This problem involves optimizing the area of a rectangular plot of land with constraints on fencing costs. Let’s break it down systematically:

Problem Restatement:

1. Plot Description:

   • One side of the rectangle (along the barn) doesn't need fencing.
   • The west side fencing cost is shared with a neighbor.

2. Cost Details:

   • Fencing costs $20 per linear foot.
   • Maximum cost the farmer is willing to spend: $5000.

3. Objective:

   • Maximize the area enclosed by the plot.

4. Variables:

   • Let the length of the plot parallel to the barn be $L$ (in feet).
   • Let the width (the perpendicular sides) of the plot be $W$ (in feet).

Constraints:

1. Cost Constraint:

   • The total fencing required is for two widths ($2W$) and one length ($L$).
   • The west-side fence cost is shared, so only $W/2$ of that side is counted.
   • Total cost:
     $\text{Cost} = 20(L + W + W/2)$ Simplify: $\text{Cost} = 20L + 30W \leq 5000$

2. Area to Maximize:

   • The area $A$ is given by: $A = L \times W$

Solve:

1. From the cost constraint: $20L + 30W = 5000$ Simplify: $L = \frac{5000 - 30W}{20} = 250 - 1.5W$

2. Substitute $L$ into the area formula: $A = (250 - 1.5W) \times W$ Expand: $A = 250W - 1.5W^2$

3. This is a quadratic equation, and the maximum area occurs at the vertex of the parabola. The vertex formula for $A = ax^2 + bx + c$ is: $W = -\frac{b}{2a}$ Here, $a = -1.5$, $b = 250$: $W = -\frac{250}{2(-1.5)} = \frac{250}{3} \approx 83.33 \, \text{feet}.$

4. Find $L$ using $L = 250 - 1.5W$: $L = 250 - 1.5(83.33) = 250 - 125 = 125 \, \text{feet}.$

Final Answer:

The dimensions for the plot are approximately: $\boxed{125, 83.33}$

Would you like a more detailed explanation or further assistance?

Related Questions:

1. How do we derive the formula for the cost in such problems?
2. Can the same approach work if the cost structure changes (e.g., different rates for different sides)?
3. What is the significance of the vertex formula in optimization problems?
4. How does sharing the cost of one side affect the maximum area?
5. Can we graph the area function to visualize the optimization?

Tip:

When maximizing an area with constraints, always express the constraint in terms of one variable before substituting it into the area formula.