Let's break down the problem step by step. From the information provided:

(a) Find a function for the area of the field in terms of $x$

Let:

• $x$ be the length of the side parallel to the river (which does not require fencing).
• $y$ be the length of the other two sides (the sides that will require fencing).

Since Martina is fencing three sides of the rectangle (two sides of length $y$ and one side of length $x$), the total length of fencing she has is given by: $x + 2y = \text{Total length of fencing available (in meters)}.$ Let this total length be $L$, so: $x + 2y = L.$ Now, solve for $y$ in terms of $x$ and $L$: $2y = L - x \quad \Rightarrow \quad y = \frac{L - x}{2}.$

The area $A$ of the rectangle is given by: $A = \text{Length} \times \text{Width} = x \times y.$ Substitute $y = \frac{L - x}{2}$ into the equation for $A$: $A(x) = x \times \frac{L - x}{2}.$ Simplify: $A(x) = \frac{x(L - x)}{2}.$

So, the function that gives the area in terms of $x$ is: $A(x) = \frac{Lx - x^2}{2}.$

(b) What side length $x$ gives the maximum area?

To find the value of $x$ that maximizes the area, we need to find the critical points of the area function. To do that, take the derivative of $A(x)$ with respect to $x$, set it equal to 0, and solve for $x$.

The derivative of $A(x)$ is: $A'(x) = \frac{d}{dx}\left( \frac{Lx - x^2}{2} \right).$ Using basic rules of differentiation: $A'(x) = \frac{L - 2x}{2}.$

Now, set the derivative equal to 0 to find the critical points: $\frac{L - 2x}{2} = 0.$ Multiply both sides by 2: $L - 2x = 0 \quad \Rightarrow \quad 2x = L \quad \Rightarrow \quad x = \frac{L}{2}.$

So, the value of $x$ that gives the maximum area is $\frac{L}{2}$.

(c) What is the maximum area?

Now that we know the value of $x$ that maximizes the area is $\frac{L}{2}$, substitute this value of $x$ into the area function to find the maximum area.

Substitute $x = \frac{L}{2}$ into the equation for $A(x)$: $A\left( \frac{L}{2} \right) = \frac{L \left( \frac{L}{2} \right) - \left( \frac{L}{2} \right)^2}{2}.$ Simplify: $A\left( \frac{L}{2} \right) = \frac{\frac{L^2}{2} - \frac{L^2}{4}}{2} = \frac{\frac{2L^2}{4} - \frac{L^2}{4}}{2} = \frac{\frac{L^2}{4}}{2} = \frac{L^2}{8}.$

So, the maximum area is $\frac{L^2}{8}$.

Summary of Answers:

• (a) The function for the area in terms of $x$ is $A(x) = \frac{Lx - x^2}{2}$.
• (b) The side length $x$ that gives the maximum area is $\frac{L}{2}$.
• (c) The maximum area is $\frac{L^2}{8}$ square meters.

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Would you like further clarification on any part of this solution? Here are some related questions to expand on the topic:

1. How does the shape of the rectangle affect its area?
2. What would happen if Martina used fencing for all four sides?
3. How could we use the concept of optimization to solve other real-world problems?
4. What are the implications of maximizing area in various fields like agriculture or construction?
5. How does the total amount of fencing (L) affect the optimal size of the rectangle?

Tip: When optimizing area, it’s often useful to first express the area as a function of one variable and then differentiate to find critical points.