To analyze the function \( A(d) = d(250 - 2d) \) that represents the area Peter can fence in for his dog, we first need to determine the reasonable ranges for the variables involved.
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Horizontal Axis (d):
- The variable \( d \) represents the width of the area. Since this is a rectangular area being fenced in and must be positive, \( d \) can be any positive value. However, since the perimeter is constrained by the available fencing (250 feet), we can derive limits for \( d \).
- The total length of fencing used will be \( d + 2d = 250 \) (which accounts for the width \( d \) and two lengths of the fence). Rearranging gives \( 3d = 250 \), leading to \( d = \frac{250}{3} \approx 83.33 \) feet. Thus, \( d \) must be less than or equal to \( 83.33 \) feet.
- Therefore, a reasonable range for \( d \) is from 0 to approximately 83.33 feet.
- A suitable scale for the horizontal axis could be increments of 10 feet:
A reasonable scale for the horizontal axis of the function \( A(d) = d(250 - 2d) \) is 10 feet.
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Vertical Axis (A(d)):
- To find the maximum area that can be achieved, we need to evaluate the function \( A(d) = d(250 - 2d) \). This is a quadratic function that opens downwards.
- The vertex of a quadratic function given in the form \( A(d) = ad^2 + bd + c \) can be found using \( d = -\frac{b}{2a} \). Here, \( a = -2 \) and \( b = 250 \), which gives \( d = \frac{250}{4} = 62.5 \) feet as the value that maximizes the area.
- To find the maximum area, we substitute \( d = 62.5 \) back into the function: \[ A(62.5) = 62.5(250 - 2 \cdot 62.5) = 62.5 \times 125 = 7812.5 \text{ square feet}. \]
- Thus, the area will range from 0 to 7812.5 square feet.
- A suitable scale for the vertical axis could also be increments of 1000 square feet:
A reasonable scale for the vertical axis of the function \( A(d) = d(250 - 2d) \) is 1000 square feet.
In summary:
- A reasonable scale for the horizontal axis is 10 feet.
- A reasonable scale for the vertical axis is 1000 square feet.
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