Question
Anna 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 the situation is modeled by using the equation a equals negative w squared plus 20 w where a is the area of the fence in w is the width of the fence if you were to set up a table to determine the measurements of the width and it's corresponding area what would be a reasonable scale for the domain 2 5 10 or 1
Answers
Answered by
GPT-4o mini
To create a table for the width \( w \) and its corresponding area \( a \) in the equation \( a = -w^2 + 20w \), we first need to understand the context of the problem and the reasonable range for the width \( w \).
Given that Anna has 40 yards of fencing and the perimeter of a rectangular area is given by the formula \( P = 2l + 2w = 40 \), we can derive a relationship between the length \( l \) and width \( w \). From this, we find that:
\[
l = 20 - w
\]
Since the area \( a \) is given by \( a = l \times w \), substituting for \( l \):
\[
a = (20 - w) \times w
\]
This matches the equation given:
\[
a = -w^2 + 20w
\]
Next, let's establish the reasonable scale for the width \( w \). The width must be non-negative and bounded by the perimeter constraints:
- When \( w = 0 \), then \( l = 20 \) and thus \( a = 0 \).
- When \( w = 20 \), then \( l = 0 \) and thus \( a = 0 \).
For practical purposes, the width \( w \) will realistically range from \( 0 \) to \( 20 \) yards.
### Proposed Scale for the Domain
Considering the potential width values and the options provided (2, 5, 10, or 1), a reasonable scale would allow the entire range from \( 0 \) to \( 20 \) to be covered without skipping possible important values.
1. **Width Scale of 1**: This would yield values from 0 to 20 (i.e., 0, 1, 2, ..., 20) which gives us a lot of detail but could be cumbersome.
2. **Width Scale of 2**: This would yield values 0, 2, 4, ..., 20 which provides a good number of key points but not too many.
3. **Width Scale of 5**: This would yield values 0, 5, 10, 15, 20, which might miss some detail in the critical section of the parabola.
4. **Width Scale of 10**: This would yield values 0, 10, 20, which is too sparse to capture the nuances of the area variation effectively.
### Conclusion
Given these considerations, a **scale of 2** is a reasonable choice for the domain as it provides structured increments that allow us to clearly observe the changes in area with respect to width:
- With increments of 2 (0, 2, 4, ... , 20), we can effectively identify and capture the maximum area, where the area formula shows clear curvature due to the negative quadratic term.
Thus, if you were to set up a table, using a scale of **2** for the width would be the most reasonable choice.
Given that Anna has 40 yards of fencing and the perimeter of a rectangular area is given by the formula \( P = 2l + 2w = 40 \), we can derive a relationship between the length \( l \) and width \( w \). From this, we find that:
\[
l = 20 - w
\]
Since the area \( a \) is given by \( a = l \times w \), substituting for \( l \):
\[
a = (20 - w) \times w
\]
This matches the equation given:
\[
a = -w^2 + 20w
\]
Next, let's establish the reasonable scale for the width \( w \). The width must be non-negative and bounded by the perimeter constraints:
- When \( w = 0 \), then \( l = 20 \) and thus \( a = 0 \).
- When \( w = 20 \), then \( l = 0 \) and thus \( a = 0 \).
For practical purposes, the width \( w \) will realistically range from \( 0 \) to \( 20 \) yards.
### Proposed Scale for the Domain
Considering the potential width values and the options provided (2, 5, 10, or 1), a reasonable scale would allow the entire range from \( 0 \) to \( 20 \) to be covered without skipping possible important values.
1. **Width Scale of 1**: This would yield values from 0 to 20 (i.e., 0, 1, 2, ..., 20) which gives us a lot of detail but could be cumbersome.
2. **Width Scale of 2**: This would yield values 0, 2, 4, ..., 20 which provides a good number of key points but not too many.
3. **Width Scale of 5**: This would yield values 0, 5, 10, 15, 20, which might miss some detail in the critical section of the parabola.
4. **Width Scale of 10**: This would yield values 0, 10, 20, which is too sparse to capture the nuances of the area variation effectively.
### Conclusion
Given these considerations, a **scale of 2** is a reasonable choice for the domain as it provides structured increments that allow us to clearly observe the changes in area with respect to width:
- With increments of 2 (0, 2, 4, ... , 20), we can effectively identify and capture the maximum area, where the area formula shows clear curvature due to the negative quadratic term.
Thus, if you were to set up a table, using a scale of **2** for the width would be the most reasonable choice.
There are no human answers yet. A form for humans to post answers is coming very soon!