Question
Am is creating a garden in her yard she needs to put up a fence to keep the animals out of it and she has 40 yards of fencing she can erect the situation is modeled using the equation a = negative w squared plus 20 w where a is the area and w is the width of the fence if you would have set up a table to determine the measurements of the width and its corresponding area what would be reasonable scale for the domain
Answers
Answered by
GPT-4o mini
To analyze the problem of Am's garden, the equation given is:
\[
a = -w^2 + 20w
\]
where \( a \) represents the area in square yards and \( w \) represents the width of the fence in yards.
### Understanding the Problem
Am has 40 yards of fencing available. Assuming the fence forms a closed rectangular shape, the perimeter \( P \) of the rectangle is given by:
\[
P = 2l + 2w = 40
\]
From this, we can derive the relationship between length \( l \) and width \( w \):
\[
l + w = 20 \implies l = 20 - w
\]
### Area Calculation
The area \( a \) of the rectangle can be expressed as:
\[
a = l \cdot w = (20 - w) \cdot w = 20w - w^2
\]
This matches the form given in the problem statement (\( a = -w^2 + 20w \)).
### Domain
To determine the reasonable scale for the domain of \( w \) (width), we should consider the physical constraints of the fence:
1. **Width Size**: \( w \) must be positive.
2. **Maximum Width**: Since the total width and length together must not exceed 20 (from the perimeter equation) for a proper rectangle, the maximum width occurs when the length becomes 0, which means \( w \) cannot exceed 20 yards.
Thus, the reasonable domain for \( w \) is:
\[
0 < w < 20
\]
### Reasonable Scale for Table
When setting up a table for the width \( w \) and its corresponding area \( a \), you might choose to represent values of \( w \) at intervals. A reasonable scale could be:
- Start from a small positive value (e.g., \( w = 1 \)) and go up to \( w = 19 \).
- Using intervals of 1 yard could be practical, resulting in a set of values like \( 1, 2, 3, \ldots, 19 \).
Thus, your table could look something like this:
| Width (w) | Area (a) |
|-----------|-----------------|
| 1 | a(1) = 19 |
| 2 | a(2) = 36 |
| 3 | a(3) = 51 |
| 4 | a(4) = 64 |
| 5 | a(5) = 75 |
| ... | ... |
| 19 | a(19) = 19 |
This would provide a complete view of how the area changes as the width increases up to its maximum feasible size while ensuring that the area is calculated properly for each corresponding width.
\[
a = -w^2 + 20w
\]
where \( a \) represents the area in square yards and \( w \) represents the width of the fence in yards.
### Understanding the Problem
Am has 40 yards of fencing available. Assuming the fence forms a closed rectangular shape, the perimeter \( P \) of the rectangle is given by:
\[
P = 2l + 2w = 40
\]
From this, we can derive the relationship between length \( l \) and width \( w \):
\[
l + w = 20 \implies l = 20 - w
\]
### Area Calculation
The area \( a \) of the rectangle can be expressed as:
\[
a = l \cdot w = (20 - w) \cdot w = 20w - w^2
\]
This matches the form given in the problem statement (\( a = -w^2 + 20w \)).
### Domain
To determine the reasonable scale for the domain of \( w \) (width), we should consider the physical constraints of the fence:
1. **Width Size**: \( w \) must be positive.
2. **Maximum Width**: Since the total width and length together must not exceed 20 (from the perimeter equation) for a proper rectangle, the maximum width occurs when the length becomes 0, which means \( w \) cannot exceed 20 yards.
Thus, the reasonable domain for \( w \) is:
\[
0 < w < 20
\]
### Reasonable Scale for Table
When setting up a table for the width \( w \) and its corresponding area \( a \), you might choose to represent values of \( w \) at intervals. A reasonable scale could be:
- Start from a small positive value (e.g., \( w = 1 \)) and go up to \( w = 19 \).
- Using intervals of 1 yard could be practical, resulting in a set of values like \( 1, 2, 3, \ldots, 19 \).
Thus, your table could look something like this:
| Width (w) | Area (a) |
|-----------|-----------------|
| 1 | a(1) = 19 |
| 2 | a(2) = 36 |
| 3 | a(3) = 51 |
| 4 | a(4) = 64 |
| 5 | a(5) = 75 |
| ... | ... |
| 19 | a(19) = 19 |
This would provide a complete view of how the area changes as the width increases up to its maximum feasible size while ensuring that the area is calculated properly for each corresponding width.
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