and draw a solid line through them because the inequality is \(\leq\), indicating that points on the line are included in the solution set.

Step 2: Shade the Solution Region for the First Inequality

Since the inequality is \(x + 2y \leq 8\), the region below or on the line of \(x + 2y = 8\) needs to be shaded. Test a point, like \((0,0)\); as it satisfies the inequality \(0 + 2(0) = 0 \leq 8\), shade the region that includes this point. The shaded region indicates all possible solutions to this inequality.

Step 3: Draw the Vertical Boundaries

Next, consider the inequality \(0 \leq x \leq 4\). This establishes vertical boundary lines at \(x = 0\) and \(x = 4\). Plot vertical lines at these x-values. Within these lines, \(x\) values including and between 0 and 4 are valid.

Step 4: Draw the Horizontal Boundaries

Similarly, consider the inequality \(0 \leq y \leq 3\). This establishes horizontal boundary lines at \(y = 0\) and \(y = 3\). Plot horizontal lines at these y-values. Within these boundaries, \(y\) values including and between 0 and 3 are valid.

Step 5: Identify the Intersection of All Regions

The solution to the system of inequalities is the region where all shaded regions and bounded areas overlap. This forms a polygonal area where the vertical and horizontal lines, along with the line from Step 1 and 2, intersect. This is the area to be shaded on your graph.

Step 6: Highlight the Solution Area

Finally, clearly shade the area which is bounded by the lines \(x = 0\), \(x = 4\), \(y = 0\), \(y = 3\), and below the line \(x + 2y = 8\). This is the solution region for the system of inequalities.

Key Concepts

Inequality Graphing

Graphing inequalities involves representing mathematical relationships visually on a coordinate plane. The main idea is to sketch the boundary lines, which define the limits of the inequalities. In our exercise, we start by transforming the inequality \(x + 2y \leq 8\) into an equation: \(x + 2y = 8\). We plot this as a solid line, showing that points on the line are part of the solution set because the inequality symbol is less than or equal to (\(\leq\)).
When dealing with inequalities, the type of boundary line matters:

• Solid Line: Used for inequalities with \(\leq\) or \(\geq\).
• Dashed Line: Used for strict inequalities \(<\) or \(>\).

After drawing the boundary line, we determine which side of the line represents solutions to the inequality by shading.
This is achieved by testing a point not on the line, typically \((0,0)\) if it doesn't lie on the boundary. If it satisfies the inequality, shade the region containing this point.

Solution Region Identification

Identifying the solution region in a graph of inequalities is like finding a common overlap area, where all conditions of the inequalities are satisfied simultaneously. After plotting each boundary, we look for areas where the shaded regions intersect. This common area is the solution region for the entire system of inequalities.
Consider the logical sequence:

• Shade below or including the line \(x + 2y = 8\) since the inequality is \(\leq\).
• The vertical region between \(x = 0\) and \(x = 4\), inclusive.
• The horizontal area between \(y = 0\) and \(y = 3\), inclusive.

Overlay these aspects on the graph, and the solution region is the area that satisfies all these conditions simultaneously.
Appropriately shading this intersection is crucial, as it visually represents the set of all possible solutions to the system.

Coordinate System Plotting

The coordinate system is a two-dimensional number line framework where every point is determined by a pair of numbers. These numbers are called coordinates and represent a point's location in terms of \(x\) and \(y\). When plotting a graph of inequalities, we use this system to organize our plot and sketch boundary lines correctly.
Steps to plot the system on a coordinate plane:

• Draw the two main axes: horizontal for \(x\) and vertical for \(y\).
• Determine the scales for \(x\) and \(y\) based on the inequalities you need to plot, such as \(0 \leq x \leq 4\) and \(0 \leq y \leq 3\).
• Plot the intercepts needed for the lines. For instance, the intercepts for \(x + 2y = 8\) are at \(x=8\) and \(y=4\).

This plotting creates a clear visual representation of all constraints on the graph, showing feasible points in the context of the inequalities.

Intercept Calculation

Intercepts are key points where a line crosses the axes on a graph, and they are crucial for drawing the boundary lines of inequalities. Calculating intercepts helps in accurately plotting the lines on a coordinate system.
Here's how to find intercepts:

• x-intercept: Set \(y = 0\) in the equation. For \(x + 2y = 8\), setting \(y = 0\) gives \(x = 8\). Therefore, the x-intercept is at \((8, 0)\).
• y-intercept: Set \(x = 0\) in the equation. For the same equation, setting \(x = 0\) gives \(2y = 8\) or \(y = 4\). Thus, the y-intercept is at \((0, 4)\).

Ensure you plot these intercepts correctly, joining them with the appropriate type of line (solid or dashed). This ensures that the graphed line accurately reflects the inequality terms. Intercepts provide a simple method for starting the graph without requiring additional points.