To solve this problem, we need to maximize the area of the rectangular pasture with one side along the canal. Since no fencing is required along the canal, we only need to enclose three sides of the rectangle: two lengths (along the width) and one width (along the length).

Step-by-Step Solution:

1. Let the dimensions of the rectangle be:

   • Length $L$ (the side along the canal, no fencing required)
   • Width $W$ (the side perpendicular to the canal, requiring fencing)

2. The total fencing used is 2000 feet, which is the perimeter of three sides of the rectangle: $2W + L = 2000$ Solving for $L$: $L = 2000 - 2W$

3. The area $A$ of the rectangle is given by: $A = L \times W$ Substituting $L = 2000 - 2W$: $A = (2000 - 2W) \times W = 2000W - 2W^2$

4. To maximize the area, we need to find the value of $W$ that maximizes $A$. This requires taking the derivative of $A$ with respect to $W$, setting it equal to zero, and solving for $W$.

   The derivative of $A$ with respect to $W$ is: $\frac{dA}{dW} = 2000 - 4W$ Set this equal to zero to find the critical point: $2000 - 4W = 0$ Solving for $W$: $W = 500$

5. Now, substitute $W = 500$ into the equation for $L$: $L = 2000 - 2(500) = 1000$

6. The dimensions of the rectangle are:

   • Length $L = 1000$ feet
   • Width $W = 500$ feet

7. The maximum area is: $A = L \times W = 1000 \times 500 = 500,000 \text{ square feet}$

Conclusion:

The largest area that can be enclosed is 500,000 square feet, and the dimensions of the pasture are 1000 feet by 500 feet.

Would you like more details on this solution, or do you have any other questions? Here's a list of related questions you might find interesting:

1. What would the area be if the fence enclosed all four sides of the rectangle (no canal)?
2. How can we verify that this is the maximum area using the second derivative test?
3. What happens if we have more fencing material, say 3000 feet, instead of 2000 feet?
4. How can this problem be generalized to irregular shapes or other fencing arrangements?
5. How would this problem change if we also had a limited budget for fencing?

Tip: When solving optimization problems, setting the derivative of the area function equal to zero helps find the maximum or minimum points. Always check if the result corresponds to a maximum (using the second derivative test or other verification).