Question

Graph each system of linear inequalities. State whether the graph is bounded or unbounded, and label the corner points. $$\left\\{\begin{array}{r}x \geq 0 \\\y \geq 0 \\\3 x+y \leq 6 \\\2 x+y \leq 2\end{array}\right.$$

Short answer

The graph is bounded with corner points at (0,2) and (1,0).

Step-by-step solution

- Identify the inequalities

The system of inequalities to graph is: 1. \(x \geq 0\)2. \(y \geq 0\)3. \(3x + y \leq 6\)4. \(2x + y \leq 2\).

- Graph the boundary lines

Draw the boundary lines for each inequality: 1. \(x = 0\) (vertical line on the y-axis)2. \(y = 0\) (horizontal line on the x-axis)3. \(3x + y = 6\) (line intersects the axes at (2, 0) and (0, 6))4. \(2x + y = 2\) (line intersects the axes at (1, 0) and (0, 2)).

- Shade the feasible region

Identify the region that satisfies all inequalities: 1. \(x \geq 0\) implies the region to the right of the y-axis.2. \(y \geq 0\) implies the region above the x-axis.3. \(3x + y \leq 6\) implies the region under the line \(3x + y = 6\).4. \(2x + y \leq 2\) implies the region under the line \(2x + y = 2\). The feasible region is where these shaded regions overlap.

- Identify corner points

Determine the vertices of the feasible region by solving intersections: 1. Intersection of \(x = 0\) and \(2x+y=2\) is \( (0,2)\).2. Intersection of \(y = 0\) and \(2x+y=2\) is \( (1,0)\).3. Intersection of \(3x+y=6\) and \(2x+y=2\) is solved by subtracting the second equation from the first: \(x=4\) and substituting to find \(y=-6\). There are no intersection points in the feasible region, so we discard this point.

- Determine boundedness

The feasible region is bounded by the lines and lies in the first quadrant. It has finite intersection points forming a closed, bounded region.

Key concepts

Feasible Region

In graphing linear inequalities, the feasible region is the area that satisfies all given inequalities at the same time. This region represents the set of potential solutions to the system. To find the feasible region:

• Plot each inequality on a graph
• Shade the area where each inequality is true
• Identify the overlap of these shaded areas, which forms the feasible region.

In the example given, the inequalities are:
1. \(x \geq 0\)
2. \(y \geq 0\)
3. \(3x \+ y \leq 6\)
4. \(2x \+ y \leq 2\).
Once plotted, the feasible region where these conditions overlap is confined within the bounds laid out by these lines.

Boundary Lines

Boundary lines are the lines that define the edges of the feasible region. Each inequality in the system corresponds to a boundary line, which is the line form of the inequality before adding the shading. The boundary lines of our example are:
1. \(x = 0\) (a vertical line along the y-axis)
2. \(y = 0\) (a horizontal line along the x-axis)
3. \(3x + y = 6\) (a diagonal line intersecting the axes at (2, 0) and (0, 6))
4. \(2x + y = 2\) (another diagonal line intersecting the axes at (1, 0) and (0, 2)).
These boundary lines help define where the feasible region lies by showing the limits within which the solutions must fall.

Intersection Points

Intersection points are the points where two boundary lines meet. These points are important because they often form the vertices of the feasible region. To find the intersection points in a system of inequalities, you need to solve simultaneously for the variables where two lines cross. In our example, the key intersection points are calculated as follows:

• Intersection of \(x = 0\) and \(2x + y = 2\) is \(0,2\)
• Intersection of \(y = 0\) and \(2x + y = 2\) is \(1,0\)
• Intersection of \(3x + y = 6\) and \(2x + y = 2\) would result in \(x = -4\y = -6\), a point which is discarded since it falls outside the feasible region.

By plotting these points, you'll have defined corners of your feasible region.

Bounded Region

A bounded region indicates that the feasible region is enclosed within a limited space on the graph. This means that the solutions are confined within finite boundaries, which form a closed shape, such as a triangle or quadrilateral. In the given example, you determine the bounded region by looking for the points where the inequalities confine the area. For the system:

• \(x \geq 0\)
• \(y \geq 0\)
• \(3x \+ y \leq 6\)
• \(2x \+ y \leq 2\)

The feasible region is bounded because the shading overlaps within the first quadrant, forming a finite, enclosed area. The vertices form the bounded region, ensuring that all solutions lie within this closed boundary.