To determine a reasonable scale for the axes of the function \( A(d) = d(250 - 2d) \), we first need to understand the dimensions of the fenced area.
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Understanding the function and dimensions:
- \( d \) represents the width of the area (the dimension perpendicular to the house).
- The length of the area is \( 250 - 2d \) (because fencing on the other two sides will take up \( 2d \)).
- Therefore, the width can range from 0 to a maximum where \( 250 - 2d \) is still positive, giving us the inequality: \[ 250 - 2d > 0 \implies d < 125 \]
- So \( d \) ranges from 0 to 125 feet.
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Calculating the area:
- We can express the area as: \[ A(d) = d(250 - 2d) = 250d - 2d^2 \]
- This is a quadratic function that opens downwards. The vertex of this parabola will give the maximum area, and is found at: \[ d = -\frac{b}{2a} = -\frac{250}{-4} = 62.5 \text{ feet} \]
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Finding the maximum area:
- To calculate the maximum area, substitute \( d = 62.5 \) into the area function: \[ A(62.5) = 62.5(250 - 2(62.5)) = 62.5(250 - 125) = 62.5(125) = 7812.5 \text{ square feet} \]
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Choosing scales for the axes:
- Horizontal axis (d):
- Since \( d \) ranges from 0 to 125, a reasonable scale could be intervals of 10 feet. This would include the points 0, 10, 20, ..., 120, 125.
- Vertical axis (A):
- The area \( A(d) \) ranges from 0 to the maximum area of about 7812.5 square feet. A reasonable scale could be 500 square feet. This gives labels such as 0, 500, 1000, ..., 7500, 8000.
- Horizontal axis (d):
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Conclusion:
- A reasonable scale for the horizontal axis (d) is 10 feet.
- A reasonable scale for the vertical axis (A) is 500 square feet.
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