Sienna has 800 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area? A rectangle that maximizes the enclosed area has a length of $$ \square $$ yards and a width of $$ \square $$ yards. 0 of 2

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To determine the dimensions of the rectangle that maximize the enclosed area with 800 yards of fencing, follow these steps:

### 1. Understanding the Problem

- **Perimeter (P)**: The total length of fencing available is 800 yards.
- **Rectangle Dimensions**: Let’s denote the length as $$ L $$ and the width as $$ W $$.

### 2. Relationship Between Length and Width

The perimeter of a rectangle is given by:
$$
P = 2L + 2W
$$
Given $$ P = 800 $$ yards:
$$
2L + 2W = 800
$$
Simplifying:
$$
L + W = 400 \quad \Rightarrow \quad W = 400 - L
$$

### 3. Expressing Area in Terms of One Variable

The area ($$ A $$) of the rectangle is:
$$
A = L \times W
$$
Substituting $$ W $$:
$$
A = L \times (400 - L) = 400L - L^2
$$

### 4. Maximizing the Area

To find the maximum area, we can take the derivative of $$ A $$ with respect to $$ L $$ and set it to zero:
$$
\frac{dA}{dL} = 400 - 2L = 0
$$
Solving for $$ L $$:
$$
2L = 400 \quad \Rightarrow \quad L = 200 \text{ yards}
$$
Substituting back to find $$ W $$:
$$
W = 400 - 200 = 200 \text{ yards}
$$

### 5. Maximum Area

$$
A = 200 \times 200 = 40,000 \text{ square yards}
$$

### **Final Answer**

- **Length**: $$ 200 $$ yards
- **Width**: $$ 200 $$ yards
- **Maximum Area**: $$ 40,\!000 $$ square yards
To maximize the enclosed area with 800 yards of fencing, the rectangle should be a square with each side measuring 200 yards. The maximum area is 40,000 square yards.

Sienna has 800 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area? A rectangle that maximizes the enclosed area has a length of \( \square \) yards and a width of \( \square \) yards. 0 of 2

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