Amber and liam are building a fence at the side of a trailer. They have 60 ft of fencing and plan to use the side of the trailer as one side of the rectangular fenced area. What is the maximum area they can enclose?

To determine the maximum area Amber and Liam can enclose using the side of the trailer as one side of the rectangular fenced area, we can use calculus and optimization principles. 

Let's denote the length of the side of the fenced area parallel to the trailer as \( x \) and the width of the other two sides as \( y \).

Given:
- The total length of the fencing available is 60 feet.
- The side of the trailer serves as one side of the rectangle, so only three sides of fencing are needed.

Thus, the total length of the fencing can be written as:
\[ x + 2y = 60 \]

We want to maximize the area \( A \) of the rectangle, which is given by:
\[ A = x \cdot y \]

First, express \( y \) in terms of \( x \) using the fencing constraint:
\[ x + 2y = 60 \]
\[ 2y = 60 - x \]
\[ y = \frac{60 - x}{2} \]

Substitute this expression for \( y \) into the area formula:
\[ A = x \cdot \frac{60 - x}{2} \]
\[ A = \frac{x(60 - x)}{2} \]
\[ A = \frac{60x - x^2}{2} \]
\[ A = 30x - \frac{x^2}{2} \]

To find the maximum area, we need to take the derivative of \( A \) with respect to \( x \) and set it to zero:
\[ \frac{dA}{dx} = 30 - x = 0 \]
\[ x = 30 \]

To confirm this is a maximum, we check the second derivative:
\[ \frac{d^2A}{dx^2} = -1 \]
Since the second derivative is negative, \( x = 30 \) gives a maximum area.

Now, calculate \( y \) using \( x = 30 \):
\[ y = \frac{60 - x}{2} = \frac{60 - 30}{2} = 15 \]

Thus, the maximum area \( A \) is:
\[ A = x \cdot y = 30 \cdot 15 = 450 \text{ square feet} \]

So, the maximum area they can enclose is **450 square feet**.

Would you like more details or have any questions? Here are some additional related questions:

1. How would the maximum area change if the length of the fencing available was different?
2. What is the general formula for the maximum area enclosed given any length of fencing?
3. How can this problem be solved without calculus?
4. What if the side of the trailer wasn't straight, how would that affect the maximum area calculation?
5. How can we determine the dimensions if the trailer forms two sides of the rectangle?
6. What are some real-life considerations that might affect the dimensions of the fenced area?
7. How does the concept of optimization apply to other areas of mathematics?
8. What is the significance of the second derivative test in determining maximum and minimum values?

**Tip:** When solving optimization problems, always verify if the critical points are maximum or minimum by using the second derivative test or analyzing the context of the problem.

Explore how to maximize the fenced area using 60 feet of fencing with one side along a trailer. Understand the calculus of optimization, including the derivative and second derivative tests. Suitable for high school students studying calculus and optimization.

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