To select reasonable scales for the axes of the function \( A(d) = d(450 - 2d) \), we will first analyze the function itself.
Horizontal Axis (d)
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Understanding the function: The function is quadratic and opens downwards. The maximum area will occur at the vertex of the parabola. We can find the vertex using the formula for the vertex of a quadratic \( d = -\frac{b}{2a} \), where \( A = -2 \) and \( B = 450 \).
\[ d = -\frac{450}{2 \cdot -2} = \frac{450}{4} = 112.5, \text{ft} \]
Since \( d \) represents a dimension in feet, it should logically range from 0 to a little above 112.5 ft for fencing in an area.
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Choosing a reasonable scale: Considering that we've determined the maximum \( d \) is around 112.5 ft, we would want tick marks on the horizontal axis that allow us to clearly display this value, hence a scale of 10 ft or 15 ft would be reasonable.
Vertical Axis (A(d))
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Calculating the maximum area: To determine the maximum area, we can substitute \( d = 112.5 \) back into the function \( A(d) \).
\[ A(112.5) = 112.5(450 - 2 \cdot 112.5) = 112.5(450 - 225) = 112.5 \cdot 225 = 25312.5, \text{square ft} \]
Therefore, the maximum area is approximately \( 25312.5 \) square feet.
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Choosing a reasonable scale: Given that the maximum area is approximately 25312.5 square feet, we should look at the options for a suitable scale. A practical scale might be in intervals of a couple of hundred square feet or more due to the large maximum area.
Based on the choices:
- Horizontal Axis: The best choice is B. 10 ft (since we want to cover up to about 112.5 ft).
- Vertical Axis: The appropriate choice would be E. 900 square ft (as it allows for intervals when viewing values approaching \( 25312.5 \) square feet).
Conclusion:
- Reasonable scale for the horizontal axis: B. 10 ft
- Reasonable scale for the vertical axis: E. 900 square ft
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