To maximize the area of a rectangular field with one side along a river (where no fencing is needed on that side), we need to determine the optimal dimensions given a limited amount of fencing material.

Let:

• $x$ represent the width of the field perpendicular to the river.
• $y$ represent the length of the field parallel to the river.

Since fencing is required for only three sides (two widths and one length), the fencing constraint can be set up as follows: $2x + y = 600$

Our goal is to maximize the area $A$ of the field, which can be represented as: $A = x \cdot y$

Step 1: Express $y$ in terms of $x$ using the fencing constraint.

From $2x + y = 600$, solve for $y$: $y = 600 - 2x$

Step 2: Substitute $y$ into the area formula.

Substitute $y = 600 - 2x$ into $A = x \cdot y$: $A = x \cdot (600 - 2x)$ $A = 600x - 2x^2$

Step 3: Maximize the area function.

To find the maximum area, take the derivative of $A$ with respect to $x$ and set it to zero: $\frac{dA}{dx} = 600 - 4x = 0$ Solving for $x$: $4x = 600$ $x = 150$

Step 4: Find $y$ using $x = 150$.

Substitute $x = 150$ back into the equation $y = 600 - 2x$: $y = 600 - 2(150) = 600 - 300 = 300$

Conclusion

To maximize the area, the farmer should use the dimensions:

• Width $x = 150$ meters
• Length $y = 300$ meters

Maximum Area

The maximum area achievable with these dimensions is: $A = x \cdot y = 150 \cdot 300 = 45000 \, \text{square meters}$

Would you like further details or have any questions?

Here are five related questions to explore further:

1. How would the solution change if fencing was needed along all four sides?
2. What if the available fencing was different? How would that affect the maximum area?
3. How would the dimensions change if the field was triangular?
4. What is the effect of doubling the available fencing material?
5. How would the solution be altered if the field had to be a square?

Tip: Remember that in optimization problems, using constraints to express variables in terms of one another simplifies finding maximum or minimum values.