Farmer Brown has 1400 feet of chain-link fence to use to construct six animal cages, as shown in the figure below, (c) Find the dimensions that maximize the enclosed area.


To solve this problem, we need to find the dimensions of the six cages that maximize the enclosed area given that Farmer Brown has 1400 feet of fence.

### Step 1: Understand the setup
The setup involves six rectangular animal cages arranged in two rows, sharing some fence segments. We can assume the following:

- Let the width of each cage be \( w \) (all cages have the same width).
- Let the length of each cage be \( l \) (all cages have the same length).

We can break down the problem to understand how the fencing is distributed.

#### Fence requirements:
- There are three vertical segments of length \( l \) in each row.
- There are five horizontal segments of length \( w \) (three between the cages and two for the outer sides).

So, the total fencing used is the sum of the vertical and horizontal segments.

### Step 2: Total fencing
The total amount of fencing used includes:
- 4 horizontal lengths (since there are two rows and two horizontal boundaries),
- 3 internal vertical sections separating the cages in each row,
- 2 external vertical boundaries (one at each end of the rows).

Thus, the total length of fencing used is:
\[
6l + 5w = 1400
\]
This is the constraint equation.

### Step 3: Maximize the area
The total enclosed area \( A \) is the combined area of the six cages:
\[
A = 6lw
\]
We need to maximize \( A = 6lw \) subject to the constraint \( 6l + 5w = 1400 \).

### Step 4: Solve the constraint for \( l \)
Solving for \( l \) in terms of \( w \) from the constraint equation:
\[
l = \frac{1400 - 5w}{6}
\]

### Step 5: Substitute \( l \) into the area equation
Now substitute \( l \) into the area equation:
\[
A(w) = 6 \left( \frac{1400 - 5w}{6} \right) w
\]
Simplifying the expression:
\[
A(w) = (1400 - 5w) w
\]
\[
A(w) = 1400w - 5w^2
\]

### Step 6: Maximize the area
To maximize the area, take the derivative of \( A(w) \) with respect to \( w \) and set it equal to zero:
\[
\frac{dA}{dw} = 1400 - 10w = 0
\]
Solving for \( w \):
\[
10w = 1400
\]
\[
w = 140
\]

### Step 7: Find \( l \)
Now substitute \( w = 140 \) into the constraint equation to find \( l \):
\[
6l + 5(140) = 1400
\]
\[
6l + 700 = 1400
\]
\[
6l = 700
\]
\[
l = \frac{700}{6} \approx 116.67
\]

### Conclusion:
The dimensions that maximize the enclosed area are:
- \( w = 140 \) feet (width of each cage),
- \( l \approx 116.67 \) feet (length of each cage).

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

Here are 5 related questions to explore further:
1. How would the solution change if Farmer Brown only wanted to construct 4 cages?
2. What happens to the enclosed area if we use 1600 feet of fencing instead?
3. How would the problem change if each cage had a different width?
4. Can you derive a general formula for any number of cages using the same amount of fencing?
5. How would the result differ if the cages were circular instead of rectangular?

**Tip:** Always check the constraint before maximizing an area or volume to ensure the problem setup aligns with real-world conditions.

Farmer Brown has 1400 feet of chain-link fence to use to construct six animal cages. Find the dimensions that maximize the enclosed area.

Find out how to maximize the enclosed area for six animal cages using 1400 feet of fencing. This problem involves key optimization techniques, quadratic equations, and geometry. Suitable for high school students in Grades 10-12.

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