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:

$A(W) = 220W - W^2$

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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.

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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.

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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.