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Let's break down the problem step by step.

### Problem Summary:
- Skyler has 600 feet of fencing to build a rectangular enclosure, with one side along the barn (so no fencing is needed on that side).
- **Goal:** Maximize the area of the enclosure using the 600 feet of fencing.
- **Variables:**
  - \( w \): the width of the enclosure (perpendicular to the barn).
  - \( l \): the length of the enclosure (parallel to the barn).

### Part a) Write an equation for the area \( A(w) \) in terms of \( w \).

- The total fencing available is 600 feet, and it only needs to enclose three sides (two widths and one length). Thus, the fencing constraint is:
  \[
  2w + l = 600 \quad \text{(Equation 1: Fencing constraint)}
  \]
  
  We want to express the area \( A \) in terms of \( w \). The area of a rectangle is:
  \[
  A = w \cdot l
  \]
  From Equation 1, solve for \( l \) in terms of \( w \):
  \[
  l = 600 - 2w
  \]
  Now substitute this expression for \( l \) into the area formula:
  \[
  A(w) = w \cdot (600 - 2w) = 600w - 2w^2
  \]
  So the area function is:
  \[
  A(w) = 600w - 2w^2
  \]

### Part b) What width \( w \) would maximize the area?

To find the value of \( w \) that maximizes the area, take the derivative of \( A(w) \) with respect to \( w \) and set it equal to zero:
\[
A'(w) = 600 - 4w
\]
Setting \( A'(w) = 0 \):
\[
600 - 4w = 0 \quad \Rightarrow \quad 4w = 600 \quad \Rightarrow \quad w = 150
\]
So, the width that maximizes the area is **150 feet**.

### Part c) What is the maximum area?

Substitute \( w = 150 \) into the area function to find the maximum area:
\[
A(150) = 600(150) - 2(150)^2 = 90000 - 45000 = 45000 \, \text{square feet}
\]
Thus, the maximum area is **45,000 square feet**.

---

Would you like further details or have any questions on this? Here are five follow-up questions:

1. How would the problem change if fencing was required on all four sides?
2. What if Skyler had only 500 feet of fencing? How would the optimal dimensions change?
3. How does the derivative help us find the maximum area in this problem?
4. What would happen if we solve this using a graph of \( A(w) \)?
5. How could you confirm that \( w = 150 \) gives a maximum and not a minimum?

**Tip:** When optimizing an area or volume problem with constraints, always express the area or volume in terms of one variable and then use calculus to find the critical points!

Skyler wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn, so he needs no fence on that side. The other three sides will be enclosed with wire fencing. If Skyler has 600 feet of fencing, you can find the dimensions that maximize the area of the enclosure.

Learn how to optimize the area of a rectangular enclosure using 600 feet of fencing, applying quadratic functions and derivatives. This problem involves solving for the dimensions that maximize area when fencing is only required for three sides. Suitable for Grades 10-12.

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