Let me analyze the image and explain the solution to the problem.

Problem Analysis:

Part a:

We have a rectangular pen with one side against a barn, and the fencing is used for the other three sides. The total fencing available is 500 m, and the goal is to maximize the area of the pen.

• Variables:

  • Let the side parallel to the barn be $x$.
  • Let the two sides perpendicular to the barn be $y$.

• Constraint: The total fencing used for the three sides is: $x + 2y = 500$

• Area: The area of the rectangle is: $A = x \cdot y$

• Solution Approach: Solve for $y$ in terms of $x$ using the fencing constraint: $y = \frac{500 - x}{2}$

  Substitute $y$ into the area formula: $A(x) = x \cdot \frac{500 - x}{2}$

  Simplify: $A(x) = \frac{500x - x^2}{2}$

  To maximize $A(x)$, take the derivative and find the critical points: $\frac{dA}{dx} = \frac{500 - 2x}{2}$

  Set $\frac{dA}{dx} = 0$: $500 - 2x = 0 \quad \Rightarrow \quad x = 250$

  Corresponding $y$: $y = \frac{500 - 250}{2} = 125$

  Optimal Dimensions:

  • Parallel side: $x = 250 \, \text{m}$
  • Perpendicular sides: $y = 125 \, \text{m}$

Part b:

The rancher plans to build four identical and adjacent rectangular pens against the barn, each with an area of $400 \, \text{m}^2$. The goal is to minimize the amount of fencing required.

• Variables:

  • Let the side perpendicular to the barn be $w$.
  • Let the side parallel to the barn (total length for all pens) be $l$.

• Constraints: Each pen has an area of $400 \, \text{m}^2$: $w \cdot \frac{l}{4} = 400 \quad \Rightarrow \quad l = \frac{1600}{w}$

  Total fencing required: $F = 5w + l$

  Substitute $l$ into $F$: $F(w) = 5w + \frac{1600}{w}$

• Solution Approach: Minimize $F(w)$ by taking the derivative: $\frac{dF}{dw} = 5 - \frac{1600}{w^2}$

  Set $\frac{dF}{dw} = 0$: $5 - \frac{1600}{w^2} = 0 \quad \Rightarrow \quad w^2 = 320 \quad \Rightarrow \quad w = 4\sqrt{5}$

  Corresponding $l$: $l = \frac{1600}{4\sqrt{5}} = 80\sqrt{5}$

  Optimal Dimensions:

  • Perpendicular sides: $w = 4\sqrt{5} \, \text{m}$
  • Parallel side: $l = 80\sqrt{5} \, \text{m}$

Final Answer:

• Part a: The dimensions are $125 \, \text{m}$ and $250 \, \text{m}$.
• Part b: The dimensions are $4\sqrt{5} \, \text{m}$ and $80\sqrt{5} \, \text{m}$.

Would you like me to explain any step in more detail or compute further? Here are related questions for further exploration:

1. Why does the derivative give the maximum or minimum value for a function?
2. How is fencing optimization used in real-world applications?
3. What are other methods to solve optimization problems without calculus?
4. Can the solutions be verified graphically using a plot?
5. How would the solution change if there were no barn wall?

Tip: In optimization problems, always double-check the constraints before solving for the critical points.