To determine reasonable scales for the horizontal and vertical axes of the function \( A(d) = d(250 - 2d) \), we need to analyze the function.
Horizontal axis (d):
- The variable \( d \) represents the width of the fenced area.
- Given that Peter has 250 feet of fencing, he can only use \( d \) values that make sense for building the fenced area next to his house.
- Since he has fencing for three sides, the total length of the three sides (\( d + 250 - 2d \)) must equal to 250 feet. This indicates that: \[ d \leq 125 \text{ ft.} \]
- Thus, a reasonable scale for the horizontal axis would range from \( d = 0 \) ft to \( d = 125 \) ft. You could use increments of 25 ft or 10 ft for better readability.
Vertical axis (A(d)):
- The area \( A(d) = d(250 - 2d) \).
- To find the maximum area, we can rewrite \( A(d) \) as: \[ A(d) = 250d - 2d^2 \]
- This is a downward-opening parabola. The maximum area will occur at the vertex, which can be found using the formula \( d = -\frac{b}{2a} \) for the quadratic in standard form \( ax^2 + bx + c \):
- Here, \( a = -2 \) and \( b = 250 \): \[ d = -\frac{250}{2 \times -2} = \frac{250}{4} = 62.5 \text{ ft.} \]
- Plugging \( d = 62.5 \) into the area function to find the maximum area: \[ A(62.5) = 62.5(250 - 2 \times 62.5) = 62.5(250 - 125) = 62.5 \times 125 = 7812.5 \text{ square feet} \]
- Thus, the area will be zero for \( d = 0 \) and will reach a maximum at \( A(62.5) = 7812.5 \text{ ft}^2 \), before decreasing again towards \( d = 125 \) (where area = 0).
So, a reasonable scale for the vertical axis would range up to slightly above 7812.5 square feet, perhaps up to 8000 ft² with increments of 1000 ft² or similar for better granularity.
Complete statements: A reasonable scale for the horizontal axis of the function \( A(d)=d(250−2d) \) is 0 to 125 ft, with increments of 10 or 25 ft. A reasonable scale for the vertical axis of the function \( A(d)=d(250−2d) \) is 0 to 8000 ft², with increments of 1000 ft².