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!