Question
Marilee is creating a garden in her yard. She needs to put up a fence to keep the animals out of it. She has 40 yards of fencing she can erect. This situation is modeled using the equation [Math Processing Error], where [Math Processing Error] is the area of the fence and w is the width of the fence. If you were to set up a table to determine the measurements of the width and its corresponding area, what would be a reasonable scale for the domain?(1 point)
Responses
5
5
2
2
10
10
1
Responses
5
5
2
2
10
10
1
Answers
Answered by
GPT-4o mini
To determine a reasonable scale for the domain when Marilee is setting up her garden with 40 yards of fencing, we first consider the constraints based on the fencing and the geometry of the garden.
Given that the total perimeter of the fence is 40 yards, if we assume the garden is rectangular, the perimeter \( P \) can be expressed using the equation:
\[
P = 2l + 2w = 40
\]
where \( l \) is the length and \( w \) is the width of the garden. Rearranging this gives:
\[
l + w = 20
\]
Now, the area \( A \) of the garden can be defined as:
\[
A = l \cdot w
\]
By substituting \( l \) from the perimeter equation, we get:
\[
A = w(20 - w) = 20w - w^2
\]
To find a reasonable scale for the domain (width \( w \)), we note that \( w \) must be positive and can go up to half the perimeter divided by 2 (since \( l + w = 20 \)). Therefore, the maximum width \( w \) could take is 20 yards (in which case the length \( l \) would be 0, which is not practical).
Considering reasonable increments, we can use a scale based on common garden dimensions, ensuring we can capture a range of widths effectively. The options provided are:
- 5
- 2
- 10
- 1
The most reasonable and manageable scale for testing different widths while still allowing for practical dimensions in a garden (which will often be larger than a few yards) is **5** yards increments. This allows the widths to be effectively captured as 0, 5, 10, 15, and 20 yards, and it avoids overwhelming detail while still being useful for calculations of area.
Therefore, the answer is **5**.
Given that the total perimeter of the fence is 40 yards, if we assume the garden is rectangular, the perimeter \( P \) can be expressed using the equation:
\[
P = 2l + 2w = 40
\]
where \( l \) is the length and \( w \) is the width of the garden. Rearranging this gives:
\[
l + w = 20
\]
Now, the area \( A \) of the garden can be defined as:
\[
A = l \cdot w
\]
By substituting \( l \) from the perimeter equation, we get:
\[
A = w(20 - w) = 20w - w^2
\]
To find a reasonable scale for the domain (width \( w \)), we note that \( w \) must be positive and can go up to half the perimeter divided by 2 (since \( l + w = 20 \)). Therefore, the maximum width \( w \) could take is 20 yards (in which case the length \( l \) would be 0, which is not practical).
Considering reasonable increments, we can use a scale based on common garden dimensions, ensuring we can capture a range of widths effectively. The options provided are:
- 5
- 2
- 10
- 1
The most reasonable and manageable scale for testing different widths while still allowing for practical dimensions in a garden (which will often be larger than a few yards) is **5** yards increments. This allows the widths to be effectively captured as 0, 5, 10, 15, and 20 yards, and it avoids overwhelming detail while still being useful for calculations of area.
Therefore, the answer is **5**.
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