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The problem in the image asks for the dimensions of another rectangular park with the same perimeter but a larger area compared to the current rectangular park.

### Current Rectangular Park:
- **Width** = 55 yards
- **Length** = 85 yards

#### Step 1: Find the perimeter of the current park.
The perimeter \( P \) of a rectangle is given by:
\[
P = 2 \times (\text{width} + \text{length})
\]
For the given dimensions:
\[
P = 2 \times (55 + 85) = 2 \times 140 = 280 \text{ yards}
\]

#### Step 2: Express the area of the park.
The area \( A \) of a rectangle is given by:
\[
A = \text{width} \times \text{length}
\]
For the given dimensions:
\[
A = 55 \times 85 = 4675 \text{ square yards}
\]

#### Step 3: Find the dimensions of the new park with the same perimeter but larger area.
We need a new set of dimensions that gives a **larger area** while maintaining the same **perimeter of 280 yards**.

Let the new dimensions be:
- Width = \( w \)
- Length = \( l \)

The perimeter constraint remains:
\[
P = 2 \times (w + l) = 280
\]
This simplifies to:
\[
w + l = 140
\]
Thus, \( l = 140 - w \).

#### Step 4: Maximize the area.
The area \( A \) of the new park is:
\[
A = w \times l = w \times (140 - w) = 140w - w^2
\]
This is a quadratic equation, and to maximize the area, we find the vertex of the parabola by using the formula \( w = -\frac{b}{2a} \), where \( A = aw^2 + bw + c \).

Here, \( a = -1 \) and \( b = 140 \), so:
\[
w = \frac{-140}{2 \times (-1)} = 70
\]
Thus, the width \( w = 70 \) yards. The length is:
\[
l = 140 - 70 = 70 \text{ yards}
\]

This results in a **square** with both dimensions equal to 70 yards.

#### Step 5: Calculate the area of the new park.
The area of the square park is:
\[
A = 70 \times 70 = 4900 \text{ square yards}
\]
This area (4900 square yards) is larger than the original area (4675 square yards), and the perimeter is still 280 yards.

### Final Answer:
- **Width** = 70 yards
- **Length** = 70 yards

Would you like more details or have any questions about this?

### Related Questions:
1. How is the perimeter of a rectangle calculated in general?
2. What would happen to the area if we reduce the width but increase the length while keeping the perimeter the same?
3. What other dimensions could give a larger area besides a square?
4. Can we determine if a square always gives the maximum area for a given perimeter?
5. How can we derive the general formula for maximizing the area of a rectangle given its perimeter?

### Tip:
When dealing with optimization problems, converting the situation into an equation and then using derivatives (or the vertex of a quadratic equation) can help find maximum or minimum values efficiently.

A rectangular park is 55 yards wide and 85 yards long. Give the length and width of another rectangular park that has the same perimeter but a larger area.

Learn how to find the dimensions of a rectangular park with the same perimeter as a given one but a larger area. The original park has a width of 55 yards and a length of 85 yards. This problem involves geometry and quadratic equations to optimize the area.

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