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To solve this problem, let's break down each part systematically. We are given:

- A rectangular field bordering a river, with no fence on the river side.
- The total fencing available is 1800 m.
- Let \( x \) represent the width of the field (the side perpendicular to the river).

Since there is no fencing along the river, the fencing will only be used for the two widths (\( 2x \)) and one length (\( y \)) of the rectangle.

### (a) Expression for the length \( y \) in terms of \( x \)
The perimeter constraint is:
\[
2x + y = 1800
\]
Solving for \( y \):
\[
y = 1800 - 2x
\]

### (b) Area of the field as a function of \( x \)
The area \( A \) of a rectangle is given by:
\[
A = \text{length} \times \text{width}
\]
Substitute \( y = 1800 - 2x \):
\[
A(x) = x \cdot (1800 - 2x)
\]
Expanding this:
\[
A(x) = 1800x - 2x^2
\]

### (c) Value of \( x \) that maximizes the area
To find the maximum area, we take the derivative of \( A(x) \) with respect to \( x \) and set it equal to zero.

1. **Differentiate** \( A(x) = 1800x - 2x^2 \):
   \[
   A'(x) = 1800 - 4x
   \]
2. **Set the derivative equal to zero**:
   \[
   1800 - 4x = 0
   \]
3. **Solve for \( x \)**:
   \[
   4x = 1800
   \]
   \[
   x = 450
   \]

Thus, \( x = 450 \) m leads to the maximum area.

### (d) Maximum area
Now, substitute \( x = 450 \) back into the area function to find the maximum area:
\[
A(450) = 1800 \times 450 - 2 \times (450)^2
\]
Calculating each term:
\[
A(450) = 810000 - 2 \times 202500
\]
\[
A(450) = 810000 - 405000
\]
\[
A(450) = 405000
\]

Therefore, the maximum area is **405,000 square meters**.

---

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### Related Questions
1. How would the solution change if there was fencing along all four sides?
2. What would the maximum area be if only 1000 m of fencing were available?
3. How does changing \( x \) affect the area in this problem?
4. How could we confirm that \( x = 450 \) is a maximum using the second derivative?
5. What other shapes could maximize the area with the same perimeter constraints?

### Tip
When maximizing or minimizing areas with constraints, translating all variables into a single variable can simplify differentiation and finding extremal values.

A campground owner has 1800 m of fencing. He wants to enclose a rectangular field bordering a river, with no fencing along the river. (a) Write an expression for the length of the field as a function of x. (b) Find the area of the field (area = length × width) as a function of x. (c) Find the value of x leading to the maximum area. (d) Find the maximum area.

Explore how to maximize the area of a rectangular field bordering a river with 1800 m of fencing. This problem involves setting up equations for the perimeter and area, using derivatives to find the optimal width, and calculating the maximum area. Ideal for students in Grades 11-12.

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