To determine reasonable scales for the axes of the function \( A(d) = d(450 - 2d) \), we can analyze both the input \( d \) (the width of the area) and the output \( A(d) \) (the area in square feet).
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Examining \( d \):
- The variable \( d \) in this context represents the width or one dimension of the rectangular area fenced for the ostrich.
- Since the problem relates to fencing, \( d \) must be a non-negative value. Additionally, for a valid rectangular area, we must ensure that \( 450 - 2d \) is also non-negative.
- To find the maximum feasible value for \( d \): \[ 450 - 2d \geq 0 \implies 2d \leq 450 \implies d \leq 225 \]
- Therefore, \( d \) ranges from \( 0 \) to \( 225 \).
A reasonable scale for the horizontal axis could be from \( 0 \) to \( 225 \) with intervals of \( 25 \) (e.g., \( 0, 25, 50, 75, 100, 125, 150, 175, 200, 225 \)).
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Examining \( A(d) \):
- Now, we analyze the output of the function \( A(d) \). We know that \( A(d) = d(450 - 2d) \) is a quadratic function that opens downwards (since the coefficient of \( d^2 \) is negative).
- To find the maximum area, we can complete the square or use the vertex formula \( d = -\frac{b}{2a} \) for the standard form \( A(d) = -2d^2 + 450d \). Here, \( a = -2 \) and \( b = 450 \): \[ d = -\frac{450}{2(-2)} = \frac{450}{4} = 112.5 \]
- Now, we calculate \( A(112.5) \): \[ A(112.5) = 112.5(450 - 2(112.5)) = 112.5(450 - 225) = 112.5 \cdot 225 = 25312.5 \text{ square feet} \]
- Since \( A(d) \) will be \( 0 \) at \( d = 0 \) and \( d = 225 \), the maximum output occurs at \( 112.5 \) with an area of approximately \( 25312.5 \). Thus, \( A(d) \) ranges from \( 0 \) to \( 25312.5 \).
A reasonable scale for the vertical axis could be from \( 0 \) to \( 25312.5 \) with intervals of \( 5000 \) (e.g., \( 0, 5000, 10000, 15000, 20000, 25000 \)).
Final Scales:
- Horizontal axis (for \( d \)): from \( 0 \) to \( 225 \)
- Vertical axis (for \( A(d) \)): from \( 0 \) to \( 25312.5 \)