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To solve this problem, let’s break it down step-by-step.

### Problem Setup:
You have 240 meters of fencing and want to enclose a rectangular area adjacent to a straight wall. Since one side of the rectangle is along the wall, you only need to fence the other three sides. Let:
- \( x \) = the length of the side parallel to the wall.
- \( y \) = the lengths of the two sides perpendicular to the wall.

### Step 1: Set Up the Perimeter Equation
Since you only need fencing for three sides, the perimeter constraint is:
\[
x + 2y = 240
\]

### Step 2: Express \( y \) in Terms of \( x \)
Rearrange the equation to express \( y \) in terms of \( x \):
\[
y = \frac{240 - x}{2}
\]

### Step 3: Write the Area Equation
The area \( A \) of the rectangle is:
\[
A = x \cdot y
\]
Substitute \( y \) from the previous step:
\[
A = x \cdot \frac{240 - x}{2}
\]
\[
A = \frac{240x - x^2}{2}
\]
\[
A = 120x - \frac{x^2}{2}
\]

### Step 4: Maximize the Area
To find the maximum area, take the derivative of \( A \) with respect to \( x \) and set it to zero.
\[
\frac{dA}{dx} = 120 - x = 0
\]
Solving this gives:
\[
x = 120
\]

### Step 5: Substitute \( x = 120 \) to Find \( y \)
Now, plug \( x = 120 \) back into the equation for \( y \):
\[
y = \frac{240 - 120}{2} = 60
\]

### Step 6: Calculate the Maximum Area
Finally, calculate the area using \( x = 120 \) and \( y = 60 \):
\[
A = 120 \times 60 = 7200 \text{ square meters}
\]

### Answer
The largest area you can enclose is:
\[
\boxed{7200 \text{ square meters}}
\]

Would you like more details or have any questions?

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Here are some related questions for further practice:
1. What if the total fencing length was different, say 300 meters?
2. How would the answer change if the rectangle needed fencing on all four sides?
3. Can you find the dimensions that maximize the area if the wall isn't straight but instead has a curve?
4. How does calculus help in maximizing area in this type of problem?
5. What is the effect on the maximum area if the wall is not aligned with the sides of the rectangle?

**Tip:** In optimization problems, using calculus to find critical points helps determine maximum or minimum values efficiently.

If you have 240 meters of fencing and want to enclose a rectangular area up against a long, straight wall, what is the largest area you can enclose?

Explore how to maximize the area you can enclose with 240 meters of fencing along a straight wall. Learn key optimization techniques using calculus and algebra, including setting up perimeter and area equations and solving for maximum area.

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