To find the domain when the maximum range is achieved for the given equation, we first need to understand the context of the problem.
The equation provided is \( A = -w^2 + 20w \), where \( A \) is the area and \( w \) is the width.
To determine the dimensions of the fence, we need to consider that Marilee has 40 yards of fencing, which forms the perimeter of a rectangle. If we denote the length as \( l \) and the width as \( w \), the perimeter \( P \) can be expressed as \( P = 2l + 2w = 40 \). Solving for \( l \), we find:
\[ l = 20 - w \]
The area \( A \) of the rectangle can then be expressed as:
\[ A = l \times w = w(20 - w) = 20w - w^2 \]
This confirms that the equation \( A = -w^2 + 20w \) is indeed the quadratic function representing the area.
To find the maximum area of the garden, we can find the vertex of the quadratic equation given by \( A = -w^2 + 20w \). The vertex \( w \) of a quadratic in standard form \( A = ax^2 + bx + c \) occurs at:
\[ w = -\frac{b}{2a} \]
Here, \( a = -1 \) and \( b = 20 \):
\[ w = -\frac{20}{2(-1)} = \frac{20}{2} = 10 \]
Therefore, the maximum area occurs when the width \( w \) is 10 yards.
Now, checking the domain, since \( w \) represents the width and must be a positive quantity that satisfies the constraint of the perimeter (and must also result in a positive length), the practical range of \( w \) is between 0 yards and 20 yards (exclusive of the endpoints since we cannot have an area of zero). Thus, the domain for \( w \) is:
\[ 0 < w < 20 \]
Since we are asked specifically about the maximum range achieved, when \( w = 10 \), it corresponds to an area calculation:
\[ A(10) = -(10^2) + 20(10) = -100 + 200 = 100 \quad \text{square yards} \]
Now, relating your choices to this information:
- 10 yards (this refers to the width for max area)
- 5 yards (not relevant to max area)
- 10 square yards (wrong; this is not the area)
- 100 yards (this doesn’t match the units needed)
Based on the analysis, the choice that directly corresponds to achieving maximum area is 10 yards (the width).
So, the answer is 10 yards — this represents the width that maximizes the area.