The problem in the image describes a situation where a farmer has 7,000 meters of fencing and wants to enclose a rectangular plot of land that borders a river. The farmer doesn't fence the side along the river, meaning only three sides are fenced.

Let’s define the variables based on the problem:

• Let $x$ be the length of the two fenced sides perpendicular to the river.
• Let $y$ be the length of the side parallel to the river, which requires no fencing.

Step 1: Express the fencing constraint

The total amount of fencing is limited to 7,000 meters, and the farmer uses it for three sides of the rectangle (two $x$-sides and one $y$-side):

$2x + y = 7,000$

Solving for $y$:

$y = 7,000 - 2x$

Step 2: Express the area

The area $A$ of the rectangular plot is given by:

$A = x \cdot y$

Substitute the expression for $y$ from Step 1:

$A(x) = x \cdot (7,000 - 2x)$

Simplifying:

$A(x) = 7,000x - 2x^2$

Step 3: Maximize the area

To find the value of $x$ that maximizes the area, we take the derivative of $A(x)$ with respect to $x$ and set it equal to zero:

$\frac{dA}{dx} = 7,000 - 4x$

Set the derivative equal to zero:

$7,000 - 4x = 0$

Solving for $x$:

$x = 1,750 \, \text{meters}$

Step 4: Calculate $y$

Now substitute $x = 1,750$ into the equation for $y$:

$y = 7,000 - 2(1,750) = 3,500 \, \text{meters}$

Step 5: Calculate the maximum area

The maximum area $A$ is:

$A = 1,750 \times 3,500 = 6,125,000 \, \text{square meters}$

Conclusion:

The largest area that can be enclosed is 6,125,000 square meters.

Would you like more details or have any questions about this solution?

Additional Questions:

1. What would happen if the length of the fencing was doubled?
2. How would the problem change if the farmer also fenced the side along the river?
3. Can this problem be solved using a different method, such as Lagrange multipliers?
4. What happens to the area if the fenced side length $x$ is reduced by half?
5. What other shapes could provide a larger area with the same perimeter?

Tip:

When optimizing problems like this, always express everything in terms of one variable before using calculus to find maximum or minimum values.