Question
A rectangular lot is 115 meters long and 80 meters wide. Give the length and width of another rectangular lot that has the same perimeter but a larger area
Answers
GPT-4o mini
To find a rectangular lot that has the same perimeter as the given lot but a larger area, we start by calculating the perimeter of the original lot.
The formula for the perimeter \( P \) of a rectangle is:
\[
P = 2(\text{length} + \text{width})
\]
For the given lot:
- Length = 115 meters
- Width = 80 meters
Calculating the perimeter:
\[
P = 2(115 + 80) = 2(195) = 390 \text{ meters}
\]
Next, we can use the perimeter to set up an equation for a new rectangle that has the same perimeter but different dimensions. Let the new rectangle have dimensions \( L \) (length) and \( W \) (width). The equation for the perimeter is:
\[
2(L + W) = 390
\]
This simplifies to:
\[
L + W = 195
\]
From here, we can express \( W \) in terms of \( L \):
\[
W = 195 - L
\]
The area \( A \) of a rectangle is given by:
\[
A = L \times W = L \times (195 - L) = 195L - L^2
\]
This is a quadratic function in terms of \( L \) and it opens downwards (as the coefficient of \( L^2 \) is negative), which means it has a maximum area at its vertex. The vertex \( L \) of the quadratic \( A = -L^2 + 195L \) can be found using the vertex formula:
\[
L = -\frac{b}{2a} = -\frac{195}{2 \times -1} = \frac{195}{2} = 97.5
\]
If \( L = 97.5 \), then:
\[
W = 195 - 97.5 = 97.5
\]
Thus, one solution would be a square:
- Length = 97.5 meters
- Width = 97.5 meters
Now we can check:
1. The perimeter:
\[
P = 2(97.5 + 97.5) = 2(195) = 390 \text{ meters} \quad \text{(same perimeter)}
\]
2. The area of the original lot:
\[
A_{\text{original}} = 115 \times 80 = 9200 \text{ square meters}
\]
3. The area of the new lot:
\[
A_{\text{new}} = 97.5 \times 97.5 = 9506.25 \text{ square meters} \quad \text{(larger area)}
\]
Thus, the length and width of the new rectangular lot, with the same perimeter and a larger area, are both **97.5 meters**.
The formula for the perimeter \( P \) of a rectangle is:
\[
P = 2(\text{length} + \text{width})
\]
For the given lot:
- Length = 115 meters
- Width = 80 meters
Calculating the perimeter:
\[
P = 2(115 + 80) = 2(195) = 390 \text{ meters}
\]
Next, we can use the perimeter to set up an equation for a new rectangle that has the same perimeter but different dimensions. Let the new rectangle have dimensions \( L \) (length) and \( W \) (width). The equation for the perimeter is:
\[
2(L + W) = 390
\]
This simplifies to:
\[
L + W = 195
\]
From here, we can express \( W \) in terms of \( L \):
\[
W = 195 - L
\]
The area \( A \) of a rectangle is given by:
\[
A = L \times W = L \times (195 - L) = 195L - L^2
\]
This is a quadratic function in terms of \( L \) and it opens downwards (as the coefficient of \( L^2 \) is negative), which means it has a maximum area at its vertex. The vertex \( L \) of the quadratic \( A = -L^2 + 195L \) can be found using the vertex formula:
\[
L = -\frac{b}{2a} = -\frac{195}{2 \times -1} = \frac{195}{2} = 97.5
\]
If \( L = 97.5 \), then:
\[
W = 195 - 97.5 = 97.5
\]
Thus, one solution would be a square:
- Length = 97.5 meters
- Width = 97.5 meters
Now we can check:
1. The perimeter:
\[
P = 2(97.5 + 97.5) = 2(195) = 390 \text{ meters} \quad \text{(same perimeter)}
\]
2. The area of the original lot:
\[
A_{\text{original}} = 115 \times 80 = 9200 \text{ square meters}
\]
3. The area of the new lot:
\[
A_{\text{new}} = 97.5 \times 97.5 = 9506.25 \text{ square meters} \quad \text{(larger area)}
\]
Thus, the length and width of the new rectangular lot, with the same perimeter and a larger area, are both **97.5 meters**.