To solve this problem, let's analyze the information given in the image:

1. Problem Setup: The farmer wants to fence a rectangular area with one side using the barn wall (meaning only three sides of the rectangle need fencing). The total length of the fence is 64 feet.

2. Variables:

   • Let $W$ represent the width of the rectangle (parallel to the barn wall).
   • Let $L$ represent the length of the rectangle (perpendicular to the barn wall).

3. Equation for Fence: Since the barn wall forms one side, the fencing will only cover two widths and one length. Thus: $64 = 2W + L$

4. Maximizing Area:

   • The area $A$ of the rectangle can be expressed as: $A = W \times L$
   • Substitute $L$ from the fencing equation: $L = 64 - 2W$
   • Now, the area in terms of $W$ is: $A = W \times (64 - 2W) = 64W - 2W^2$
   • This equation $A = 64W - 2W^2$ is a quadratic function of $W$, and we can find the maximum area by completing the square or using the vertex formula for parabolas.

5. Finding the Optimal Width $W$:

   • The vertex of a quadratic function $ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$.
   • Here, $A = -2W^2 + 64W$, so $a = -2$ and $b = 64$.
   • Thus: $W = -\frac{64}{2 \times -2} = \frac{64}{4} = 16$
   • Substitute $W = 16$ back into the fencing equation to find $L$: $L = 64 - 2(16) = 64 - 32 = 32$

6. Solution:

   • The maximum area is achieved when $W = 16$ feet and $L = 32$ feet.
   • The maximum area is: $A = 16 \times 32 = 512 \, \text{square feet}$

Summary of Answers:

• Correct Equation: $64 = 2W + L$ (Option A).
• Maximum Area: 512 square feet.
• Dimensions:
  • Larger dimension: 32 feet
  • Smaller dimension: 16 feet

Would you like further details on any step or have additional questions?

Follow-up Questions:

1. What is the process to maximize the area for a given perimeter in general?
2. How does changing the fence length affect the dimensions of the maximum area?
3. What would happen if both lengths were to be fenced (not using the barn)?
4. How can we generalize this method to any perimeter length?
5. Why does a quadratic function represent area in this scenario?

Tip:

For optimization problems involving fencing, it’s useful to express one dimension in terms of the other using the perimeter constraint before setting up the area equation.