Problem 67
Question
Graph the solution set of each system of inequalities by hand. $$\begin{array}{r}x \leq 4 \\\x \geq 0 \\\y \geq 0 \\\x+2 y \geq 2\end{array}$$
Step-by-Step Solution
Verified Answer
Shade the region bounded by \(0 \leq x \leq 4\), \(y \geq 0\), and \(x + 2y \geq 2\).
1Step 1: Understand the Inequalities
To graph the solution set, we need to understand each inequality: \(x \leq 4\) represents all points to the left of or on the vertical line \(x = 4\); \(x \geq 0\) represents all points to the right of or on the vertical line \(x = 0\); \(y \geq 0\) includes all points above or on the horizontal line \(y = 0\); \(x + 2y \geq 2\) is a line with the equation \(y \geq -\frac{1}{2}x + 1\). Points above or on this line satisfy the inequality.
2Step 2: Graph the Line x = 4
Draw a vertical line at \(x = 4\) on the coordinate plane. This line acts as a boundary for the inequality \(x \leq 4\), shading to the left of the line as this is the region of interest.
3Step 3: Graph the Line x = 0
Draw a vertical line at \(x = 0\) on the coordinate plane. Shade to the right of the line, as this satisfies \(x \geq 0\). This line corresponds to the y-axis.
4Step 4: Graph the Line y = 0
Draw a horizontal line along the x-axis, which corresponds to \(y = 0\). We shade above this line because \(y \geq 0\).
5Step 5: Graph the Line x + 2y = 2
To draw this line, find two points on the line by setting \(x = 0\) and solving for \(y\) which gives \(y = 1\), and \(y = 0\) to find \(x = 2\). Plot both points (0,1) and (2,0), and connect them to form the line. Shade above this line as this region satisfies \(x + 2y \geq 2\).
6Step 6: Identify the Feasible Region
The feasible region is where all the shaded areas overlap. Check that this includes the intersection of all the boundaries set by the inequalities: within \(0 \leq x \leq 4\), \(y \geq 0\), and above or on the line \(y \geq -\frac{1}{2}x + 1\).
7Step 7: Shade the Solution Set
Shade the region where \(0 \leq x \leq 4\), \(y \geq 0\), and \(x + 2y \geq 2\) overlap. This is the area that satisfies all given inequalities.
Key Concepts
Graphical SolutionsFeasible RegionShading Techniques
Graphical Solutions
In order to solve a system of inequalities graphically, you must translate each inequality into a visual boundary on a coordinate plane. Each inequality defines a specific region. The goal is to illustrate these regions on the graph and identify their intersections.
- Start by analyzing each inequality separately. Notice whether it describes a vertical or horizontal line, or if it forms an oblique line.
- Plot each line using standard graphing techniques like finding intercepts or using slope-intercept form.
- Remember that each inequality represents a half-plane, meaning that one side of the plotted line will satisfy the inequality.
Feasible Region
The feasible region is a crucial concept when graphing systems of inequalities. It represents the area on a graph that satisfies all inequalities at the same time. To find the feasible region:
- Graph each inequality following the steps for graphical solutions, including plotting lines and shading the respective regions.
- Observe the areas of overlap. This intersection is the feasible region, where all conditions of the system of inequalities coincide.
- This region may be bounded if all inequalities restrict it to a finite area, or unbounded if some inequalities extend indefinitely.
Shading Techniques
Shading techniques are fundamental in visualizing systems of inequalities. Shading specifies which side of a boundary line is part of the solution set for an inequality. Here's how you can effectively use shading techniques:
- After graphing a line, decide whether to shade above, below, to the left, or right based on the inequality's greater than or less than sign.
- For a condition like \(x \leq 4\), shade everything leftward from the line \(x = 4\). For \(y \geq 0\), shade above the \(x\)-axis.
- Use consistent shading patterns to avoid confusion, especially when multiple boundaries intersect. Cross-hatch or use different colors for clarity.
- Where all shadings meet is the feasible region, important for identifying valid solutions.
Other exercises in this chapter
Problem 66
Find each matrix product if possible. $$\left[\begin{array}{lll}0 & 3 & -4\end{array}\right]\left[\begin{array}{rrr}-2 & 6 & 3 \\ 0 & 4 & 2 \\ -1 & 1 & 4\end{ar
View solution Problem 66
Solve each system graphically. Check your solutions. Do not use a calculator. $$\begin{aligned}&x-y^{2}=1\\\&x^{2}+y^{2}=5\end{aligned}$$
View solution Problem 67
Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to complete the solution. $$\begin{aligned}2 x-y+4 z &=-2 \\\3 x+2 y-z &=-3
View solution Problem 67
Solve each system. Write solutions in terms of \(z\) if necessary. $$\begin{aligned} x+2 y-z &=0 \\ 3 x-y+z &=6 \\ -2 x-4 y+2 z &=0 \end{aligned}$$
View solution