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Solve the system graphically and indicate whether the solution region is bounded or unbounded. Find the coordinates of each corner point.

[tex]\[
\begin{array}{l}
x + 2y \leq 4 \\
x \geq 0 \\
y \geq 0
\end{array}
\][/tex]

Use the graphing tool to graph the system of inequalities.

The solution region is \_\_\_\_\_\_\_\_

What are the coordinates of each corner point? (Type an ordered pair. Use a comma to separate answers as needed.)

\_\_\_\_\_\_\_\_

Sagot :

GinnyAnswer
To solve the system of inequalities graphically and determine whether the solution region is bounded or unbounded, follow these steps:

### Step 1: Understanding the Inequalities

We have the following system of inequalities:

1. [tex]\( x + 2 y \leq 4 \)[/tex]
2. [tex]\( x \geq 0 \)[/tex]
3. [tex]\( y \geq 0 \)[/tex]

### Step 2: Plotting the Boundary Lines

First, consider the equality for each inequality:

1. [tex]\( x + 2 y = 4 \)[/tex]
2. [tex]\( x = 0 \)[/tex]
3. [tex]\( y = 0 \)[/tex]

- [tex]\( x + 2 y = 4 \)[/tex]:

This is a straight line. We can find the intercepts:

- For [tex]\( x \)[/tex]-intercept ([tex]\(y = 0\)[/tex]):
[tex]\( x + 2(0) = 4 \Rightarrow x = 4 \)[/tex]

- For [tex]\( y \)[/tex]-intercept ([tex]\(x = 0\)[/tex]):
[tex]\( 0 + 2 y = 4 \Rightarrow 2 y = 4 \Rightarrow y = 2 \)[/tex]

So, the line passes through points [tex]\( (4, 0) \)[/tex] and [tex]\( (0, 2) \)[/tex].

- [tex]\( x = 0 \)[/tex]:

This is the [tex]\( y \)[/tex]-axis.

- [tex]\( y = 0 \)[/tex]:

This is the [tex]\( x \)[/tex]-axis.

### Step 3: Shading the Regions

- For [tex]\( x + 2 y \leq 4 \)[/tex]: We shade the region below the line [tex]\( x + 2 y = 4 \)[/tex].

- For [tex]\( x \geq 0 \)[/tex]: We shade the region to the right of the [tex]\( y \)[/tex]-axis ([tex]\( x = 0 \)[/tex]).

- For [tex]\( y \geq 0 \)[/tex]: We shade the region above the [tex]\( x \)[/tex]-axis ([tex]\( y = 0 \)[/tex]).

### Step 4: Determining the Solution Region

The solution region is the intersection of all these shaded regions. The feasible region is bounded by the lines [tex]\( x + 2 y = 4 \)[/tex], [tex]\( x = 0 \)[/tex], and [tex]\( y = 0 \)[/tex].

### Step 5: Identifying the Corner Points

The corner points of the feasible solution region (where lines intersect) can be calculated:

1. Intersection of [tex]\( x + 2 y = 4 \)[/tex] and [tex]\( x = 0 \)[/tex]:
- Set [tex]\( x = 0 \)[/tex] in [tex]\( x + 2 y = 4 \)[/tex]:
[tex]\( 0 + 2 y = 4 \Rightarrow y = 2 \)[/tex]
- Coordinate: [tex]\( (0, 2) \)[/tex]

2. Intersection of [tex]\( x + 2 y = 4 \)[/tex] and [tex]\( y = 0 \)[/tex]:
- Set [tex]\( y = 0 \)[/tex] in [tex]\( x + 2 y = 4 \)[/tex]:
[tex]\( x + 2(0) = 4 \Rightarrow x = 4 \)[/tex]
- Coordinate: [tex]\( (4, 0) \)[/tex]

3. Intersection of [tex]\( x = 0 \)[/tex] and [tex]\( y = 0 \)[/tex]:
- Coordinate: [tex]\( (0, 0) \)[/tex]

### Step 6: Conclusion

- The solution region is bounded (as it is restricted by the lines and does not extend infinitely).

- The coordinates of the corner points are:
[tex]\[ (0, 0), (0, 2), (4, 0) \][/tex]

So, to summarize:

- The solution region is: Bounded
- The coordinates of each corner point: [tex]\((0, 0)\)[/tex], [tex]\((0, 2)\)[/tex], [tex]\((4, 0)\)[/tex]