To complete the statements regarding the reasonable scales for the horizontal and vertical axes based on the function \( A(d) = d(250 - 2d) \):
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Horizontal Axis Scale: The variable \( d \) represents the width of the fenced area. Since Peter has 250 feet of fencing and needs to account for the 2 widths (since he needs fencing on 2 sides), the maximum width \( d \) can be is 125 feet (because \( 250 - 2d \geq 0 \) leads to \( d \leq 125 \)). A reasonable scale for the horizontal axis could indeed be 2 ft, allowing for clear representation of the width as it increases from 0 to 125 feet.
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Vertical Axis Scale: The area \( A(d) \) is given in square feet. To find the maximum area, we can complete the square for the function \( A(d) = d(250 - 2d) \). This is a quadratic function that opens downward. The vertex of this parabola will indicate the maximum area.
To find the vertex, we can use the formula for the vertex \( d = -\frac{b}{2a} \) where \( A(d) = -2d^2 + 250d \). Here, \( a = -2 \) and \( b = 250 \):
\[ d = -\frac{250}{2 \times -2} = \frac{250}{4} = 62.5 \text{ ft} \]
Now, substituting \( d = 62.5 \) back into the function \( A(d) \):
\[ A(62.5) = 62.5(250 - 2 \times 62.5) = 62.5(250 - 125) = 62.5 \times 125 = 7812.5 \text{ sq ft} \]
Thus, the maximum area \( A(d) \) is 7812.5 square feet. A reasonable vertical scale could be \( 500 \) sq ft, as this allows for clear representation of areas up to the maximum of approximately 7800 sq ft when increasing by 500 sq ft increments.
Summary
- Horizontal Axis Scale: \( 2 \) ft
- Vertical Axis Scale: \( 500 \) sq ft
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