To find the domain when the maximum area (range) is achieved using the equation \( A = -w^2 + 20w \), we first need to understand the context of the problem.
The equation represents a quadratic function where:
- \( A \) is the area of the rectangular garden.
- \( w \) is the width of the fence.
Since Marilee has 40 yards of fencing, the perimeter \( P \) is given by \( 2l + 2w = 40 \), where \( l \) is the length. This can be rearranged to express \( l \) in terms of \( w \): \[ l = 20 - w. \]
Thus, the area can also be expressed as: \[ A = l \times w = (20 - w)w = -w^2 + 20w, \] which is the same as the given equation.
To find the maximum area, we recognize that this is a downward-opening parabola. The maximum value occurs at the vertex. The vertex \( w \) of a parabola described by \( A = -w^2 + 20w \) can be found using the formula: \[ w = -\frac{b}{2a} \] where \( a = -1 \) and \( b = 20 \).
Calculating this gives: \[ w = -\frac{20}{2(-1)} = \frac{20}{2} = 10. \]
Now, we know that the width \( w \) should be in the range \( 0 < w < 20 \) (since \( w \) cannot be 0 and cannot exceed 20 yards). The maximum area occurs when \( w = 10 \).
At this value of \( w \):
- The width of the fence is \( 10 \) yards.
- The corresponding length \( l \) is \( 20 - 10 = 10 \) yards, making a square garden of area \( 100 \) square yards.
In terms of the domain when the maximum area (or range) is achieved, it corresponds to the width of the garden, which is:
10 yards.