Please only use grade 7-11 methods. No high level math. I beg you. keep it insanely simple. I have to draw all this out and write it out

Let's solve the problem using simpler methods suitable for grades 7-11.

We have an enclosure made up of three equal sections using 160 meters of rope. We want to find the largest area that Alice can enclose and the dimensions of the enclosure.

Let's denote:
- \( l \) as the length of each section
- \( w \) as the width of each section

The total perimeter (rope) is given by the sum of all the lengths and widths. Since the enclosure has three sections, we have:
- Two lengths (one at the top and one at the bottom): \( 2l \)
- Four widths (two for the sides and two for the dividers between the sections): \( 4w \)

So, we can write the equation for the perimeter:
\[ 2l + 4w = 160 \]

### Step 1: Simplify the equation
First, we divide everything by 2 to simplify:
\[ l + 2w = 80 \]

### Step 2: Express \( l \) in terms of \( w \)
We solve for \( l \):
\[ l = 80 - 2w \]

### Step 3: Calculate the area
The total area \( A \) of the three sections is:
\[ A = 3lw \]
Substitute \( l \) from the above equation:
\[ A = 3w(80 - 2w) \]
\[ A = 240w - 6w^2 \]

### Step 4: Find the dimensions that give the maximum area
We need to find the values of \( w \) and \( l \) that give the largest area.

Let's try some values to keep it simple:

**Example:**
1. Let \( w = 10 \):
   \[ l = 80 - 2(10) = 60 \]
   \[ A = 3(10)(60) = 1800 \text{ square meters} \]

2. Let \( w = 20 \):
   \[ l = 80 - 2(20) = 40 \]
   \[ A = 3(20)(40) = 2400 \text{ square meters} \]

3. Let \( w = 30 \):
   \[ l = 80 - 2(30) = 20 \]
   \[ A = 3(30)(20) = 1800 \text{ square meters} \]

From these calculations, we see that the largest area is 2400 square meters when the width \( w = 20 \) meters and the length \( l = 40 \) meters.

### Summary:
- Largest area: 2400 square meters
- Dimensions: 20 meters (width) by 40 meters (length)

Would you like more details or have any questions?

Here are some additional questions you might find interesting:
1. What if there were four sections instead of three?
2. How does changing the total rope length affect the dimensions?
3. What happens if the sections are not all the same width?
4. Can we use a different shape to enclose a larger area?
5. What if we use the rope to make a circular enclosure?

**Tip:** When dealing with optimization problems like this, it's often helpful to test different values within the given constraints to find the best solution.

Explore how to maximize the area of a three-section enclosure using 160 meters of rope. Learn about perimeter and area calculations in a simple, step-by-step manner suitable for grades 7-11 students.

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