Question
Mr. Jones needs to build a rectangular fence in his backyard to keep their dog safe. The length of the fence (y) should be at least 60 feet, and the distance around (perimeter) should be no more than 260 feet. Which system of inequalities and graph represent the possible dimensions of the pen?(1 point)
Responses
Responses
Answers
GPT-4o mini
To establish a system of inequalities representing the conditions given for Mr. Jones's fence, we need to consider the two main constraints: the length of the fence and the perimeter.
1. **Length Constraint**: The length \( y \) of the fence should be at least 60 feet.
\[
y \geq 60
\]
2. **Perimeter Constraint**: The perimeter \( P \) of a rectangle is given by the formula \( P = 2x + 2y \), where \( x \) is the width. The total perimeter should be no more than 260 feet.
\[
2x + 2y \leq 260
\]
Simplifying this, we divide by 2:
\[
x + y \leq 130
\]
Now we have the following system of inequalities:
\[
\begin{cases}
y \geq 60 \\
x + y \leq 130
\end{cases}
\]
### Graphing the inequalities:
1. **For \( y \geq 60 \)**: This is a horizontal line at \( y = 60 \), and the solution area is above this line since \( y \) must be greater than or equal to 60.
2. **For \( x + y \leq 130 \)**: To graph this line, we can find the x-intercept and the y-intercept:
- If \( y = 0 \), then \( x = 130 \).
- If \( x = 0 \), then \( y = 130 \).
Plotting these points creates a line passing through (130, 0) and (0, 130). The area below this line (including the line itself) is where \( x + y \leq 130 \).
### Feasible Region:
- The feasible region is the area that satisfies both inequalities:
- Above the line \( y = 60 \).
- Below or on the line \( x + y = 130 \).
### Summary
The system of inequalities is:
\[
\begin{cases}
y \geq 60 \\
x + y \leq 130
\end{cases}
\]
Graphically, you would shade the area above \( y = 60 \) and below or on the line \( x + y = 130 \) to find the possible dimensions for the fence.
1. **Length Constraint**: The length \( y \) of the fence should be at least 60 feet.
\[
y \geq 60
\]
2. **Perimeter Constraint**: The perimeter \( P \) of a rectangle is given by the formula \( P = 2x + 2y \), where \( x \) is the width. The total perimeter should be no more than 260 feet.
\[
2x + 2y \leq 260
\]
Simplifying this, we divide by 2:
\[
x + y \leq 130
\]
Now we have the following system of inequalities:
\[
\begin{cases}
y \geq 60 \\
x + y \leq 130
\end{cases}
\]
### Graphing the inequalities:
1. **For \( y \geq 60 \)**: This is a horizontal line at \( y = 60 \), and the solution area is above this line since \( y \) must be greater than or equal to 60.
2. **For \( x + y \leq 130 \)**: To graph this line, we can find the x-intercept and the y-intercept:
- If \( y = 0 \), then \( x = 130 \).
- If \( x = 0 \), then \( y = 130 \).
Plotting these points creates a line passing through (130, 0) and (0, 130). The area below this line (including the line itself) is where \( x + y \leq 130 \).
### Feasible Region:
- The feasible region is the area that satisfies both inequalities:
- Above the line \( y = 60 \).
- Below or on the line \( x + y = 130 \).
### Summary
The system of inequalities is:
\[
\begin{cases}
y \geq 60 \\
x + y \leq 130
\end{cases}
\]
Graphically, you would shade the area above \( y = 60 \) and below or on the line \( x + y = 130 \) to find the possible dimensions for the fence.