A farmer wants to fence in a rectangular plot of land adjacent to the north wall of his barn. No fencing is needed along the barn, and the fencing along the west side of the plot is shared with a neighbor who will split the cost of that portion of the fence. If the fencing costs $12 per linear foot to install and the farmer is not willing to spend more than $3000, find the dimensions for the plot that would enclose the most area. (Enter the dimensions as a comma separated list.)
  
Incorrect: Your answer is incorrect.

To solve this problem, let's define the variables and set up the equations.

### Variables:
- Let $ x $ be the length of the plot along the north wall of the barn (the side parallel to the barn).
- Let $ y $ be the width of the plot (the side perpendicular to the barn).

### Problem Setup:
- The fencing cost is $12 per linear foot.
- The total cost of the fencing must not exceed $3000.
- Fencing is required on three sides: the west side $ y $, the east side $ y $, and the north side $ x $.
- The cost for the west side $ y $ is shared with the neighbor, so the farmer only pays half of it.

### Cost Equation:
The cost of the fencing can be expressed as:
\[
\text{Total Cost} = 12 \times \left( \frac{y}{2} + y + x \right)
\]
Simplifying:
\[
\text{Total Cost} = 12 \times \left( \frac{3y}{2} + x \right)
\]
\[
\text{Total Cost} = 18y + 12x
\]
Given that the total cost is limited to $3000:
\[
18y + 12x \leq 3000
\]
Dividing the entire inequality by 6:
\[
3y + 2x \leq 500
\]
This is the constraint equation.

### Area to Maximize:
The area $ A $ of the plot is:
\[
A = x \times y
\]
We want to maximize this area subject to the constraint $ 3y + 2x \leq 500 $.

### Solving the Optimization Problem:
We can express $ x $ in terms of $ y $ using the constraint:
\[
2x \leq 500 - 3y
\]
\[
x \leq \frac{500 - 3y}{2}
\]
Thus:
\[
A = x \times y = y \times \frac{500 - 3y}{2} = \frac{500y - 3y^2}{2}
\]
To find the maximum area, we take the derivative of $ A(y) $ with respect to $ y $ and set it to zero:
\[
\frac{dA}{dy} = \frac{500}{2} - \frac{3y}{2} = 250 - 3y
\]
Setting the derivative equal to zero:
\[
250 - 3y = 0
\]
\[
3y = 250
\]
\[
y = \frac{250}{3} \approx 83.33 \text{ feet}
\]
Substituting $ y = 83.33 $ into the constraint equation to find $ x $:
\[
3(83.33) + 2x \leq 500
\]
\[
250 + 2x \leq 500
\]
\[
2x \leq 250
\]
\[
x \leq 125 \text{ feet}
\]
Thus, the dimensions that would enclose the most area, given the constraints, are approximately:
\[
\boxed{125 \text{ feet}, 83.33 \text{ feet}}
\]

Let me know if you need more details or have any questions.

1. How do we find the area of a rectangle in general?
2. What is the significance of maximizing the area in this problem?
3. How would the problem change if the neighbor did not share the cost of the west side?
4. How do you find the critical points in a function when maximizing or minimizing?
5. Why is it important to consider constraints in optimization problems?

**Tip:** When solving optimization problems with constraints, always check if the solution satisfies the constraint before concluding.

Learn how to maximize the area of a rectangular plot of land adjacent to a barn, with fencing constraints. Solve using linear inequalities and optimization techniques. Suitable for high school students.

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