This is a classic optimization problem where we want to maximize the area of a rectangle with a given amount of fencing. Since the plot borders a river, only three sides of the rectangle need to be fenced (two widths and one length).

Step-by-step solution:

1. Let the variables be:

   • $x$ = the width of the rectangle (perpendicular to the river, needs to be fenced on both sides),
   • $y$ = the length of the rectangle (parallel to the river, only needs to be fenced on one side).

2. Given:

   • Farmer Ed has 7,500 meters of fencing.
   • The total fencing is used for two widths and one length: $2x + y = 7500$ Solve this for $y$: $y = 7500 - 2x$

3. Area of the rectangle: The area $A$ of the rectangle is given by: $A = x \times y$ Substitute $y = 7500 - 2x$ into this equation: $A = x \times (7500 - 2x) = 7500x - 2x^2$

4. Maximizing the area: To find the value of $x$ that maximizes the area, take the derivative of $A$ with respect to $x$ and set it equal to zero: $\frac{dA}{dx} = 7500 - 4x$ Set the derivative equal to zero: $7500 - 4x = 0$ Solving for $x$: $x = 1875$

5. Find $y$: Now substitute $x = 1875$ back into the equation for $y$: $y = 7500 - 2(1875) = 3750$

6. Calculate the maximum area: The maximum area is: $A = 1875 \times 3750 = 7,031,250 \, \text{square meters}$

So, the largest area that can be enclosed is 7,031,250 square meters.

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Would you like further details or have any questions?

Here are some related questions:

1. What happens to the area if the length is also restricted by the river?
2. Can you derive a general formula for maximizing the area given any amount of fencing?
3. How would the solution change if all four sides needed to be fenced?
4. Can you find the critical points for other optimization problems in geometry?
5. What is the importance of derivative tests in finding maximum or minimum values?

Tip: Optimization problems often require setting up the relationship between variables and using calculus to find maximum or minimum values!