To determine the domain when the maximum range is achieved using the equation \( A = -w^2 + 20w \), we first need to find the values of \( w \) (the width) for which \( A \) will be maximized.
The given quadratic equation \( A = -w^2 + 20w \) represents a downward-opening parabola. The maximum area can be found at the vertex of the parabola. The vertex \( w \) for a quadratic equation in the form \( A = aw^2 + bw + c \) can be found using the formula:
\[ w = -\frac{b}{2a} \]
In this case:
- \( a = -1 \)
- \( b = 20 \)
Substituting these values into the formula gives:
\[ w = -\frac{20}{2 \cdot (-1)} = \frac{20}{2} = 10 \]
Thus, the maximum area is achieved when the width \( w \) is 10 yards.
Next, we also need to consider the overall dimensions of the garden and what values \( w \) can logically take. Since \( w \) represents the width of the fence, it must be non-negative and also should not exceed half the total amount of fencing (which would be impossible), thus:
- \( w \geq 0 \)
- \( 2w \) must not exceed the total fencing available when also accounting for the length of the garden. In our case, the total fencing is 40 yards giving this relationship \( 2l + 2w = 40 \) or effectively \( l + w = 20 \) where \( l = \) length.
This means the realistic domain for \( w \) must be within the range of [0, 20] yards.
Since the maximum area occurs at \( w = 10 \), the domain where the maximum range is achieved could be described as:
- 10 yards, since this is where the maximum area is achieved given the constraints of the problem.
Therefore, the answer is 10 yards.