Let:

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

Step 1: Formulate the cost constraint

Since no fencing is needed along the barn wall, only three sides need fencing:

• The west side (shared with the neighbor) with length $y$,
• The east side with length $y$,
• The south side with length $x$.

The cost for these sides is: $14 \times (x + 2y) \leq 7000$

Dividing both sides by 14 gives: $x + 2y \leq 500$

Step 2: Express area as a function of one variable

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

Using the cost constraint $x + 2y = 500$, solve for $x$ in terms of $y$: $x = 500 - 2y$

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

Step 3: Find the maximum area

To maximize the area, differentiate $A$ with respect to $y$ and set the derivative to zero: $\frac{dA}{dy} = 500 - 4y = 0$

Solving for $y$: $4y = 500 \Rightarrow y = 125$

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

Conclusion

The dimensions of the plot that maximize the area are: $\text{Width (} x \text{)} = 250 \text{ feet, Length (} y \text{)} = 125 \text{ feet}$

Thus, the dimensions are: 250, 125

Would you like more details or have any questions?

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Related Questions

1. How would the solution change if the cost per foot of fencing increased?
2. What if the farmer wanted fencing along all four sides?
3. How does sharing the cost impact the total fencing required for maximum area?
4. How could we generalize this approach to different shapes of land plots?
5. What would the optimal dimensions be if the farmer could spend only $5000?

Tip

When maximizing areas with budget constraints, isolating one variable in terms of the other helps reduce complexity and allows using calculus more effectively.