Let $x$ be the length of the plot parallel to the barn (which does not need fencing), and let $y$ be the width of the plot perpendicular to the barn. The objective is to maximize the area of the rectangular plot, given the constraints.

Step 1: Set up the constraints

Since the barn does not need fencing, we only need fencing along:

• The west side (which is $y$),
• The east side (which is $y$), and
• The south side (which is $x$).

Therefore, the total length of fencing needed is: $2y + x$

The fencing cost is $10 per linear foot, and the farmer does not want to spend more than $5000. So, the cost constraint is: $10(2y + x) \leq 5000$

Dividing by 10 to simplify: $2y + x \leq 500$

Thus, the fencing constraint is: $x = 500 - 2y$

Step 2: Set up the area function

The area $A$ of the rectangular plot is given by: $A = x \cdot y$

Substitute $x = 500 - 2y$ into the area equation: $A = (500 - 2y) \cdot y$ $A = 500y - 2y^2$

Step 3: Maximize the area function

The area $A = 500y - 2y^2$ is a quadratic function of $y$ and opens downward (since the $y^2$ term has a negative coefficient). To find the maximum area, we can take the derivative with respect to $y$, set it to zero, and solve for $y$.

Derivative of $A$ with respect to $y$:

$\frac{dA}{dy} = 500 - 4y$

Set the derivative equal to zero to find the critical point: $500 - 4y = 0$ $4y = 500$ $y = 125$

Now, substitute $y = 125$ back into the equation $x = 500 - 2y$ to find $x$: $x = 500 - 2(125)$ $x = 500 - 250$ $x = 250$

Final Answer

The dimensions of the plot that enclose the maximum area, within the given cost constraint, are: $x = 250, \quad y = 125$

Answer: The dimensions are 250, 125.

Would you like further clarification on any step?

---

1. How does the cost constraint affect the maximum area achievable?
2. What would the dimensions be if the cost per foot of fencing were different?
3. How can we verify that this is the maximum area using a second derivative test?
4. What if both the north and south sides required fencing?
5. How would the solution change if the fencing along the shared side was not split in cost?

Tip: When solving optimization problems, setting up the objective function and constraints clearly can simplify finding the maximum or minimum solutions.