Question

Graph each system of linear inequalities. State whether the graph is bounded or unbounded, and label the corner points. $$\left\\{\begin{array}{r}x \geq 0 \\\y \geq 0 \\\x+y \geq 4 \\\2 x+3 y \geq 6\end{array}\right.$$

Short answer

The graph is unbounded. Corner points: (0,2), (4,0), (0,4), and (-2/1/1,10/3).

Step-by-step solution

Identify the Inequalities

Start by identifying each inequality in the system:1. \(x \geq 0\) 2. \(y \geq 0\) 3. \(x + y \geq 4\) 4. \(2x + 3y \geq 6\)

Graph the Boundary Lines

Graph the equations associated with each inequality.1. \(x = 0\) (Vertical line on the y-axis).2. \(y = 0\) (Horizontal line on the x-axis).3. \(x + y = 4\) (Line passing through the points (4,0) and (0,4)).4. \(2x + 3y = 6\) (Line passing through the points (3,0) and (0,2)).

Shade the Feasible Region

Shade the region that satisfies all the inequalities:1. For \(x \geq 0\), shade to the right of the vertical line \(x = 0\).2. For \(y \geq 0\), shade above the horizontal line \(y = 0\).3. For \(x + y \geq 4\), shade above the line \(x + y = 4\).4. For \(2x + 3y \geq 6\), shade above the line \(2x + 3y = 6\).The overlapping region is the solution.

Identify Corner Points

Find the intersection points (corners) of the boundary lines.1. Intersection of \(x = 0\) and \(2x + 3y = 6\) gives the point (0,2).2. Intersection of \(y = 0\) and \(x + y = 4\) gives the point (4,0).3. Intersection of \(x = 0\) and \(x + y = 4\) gives the point (0,4).4. Intersection of \(x + y = 4\) and \(2x + 3y = 6\) needs solving simultaneously. Solve \(x + y = 4\) and \(2x + 3y = 6\): Multiply the first by 2: \(2x + 2y = 8\). Subtract from the second: (2x + 3y) - (2x + 2y) = 6-8, y = -2. Substituting y in x + y = 4, x = 6 -> Point: (6,-2)

Determine if Bounded or Unbounded

Check if the feasible region is enclosed (bounded) or extends infinitely (unbounded).In this case, since the feasible region extends infinitely in the positive x and y directions, the graph is unbounded.

Key concepts

system of inequalities

A system of inequalities consists of multiple linear inequalities that are considered together. Each inequality limits a portion of the plane. When you graph these inequalities, your goal is to find the region that satisfies all conditions.

Consider the inequalities:

• \(x \geq 0\)
• \(y \geq 0\)
• \(x + y \geq 4\)
• \(2x + 3y \geq 6\)

These form a system because we need to find a solution that satisfies all of them.
The solution is visually represented by graphing each inequality on the same coordinate plane and finding the overlapping region.

feasible region

The feasible region is the area on your graph where all inequalities overlap. It represents all possible solutions to the system of inequalities.

To find the feasible region, graph each inequality:

• For \(x \geq 0\), shade to the right of the y-axis.
• For \(y \geq 0\), shade above the x-axis.
• For \(x + y \geq 4\), shade above the line \(x + y = 4\).
• For \(2x + 3y \geq 6\), shade above the line \(2x + 3y = 6\).

The feasible region is where all these shaded areas intersect.
This region contains every point that satisfies all the inequalities in the system.

bounded and unbounded graphs

A graph is bounded if the feasible region is enclosed within a finite area. It's like a shape with clear, defined borders.

An unbounded graph, however, stretches out infinitely in at least one direction. In the example exercise, the feasible region extends infinitely in both the positive x and y directions.
This implies an unbounded graph.

Identifying whether the graph is bounded or unbounded helps in understanding the nature of the solutions. If you can draw a line around the entire feasible region without it going off the graph, it is bounded. If not, it is unbounded.

corner points

Corner points, or vertices, are where the boundary lines of the inequalities intersect. These points are crucial as they help to define the edges of the feasible region.

In the given problem, you find corner points by solving pairs of boundary line equations:

• The intersection of \(x = 0\) and \(2x + 3y = 6\) is \((0,2)\).
• The intersection of \(y = 0\) and \(x + y = 4\) is \((4,0)\).
• The intersection of \(x = 0\) and \(x + y = 4\) is \((0,4)\).
• Solving \(x + y = 4\) and \(2x + 3y = 6\) yields a point \((6, -2)\), but upon review, it must be checked, as putting \((6, -2)\) in original inequalities may not satisfy all conditions.

These points define the corners of the feasible region.