Peter is building a fenced in area for his dog. He has 250 feet (ft.) of fencing to create an area for his dog. He is building it next to his house so he will only need fencing for 3 sides of the rectangular area. The diagram shows the width, d, of the area Peter is building for his dog.

The function A(d)=d(250−2d) represents the area that will be fenced in for Peter's dog, in square feet.

Complete the statements.

(2 points)
A reasonable scale for the horizontal axis of the function A(d)=d(250−2d) is
, A reasonable scale for the vertical axis of the function A(d)=d(250−2d) is

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1 answer

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.