Question: Diana has 1600 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?
1. **State the problem:** Diana has 1600 yards of fencing to enclose a rectangular area. We want to find:
(a) The area $A$ as a function of the width $W$.
(b) The value of $W$ that maximizes the area.
(c) The maximum area.
2. **Formula and rules:**
The perimeter $P$ of a rectangle is given by:
$$P = 2L + 2W$$
where $L$ is the length and $W$ is the width.
The area $A$ is:
$$A = L \times W$$
3. **Express $L$ in terms of $W$ using the perimeter:**
Given $P = 1600$,
$$1600 = 2L + 2W$$
Divide both sides by 2:
$$\cancel{2}L + \cancel{2}W = \frac{1600}{2}$$
$$L + W = 800$$
Solve for $L$:
$$L = 800 - W$$
4. **Express area $A$ as a function of $W$:**
Substitute $L$ into the area formula:
$$A = L \times W = (800 - W)W = 800W - W^2$$
5. **Find $W$ that maximizes $A$:**
$A(W) = -W^2 + 800W$ is a quadratic function opening downward.
The vertex (maximum) occurs at:
$$W = -\frac{b}{2a}$$
Here, $a = -1$, $b = 800$, so
$$W = -\frac{800}{2 \times (-1)} = \frac{800}{2} = 400$$
6. **Calculate the maximum area:**
Substitute $W = 400$ into $A(W)$:
$$A = 800(400) - (400)^2 = 320000 - 160000 = 160000$$
**Final answers:**
(a) $$A(W) = 800W - W^2$$
(b) $$W = 400$$
(c) $$\text{Maximum area} = 160000$$