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Let's go step by step to solve this problem about enclosing a rectangular area with 480 yards of fencing.

### Given:
- The perimeter of the rectangle is 480 yards.
- Let the width of the rectangle be \( W \), and the length be \( L \).

The perimeter of the rectangle is given by:
\[
2L + 2W = 480
\]
Simplifying, we get:
\[
L + W = 240 \quad \text{or} \quad L = 240 - W
\]

### Part (a) — Express the area \( A(W) \) as a function of the width \( W \)
The area \( A \) of a rectangle is given by:
\[
A = L \times W
\]
Substitute \( L = 240 - W \) into the area formula:
\[
A(W) = W(240 - W) = 240W - W^2
\]

### Part (b) — For what value of \( W \) is the area largest?
The function \( A(W) = 240W - W^2 \) is a quadratic function of the form \( -W^2 + 240W \), which opens downwards. To find the maximum area, we need to find the vertex of the parabola. The width \( W \) that maximizes the area is given by:
\[
W = \frac{-b}{2a}
\]
where \( a = -1 \) and \( b = 240 \). Substituting these values:
\[
W = \frac{-240}{2(-1)} = \frac{240}{2} = 120
\]

### Part (c) — What is the maximum area?
To find the maximum area, substitute \( W = 120 \) into the area function:
\[
A(120) = 240(120) - 120^2 = 28800 - 14400 = 14400 \text{ square yards}
\]

### Summary of the answers:
- (a) \( A(W) = 240W - W^2 \)
- (b) The area is largest when \( W = 120 \) yards.
- (c) The maximum area is \( 14400 \) square yards.

Would you like further details or explanations?

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Here are 5 related questions:
1. What happens to the area if the width is less than 120 yards?
2. How would the area change if the perimeter was increased to 500 yards?
3. What would be the length if the width is 150 yards instead of 120?
4. Can you graph the area function \( A(W) = 240W - W^2 \) to see the maximum point visually?
5. How would the problem change if the rectangle needed to have a specific minimum width?

**Tip**: The vertex formula \( W = \frac{-b}{2a} \) is essential for finding the maximum or minimum of a quadratic function!

David has 480 yards of fencing and wishes to enclose a rectangular area. (a) Express the area A of the rectangle as a function of the width W of the rectangle. (b) For what value of W is the area largest? (c) What is the maximum area?

Learn how to express the area of a rectangle as a function of its width, and find the width that maximizes the area when given 480 yards of fencing. This problem covers algebraic concepts such as quadratic equations and optimization, ideal for high school students.

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