Let's go through the problem step-by-step.

Given Information:

• Total fencing available = 1400 m
• The field is rectangular, bordered by a river on one side, so no fencing is needed along the river.
• Let $x$ represent the width of the field.
• The length of the field (let's call it $L$) requires fencing on the opposite side and along both widths.

Problem Breakdown:

1. (a) Write an expression for the length of the field as a function of $x$:

   Since the river side does not require fencing, the perimeter to be fenced is: $2x + L = 1400$ Solving for $L$ in terms of $x$: $L = 1400 - 2x$

2. (b) Find the area of the field (area = length × width) as a function of $x$:

   The area $A$ of the field is given by: $A = L \cdot x$ Substituting $L = 1400 - 2x$: $A(x) = x(1400 - 2x) = 1400x - 2x^2$

3. (c) Find the value of $x$ leading to the maximum area:

   To maximize $A(x) = 1400x - 2x^2$, we take the derivative of $A(x)$ with respect to $x$ and set it equal to zero: $A'(x) = 1400 - 4x$ Setting $A'(x) = 0$: $1400 - 4x = 0$ $4x = 1400$ $x = 350$

4. (d) Find the maximum area:

   Substitute $x = 350$ back into the area function: $A(350) = 1400(350) - 2(350)^2$ $= 490000 - 245000 = 245000 \text{ square meters}$

Summary of Answers:

• (a) The length $L$ of the field as a function of $x$: $L = 1400 - 2x$.
• (b) The area $A(x)$ of the field as a function of $x$: $A(x) = 1400x - 2x^2$.
• (c) The width $x$ that maximizes the area is $x = 350$ meters.
• (d) The maximum area is $245000$ square meters.

Would you like further details on any step?

Follow-Up Questions:

1. What happens to the area if the available fencing is increased?
2. How would the solution change if fencing was needed on all four sides?
3. Can you derive the maximum area if the field shape was circular instead of rectangular?
4. How does changing the width affect the length in this setup?
5. How would this problem change if the fenced area was triangular?

Tip:

In optimization problems like this, expressing all variables in terms of one primary variable simplifies finding maximum or minimum values.