has available 440 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?

We are given that 440 yards of fencing is available to enclose a rectangular area. Let's work through the problem step by step.

### Part (a): Express the area \( A \) as a function of the width \( W \).

Let:
- \( W \) be the width of the rectangle.
- \( L \) be the length of the rectangle.

The perimeter \( P \) of the rectangle is the total length of fencing, which is 440 yards. The formula for the perimeter of a rectangle is:

\[
P = 2L + 2W
\]

Given that \( P = 440 \), we can write:

\[
2L + 2W = 440
\]

Dividing both sides by 2:

\[
L + W = 220
\]

Solving for \( L \):

\[
L = 220 - W
\]

Now, the area \( A \) of the rectangle is given by the formula:

\[
A = L \times W
\]

Substituting \( L = 220 - W \) into the area equation:

\[
A(W) = W \times (220 - W)
\]

Simplifying:

\[
A(W) = 220W - W^2
\]

Thus, the area \( A \) as a function of the width \( W \) is:

\[
A(W) = 220W - W^2
\]

---

### Part (b): For what value of \( W \) is the area largest?

To find the width \( W \) that maximizes the area, we need to find the critical points by taking the derivative of \( A(W) \) with respect to \( W \) and setting it equal to zero.

The derivative of \( A(W) = 220W - W^2 \) is:

\[
\frac{dA}{dW} = 220 - 2W
\]

Setting this equal to zero to find the critical points:

\[
220 - 2W = 0
\]

Solving for \( W \):

\[
2W = 220 \quad \Rightarrow \quad W = 110
\]

Thus, the width \( W = 110 \) yards will give the largest area.

---

### Part (c): What is the maximum area?

To find the maximum area, we substitute \( W = 110 \) into the area function \( A(W) = 220W - W^2 \):

\[
A(110) = 220(110) - (110)^2
\]

\[
A(110) = 24200 - 12100 = 12100 \, \text{square yards}
\]

Thus, the maximum area is **12,100 square yards**.

---

### Summary:
- (a) The area as a function of width is \( A(W) = 220W - W^2 \).
- (b) The area is largest when \( W = 110 \) yards.
- (c) The maximum area is 12,100 square yards.

Would you like further clarification on any part?

Here are 5 related questions:
1. How would the problem change if you had more fencing available?
2. What happens to the maximum area if the shape must be a square?
3. Can you graph the area function \( A(W) \)? What does its shape tell you?
4. What if the length of the rectangle was fixed at a certain value—how would that affect the width and area?
5. Can this technique be used for other shapes, like triangles or circles?

**Tip:** When maximizing areas or volumes, always consider symmetry—it often leads to optimal solutions.

A user has available 440 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?

Solve how to maximize the area of a rectangle using 440 yards of fencing. Learn how to express the area as a function of width, determine the width for the largest area using calculus, and find the maximum area of 12,100 square yards. Suitable for students in Grades 9-12.

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