To find the domain for the width \( w \) of the fence, we need to understand the constraints given by the total length of the fence, which is 40 yards. The equation for the area \( A \) is given by:
\[ A = -w^2 + 20w \]
Here, \( w \) represents the width of the fenced area. The perimeter \( P \) of a rectangular area, given by a width \( w \) and a length \( l \), can be expressed as:
\[ P = 2l + 2w \]
Since Marilee has 40 yards of fencing, we can write:
\[ 2l + 2w = 40 \]
This simplifies to:
\[ l + w = 20 \quad \Rightarrow \quad l = 20 - w \]
To ensure the width \( w \) is valid, it must satisfy:
\[ 0 < w < 20 \]
This means the width can range from greater than 0 yards to less than 20 yards to maintain a positive length for the rectangle.
Given the options provided, the key value that fits this domain constraint and represents possible dimensions for the fence when the maximum range is achieved is:
20 yards
Therefore, the width of the fence can range from values approaching 0 up to but not including 20 yards, with 20 yards being significant in the context of the problem.