To analyze the function \( A(d) = d(250 - 2d) \), we should first determine the reasonable ranges for the horizontal axis (representing \( d \)) and the vertical axis (representing \( A(d) \)).
1. Horizontal Axis (d):
The variable \( d \) represents the width of the fenced area, which must be a non-negative value. Given that Peter has 250 feet of fencing for three sides (a width \( d \) and two lengths of \( 250 - 2d \)), we need to find the feasible range of \( d \).
- Lower Bound: The width \( d \) must be greater than 0.
- Upper Bound: Since there are two lengths of \( 250 - 2d \), for \( d \) to be reasonable, the following must hold: \[ 250 - 2d > 0 \implies 250 > 2d \implies d < 125 \]
Thus, \( d \) can range from \( 0 \) to \( 125 \). Therefore, a reasonable scale for the horizontal axis could be: \[ 0 \leq d \leq 125 \]
2. Vertical Axis (A(d)):
Now we analyze the function \( A(d) \) to determine the reasonable scale for the vertical axis, which represents the area \( A(d) \).
The area function can be simplified and evaluated: \[ A(d) = d(250 - 2d) = 250d - 2d^2 \]
This is a quadratic function opening downwards (as indicated by the negative coefficient of \( d^2 \)). To find the maximum area, we can calculate the vertex of the parabola.
The vertex \( d_{vertex} \) can be found using the formula for the \( d \) coordinate of the vertex of \( A(d) = ad^2 + bd + c \): \[ d_{vertex} = -\frac{b}{2a} = -\frac{250}{-4} = 62.5 \]
Now, we can calculate the maximum area: \[ A(62.5) = 62.5(250 - 2 \cdot 62.5) = 62.5(250 - 125) = 62.5 \cdot 125 = 7812.5 \text{ square feet} \]
Conclusion:
Based on this analysis:
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A reasonable scale for the horizontal axis (for \( d \)) is: \[ 0 \leq d \leq 125 \]
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A reasonable scale for the vertical axis (for \( A(d) \)) is: \[ 0 \leq A(d) \leq 7812.5 \]
To summarize:
- Horizontal axis: \( 0 \leq d \leq 125 \)
- Vertical axis: \( 0 \leq A(d) \leq 7812.5 \)
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