To determine the domain of the function \( A = -w^2 + 20w \), we first need to understand the meaning of the function in the context of the fencing problem.
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Identify what \( w \) represents:
- \( w \) is the width of the garden. Since Marilee is using fencing to create a rectangular area, both the width and the length must be positive values.
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Understanding the constraints:
- The total perimeter is given by the total fencing available, which is 40 yards. For a rectangular area, the perimeter \( P \) is given by the formula \( P = 2l + 2w \). If we set \( l \) as the length and \( w \) as the width, we have: \[ 2l + 2w = 40 \] Simplifying this gives: \[ l + w = 20 \quad \Rightarrow \quad l = 20 - w \]
- Since both \( l \) and \( w \) must be positive, we find constraints on \( w \):
- \( w > 0 \) (width must be positive)
- \( l = 20 - w > 0 \) implies \( w < 20 \).
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Domain conclusion:
- Therefore, the domain of \( w \) is \( 0 < w < 20 \).
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Finding the maximum area:
- The area \( A = -w^2 + 20w \) is a quadratic function that opens downwards (since the coefficient of \( w^2 \) is negative), which means it has a maximum value at its vertex.
- The vertex \( w \) can be found using the formula for the vertex of a parabola, \( w = -\frac{b}{2a} \): \[ w = -\frac{20}{2(-1)} = 10. \]
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Calculate the corresponding area at this width:
- When \( w = 10 \), \[ A = -10^2 + 20(10) = -100 + 200 = 100 \text{ square yards.} \]
Thus, at \( w = 10 \), we achieve the maximum area of 100 square yards, while the width itself is constrained to be in the domain \( (0, 20) \).
In the context of the multiple-choice options provided, the most relevant response regarding the domain when the maximum area is achieved would be:
- 10 yards (which refers to the width \( w \) at the point of maximum area).
So the correct answer is:
10 yards.