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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]
### 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]
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