The problem involves a farmer with 4500 meters of fencing who wants to enclose a rectangular plot next to a river. The side along the river does not require fencing, so only three sides of the rectangle (two widths and one length) need to be enclosed.

Let's define the variables:

• Let x be the width of the rectangle (the two sides perpendicular to the river).
• The remaining fencing, which forms the length of the rectangle, is given by $4500 - 2x$.

The area A of the rectangle is: $A = \text{Length} \times \text{Width} = x \times (4500 - 2x)$ $A = 4500x - 2x^2$

This is a quadratic equation in the form of $A = -2x^2 + 4500x$. To find the value of x that maximizes the area, we need to find the vertex of this parabola. The x-coordinate of the vertex for a quadratic equation $ax^2 + bx + c$ is given by: $x = \frac{-b}{2a}$ For the equation $A = -2x^2 + 4500x$:

• $a = -2$
• $b = 4500$

Substituting these values: $x = \frac{-4500}{2(-2)} = \frac{4500}{4} = 1125$

So, the width of the rectangle that maximizes the area is 1125 meters.

To find the corresponding maximum area: $A = 4500(1125) - 2(1125)^2$ $A = 5062500 - 2(1265625)$ $A = 5062500 - 2531250 = 2531250 \, \text{square meters}$

Thus, the largest area that can be enclosed is 2,531,250 square meters.

Let me know if you need more details or have any questions!

Here are some follow-up questions:

1. How do you derive the formula for the area of a rectangle given constraints?
2. Why does the vertex formula help in finding the maximum area?
3. How would the solution change if all four sides of the rectangle required fencing?
4. Can the quadratic equation technique be applied to other shapes for optimization problems?
5. What are some real-life applications of maximizing areas in farming or land management?

Tip: The vertex formula $x = \frac{-b}{2a}$ is useful in any situation where you are trying to maximize or minimize a quadratic function.