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Part 1
David 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 David has 440 yards of fencing to enclose a rectangular area. Let the width of the rectangle be \( W \) and the length be \( L \).

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

The perimeter of the rectangle is the total amount of fencing used, which is given by:
\[
2L + 2W = 440
\]
Simplifying this equation:
\[
L + W = 220
\]
Now, solve for \( L \):
\[
L = 220 - W
\]

The area \( A \) of the rectangle is the product of the length and width:
\[
A = L \times W = (220 - W) \times W
\]
Thus, the area as a function of \( W \) is:
\[
A(W) = W(220 - W) = 220W - W^2
\]

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

To find the value of \( W \) that maximizes the area, we need to find the critical points of \( A(W) \). To do this, take the derivative of \( A(W) \) with respect to \( W \), and set it equal to zero:
\[
A'(W) = 220 - 2W
\]
Set the derivative equal to zero to find the critical points:
\[
220 - 2W = 0
\]
\[
W = 110
\]

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

Now that we know the width \( W = 110 \), substitute it back into the equation for the length:
\[
L = 220 - 110 = 110
\]

The maximum area occurs when both the length and width are 110. The maximum area is:
\[
A = 110 \times 110 = 12,100 \, \text{square yards}
\]

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

Do you want any additional details on the solution? Here are five related questions:

1. How would the area function change if David had 500 yards of fencing instead?
2. What would happen to the area if David enclosed a square instead of a rectangle?
3. How does the derivative help in finding the maximum area?
4. Can you explain why the area is largest when \( W = L \)?
5. How could the problem change if David only used fencing on three sides of the rectangle?

**Tip:** To find maximum or minimum values in any optimization problem, always look for the critical points by setting the derivative equal to zero!

David 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?

Explore how to maximize the area of a rectangular enclosure with 440 yards of fencing. This optimization problem uses algebra and derivatives to find that the maximum area occurs when the width and length are both 110 yards, resulting in a maximum area of 12,100 square yards. Suitable for Grades 10-12.

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