Ivana has 240 meters of fencing and wishes to form three sides of a rectangular field. The fourth side borders a river and will not need fencing. One of the sides has length x (in meters).

a) Find a function that gives the area A(x) of the field (in square meters) in terms of x.

b) What side length x gives the maximum area that the field can have?

c) What is the maximum area that the field can have?

Let's solve this step by step.

### Problem Recap:
- Ivana has **240 meters** of fencing.
- She wants to form **three sides** of a rectangular field, with one of the sides (the fourth) bordering a river (which does not need fencing).
- The length of one of the sides is \( x \) meters.

### Part (a): Find a function for the area \( A(x) \)

Let:
- \( x \) represent the length of one side parallel to the river.
- The total fencing available is 240 meters, which will be used for the other two sides (both of length \( x \)) and one side perpendicular to the river (width).

The fencing used for the two lengths and one width is:
\[
2x + y = 240
\]
where \( y \) is the width of the field (perpendicular to the river).

Solve for \( y \):
\[
y = 240 - 2x
\]

The area \( A(x) \) of the rectangle is the product of the length \( x \) and the width \( y \):
\[
A(x) = x \times y = x(240 - 2x) = 240x - 2x^2
\]

Thus, the area function is:
\[
A(x) = 240x - 2x^2
\]

---

### Part (b): Find the value of \( x \) that gives the maximum area

To find the value of \( x \) that maximizes the area, we take the derivative of the area function and set it equal to zero (to find the critical points):

\[
\frac{dA}{dx} = 240 - 4x
\]

Set the derivative equal to zero:
\[
240 - 4x = 0
\]
\[
4x = 240
\]
\[
x = 60
\]

Thus, the value of \( x \) that maximizes the area is **60 meters**.

---

### Part (c): Find the maximum area

Now, substitute \( x = 60 \) into the area function \( A(x) = 240x - 2x^2 \):

\[
A(60) = 240(60) - 2(60)^2
\]
\[
A(60) = 14400 - 2(3600)
\]
\[
A(60) = 14400 - 7200
\]
\[
A(60) = 7200 \text{ square meters}
\]

Thus, the maximum area that the field can have is **7200 square meters**.

---

### Summary:

- (a) The area function is \( A(x) = 240x - 2x^2 \).
- (b) The side length \( x = 60 \) meters gives the maximum area.
- (c) The maximum area is 7200 square meters.

---

Would you like more details on any part of this solution, or have any questions? Here are five related questions that could extend this concept:

1. What would happen to the area if the total fencing were less than 240 meters?
2. How would the area function change if all four sides needed to be fenced?
3. What happens to the maximum area if Ivana had 300 meters of fencing instead?
4. How do we know that the value of \( x \) we found gives a maximum, not a minimum?
5. What would be the area if \( x \) were 30 meters instead of 60?

**Tip**: The maximum or minimum of a quadratic function \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). This is useful for solving similar problems quickly!

Ivana has 240 meters of fencing and wishes to form three sides of a rectangular field. The fourth side borders a river and will not need fencing. One of the sides has length x (in meters).

(a) Find a function that gives the area A(x) of the field (in square meters) in terms of x.

(b) What side length x gives the maximum area that the field can have?

(c) What is the maximum area that the field can have?

Learn how to maximize the area of a rectangular field using 240 meters of fencing, with a step-by-step solution involving quadratic functions and optimization. Understand the role of derivatives in finding the maximum area for side length x = 60 meters, yielding a maximum area of 7200 square meters.

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