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### Problem Summary:
Peter is building a fenced area for his dog, and he has 250 feet of fencing. Since the area is adjacent to his house, only three sides of the rectangular area need fencing. The width of the area is denoted by \( d \), and the function representing the total fenced area is:
\[
A(d) = d(250 - 2d)
\]
This function gives the area enclosed by the fencing in square feet.

### Questions to answer:
- A reasonable scale for the horizontal axis (which represents \( d \), the width).
- A reasonable scale for the vertical axis (which represents \( A(d) \), the area).

### Key Insights:
- The function is quadratic, and the maximum value occurs at the vertex.
- To find a reasonable range for \( d \), we consider the constraint on the perimeter: \( 2d + l = 250 \) (since only 3 sides are fenced). This implies that \( l = 250 - 2d \), and thus, \( d \) can’t exceed 125 feet (because otherwise, there wouldn't be any fencing left for the length).
- The area \( A(d) \) is maximized when \( d \) is at half the distance where the fencing equation is balanced, which we can compute.

Would you like me to calculate the optimal value of \( d \) and \( A(d) \), or provide scales based on this quadratic function?

Peter is building a fenced area for his dog. He has 250 feet of fencing, and the area next to his house only requires fencing on 3 sides. The width of the area is denoted by d, and the function representing the total fenced area is A(d) = d(250 - 2d). Find a reasonable scale for the horizontal axis (width d) and the vertical axis (area A(d)).

Solve the problem where Peter is building a fenced area for his dog with 250 feet of fencing. The problem involves using quadratic functions to optimize the area and find reasonable scales for width and area. Suitable for Grades 9-11.

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