Farmer Brown has 1200 feet of chain-link fence to use to construct six animal cages, as shown in the figure below.

(a) Express the width y in terms of the length x.
y = 
  
Incorrect: Your answer is incorrect.


(b) Express the total enclosed area A of the cages in terms of x.
A = 
  
Incorrect: Your answer is incorrect.


(c) Find the dimensions that maximize the enclosed area.

Let's break this problem into parts and solve each one step by step.

### Part (a) Express the width \( y \) in terms of the length \( x \)

Farmer Brown is building **six** animal cages. Based on the problem, we assume the cages are arranged in a rectangular shape. Since there are six cages, let's analyze how the fencing is used.

The fencing must cover:
- **2 lengths** of the rectangular structure (each of length \( x \))
- **7 widths** (since there are six cages, there are seven vertical dividers, each of width \( y \))

Thus, the total amount of fencing used is the sum of the lengths and the widths. Given that Farmer Brown has 1200 feet of fencing, the equation for the total fencing is:

\[
2x + 7y = 1200
\]

To express \( y \) in terms of \( x \), solve for \( y \):

\[
7y = 1200 - 2x
\]

\[
y = \frac{1200 - 2x}{7}
\]

So, the width \( y \) in terms of the length \( x \) is:

\[
y = \frac{1200 - 2x}{7}
\]

---

### Part (b) Express the total enclosed area \( A \) in terms of \( x \)

The total enclosed area \( A \) of the cages is simply the area of the rectangular structure formed by the cages. The area of a rectangle is given by:

\[
A = \text{length} \times \text{width}
\]

Here, the total length is \( x \), and the total width is \( y \). From part (a), we know that:

\[
A = x \times y
\]

Substitute \( y = \frac{1200 - 2x}{7} \) into this equation:

\[
A = x \times \frac{1200 - 2x}{7}
\]

Simplifying:

\[
A = \frac{x(1200 - 2x)}{7}
\]

\[
A = \frac{1200x - 2x^2}{7}
\]

So, the total enclosed area \( A \) in terms of \( x \) is:

\[
A = \frac{1200x - 2x^2}{7}
\]

---

### Part (c) Find the dimensions that maximize the enclosed area

To maximize the area, we need to find the value of \( x \) that maximizes the function \( A(x) \). This involves finding the critical points of the function by taking the derivative of \( A(x) \), setting it equal to zero, and solving for \( x \).

The function we are maximizing is:

\[
A(x) = \frac{1200x - 2x^2}{7}
\]

First, take the derivative of \( A(x) \) with respect to \( x \):

\[
A'(x) = \frac{1}{7} \left( 1200 - 4x \right)
\]

Set the derivative equal to zero to find the critical points:

\[
\frac{1}{7} \left( 1200 - 4x \right) = 0
\]

Multiply both sides by 7:

\[
1200 - 4x = 0
\]

Solve for \( x \):

\[
4x = 1200
\]

\[
x = 300
\]

Now that we have \( x = 300 \), we can find the corresponding \( y \) using the equation for \( y \) from part (a):

\[
y = \frac{1200 - 2(300)}{7} = \frac{1200 - 600}{7} = \frac{600}{7} \approx 85.71
\]

Thus, the dimensions that maximize the enclosed area are approximately:
- Length \( x = 300 \) feet
- Width \( y \approx 85.71 \) feet

---

### Summary of Results:
- **(a)** \( y = \frac{1200 - 2x}{7} \)
- **(b)** \( A = \frac{1200x - 2x^2}{7} \)
- **(c)** The dimensions that maximize the enclosed area are:
  - Length: \( 300 \) feet
  - Width: \( 85.71 \) feet

Would you like more details or have any questions?

Here are five related questions:
1. How do you compute critical points for a function?
2. What role do derivatives play in optimization problems?
3. How can you check if a critical point is a maximum or minimum?
4. What is the importance of constraint equations in optimization problems?
5. How would the solution change if there were more or fewer cages?

**Tip:** When solving optimization problems, always check the boundaries or constraints, as they may limit possible solutions.

Farmer Brown has 1200 feet of chain-link fence to use to construct six animal cages, as shown in the figure below.

(a) Express the width y in terms of the length x.

(b) Express the total enclosed area A of the cages in terms of x.

(c) Find the dimensions that maximize the enclosed area.

Solve Farmer Brown's fencing problem, where 1200 feet of chain-link fence is used to build six animal cages. Learn how to express width in terms of length, calculate the total enclosed area, and determine the dimensions that maximize the area. This problem involves algebra and optimization with derivatives, suitable for Grades 10-12.

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