Let's break down the two parts of the problem based on the image.

Part (a)

You have 1100 meters of fencing, and one side of the rectangular pen is against the barn. To maximize the area of the pen, we need to figure out the dimensions. Let:

• $x$ be the length of the sides perpendicular to the barn,
• $y$ be the length of the side parallel to the barn.

The total amount of fencing used is for three sides (since one side is against the barn): $2x + y = 1100$ The area of the rectangle is: $A = x \cdot y$ We want to maximize $A$. First, solve for $y$ from the fencing constraint: $y = 1100 - 2x$ Substitute this into the area formula: $A = x(1100 - 2x) = 1100x - 2x^2$ This is a quadratic equation. To find the value of $x$ that maximizes the area, take the derivative of $A$ with respect to $x$ and set it equal to zero: $\frac{dA}{dx} = 1100 - 4x = 0$ $x = 275$ Substitute $x = 275$ into the equation for $y$: $y = 1100 - 2(275) = 550$ Thus, to maximize the area of the pen, the dimensions are:

• $x = 275 \, \text{m}$ (sides perpendicular to the barn),
• $y = 550 \, \text{m}$ (side parallel to the barn).

Part (b)

The rancher wants to make four identical adjacent rectangular pens, each with an area of 400 square meters. The setup shown in the diagram indicates that the fencing will share some sides between the pens.

Let:

• $x$ be the length of each pen perpendicular to the barn,
• $y$ be the total length of the side parallel to the barn (this will include the dividers between the pens).

Each pen has an area of 400 square meters, so: $x \cdot y = 400$ The fencing needs to enclose the four pens, meaning that the total length of fencing used is for the five vertical segments (since four pens have five vertical sides) and one long horizontal side: $5x + y$ We want to minimize the total amount of fencing, so first solve for $y$ from the area constraint: $y = \frac{400}{x}$ Substitute this into the fencing formula: $f = 5x + \frac{400}{x}$ Now, to minimize $f$, take the derivative of $f$ with respect to $x$ and set it equal to zero: $\frac{df}{dx} = 5 - \frac{400}{x^2} = 0$ $5 = \frac{400}{x^2}$ $x^2 = \frac{400}{5} = 80$ $x = \sqrt{80} = 4\sqrt{5} \, \text{m}$ Substitute $x = 4\sqrt{5}$ into the equation for $y$: $y = \frac{400}{4\sqrt{5}} = 20\sqrt{5} \, \text{m}$

Thus, to minimize the amount of fencing, the dimensions of each pen are:

• $x = 4\sqrt{5} \, \text{m}$ (sides perpendicular to the barn),
• $y = 20\sqrt{5} \, \text{m}$ (side parallel to the barn).

Would you like any further clarifications or details? Here are some follow-up questions:

1. Why does solving for the derivative help find the maximum or minimum values in optimization problems?
2. How would the solution change if more than one side were against the barn?
3. Could you generalize this method for any number of adjacent pens?
4. How would the fencing be affected if the barn was not perfectly straight?
5. What happens if you add more pens with different areas?

Tip: When solving optimization problems, always remember to confirm if you're maximizing or minimizing, as the methods are similar but lead to different goals.