Let's break this down and solve the parts of the question step by step.

Part 1a: Maximize the area of a single pen

You are given 400 meters of fencing to create a rectangular pen against a barn, meaning only three sides need fencing (the fourth side is the barn wall). Let:

• $x$ be the width of the pen (parallel to the barn),
• $y$ be the length of the pen (perpendicular to the barn).

Since the fencing is only required for three sides, the total amount of fencing is given by: $2y + x = 400$ Solving for $x$: $x = 400 - 2y$

The area $A$ of the rectangular pen is: $A = x \times y = (400 - 2y) \times y = 400y - 2y^2$

Now, to maximize the area, we take the derivative of $A$ with respect to $y$ and set it equal to 0: $\frac{dA}{dy} = 400 - 4y = 0$ Solving for $y$: $y = 100$

Substitute $y = 100$ back into the equation for $x$: $x = 400 - 2(100) = 200$

Thus, the dimensions of the pen that maximize the area are:

• Width $x = 200$ meters
• Length $y = 100$ meters

The maximum area is: $A = 200 \times 100 = 20,000 \text{ square meters}$

Part 1b: Minimize fencing for four adjacent pens

In this part, the rancher wants to create four adjacent pens, all sharing the side against the barn. Each pen has an area of 400 m².

Let:

• $x$ be the width of each pen (parallel to the barn),
• $y$ be the length of each pen (perpendicular to the barn).

Each pen has an area of 400 m², so: $x \times y = 400$ or equivalently: $y = \frac{400}{x}$

Since the pens are adjacent, we need fencing for:

• One long side of all four pens together (length $x \times 4 = 4x$),
• Two sides along the length of the pens (each of length $y$),
• Three internal divisions between the pens (each of length $y$).

Thus, the total length of fencing $L$ required is: $L = 4x + 5y$

Substitute $y = \frac{400}{x}$ into this equation: $L = 4x + 5\left(\frac{400}{x}\right) = 4x + \frac{2000}{x}$

To minimize $L$, we take the derivative of $L$ with respect to $x$ and set it equal to 0: $\frac{dL}{dx} = 4 - \frac{2000}{x^2} = 0$ Solving for $x$: $4x^2 = 2000 \quad \Rightarrow \quad x^2 = 500 \quad \Rightarrow \quad x = \sqrt{500} \approx 22.36$

Now, substitute $x \approx 22.36$ back into the equation for $y$: $y = \frac{400}{22.36} \approx 17.88$

Thus, the dimensions of each pen that minimize the amount of fence used are approximately:

• Width $x \approx 22.36$ meters
• Length $y \approx 17.88$ meters.

Do you have any further questions or would you like additional details on any part of the solution? Here are some related questions:

1. What happens to the area of the pen if more or less fencing is available?
2. How does the number of adjacent pens affect the fencing needed?
3. What are the implications of the barn being wider or narrower in terms of optimal fence layout?
4. Can this method be extended to non-rectangular pens for different animals?
5. How would the solution change if the pens were placed in a different arrangement?

Tip: When optimizing areas or lengths in geometry problems, always set up the equation for the variable you need to minimize or maximize, and take its derivative to find the optimal point.