The problem states:

• A farmer wants to fence a rectangular plot of land adjacent to the north wall of his barn.
• Fencing is not required along the barn side, and the west side fencing cost is shared with a neighbor.
• The fencing cost is $20 per linear foot.
• The farmer has a budget of $8000.

Let's denote:

• $x$ as the width of the plot (perpendicular to the barn wall),
• $y$ as the length of the plot (parallel to the barn wall).

Step 1: Set up the cost equation

The cost for fencing includes only three sides (two widths and one length) because one side is adjacent to the barn and doesn’t need fencing. Therefore, the fencing cost equation is: $20(2x + y) = 8000$ Solving for $y$ in terms of $x$: $2x + y = 400$ $y = 400 - 2x$

Step 2: Set up the area equation

The area $A$ of the rectangular plot is given by: $A = x \cdot y$ Substitute $y = 400 - 2x$: $A = x(400 - 2x) = 400x - 2x^2$

Step 3: Maximize the area

To find the maximum area, take the derivative of $A$ with respect to $x$ and set it to zero: $\frac{dA}{dx} = 400 - 4x$ Setting $\frac{dA}{dx} = 0$: $400 - 4x = 0$ $x = 100$

Step 4: Find $y$

Substitute $x = 100$ into $y = 400 - 2x$: $y = 400 - 2(100) = 200$

Solution

The dimensions for the plot that maximize the area, within the budget constraint, are: $\text{(width, length)} = (100, 200)$

Would you like further details on any part of this solution, or have any additional questions?

Here are some related questions to explore:

1. How would the dimensions change if the budget were increased to $10,000?
2. What if the fencing cost per foot changed to $25? How would that affect the dimensions?
3. How can we verify that the dimensions found indeed maximize the area?
4. How would the problem change if the farmer needed fencing on all four sides?
5. Could the same method be applied to find maximum area with a triangular plot?

Tip: Optimization problems often require setting up equations based on the given constraints, then using calculus to find maximum or minimum values.