A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. Two having fence are of same length x. The maximum area enclosed by the park is :-

Options

A. `sqrt(x^(3))/(8)`
B. `1/2 x^(2)`
C. `pi x^(2)`
D. `3/2 x^(2)`

Answer

To solve the problem of finding the maximum area of a triangular park enclosed by two fences of equal length \(x\) and one side along a river bank, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a triangular park \(ABC\) with \(AB = AC = x\) (the lengths of the fences) and \(BC\) along the river bank.
- We need to find the maximum area of triangle \(ABC\).

2. **Setting Up the Triangle:**
- Let \( \angle ACB = \theta \).
- The height from point \(A\) to side \(BC\) can be dropped to point \(D\), creating two right triangles.

3. **Finding the Height:**
- The height \(AD\) can be expressed in terms of \(x\) and \(\theta\):
\[ AD = x \sin(\theta) \]

4. **Finding the Base:**
- The base \(DC\) can also be expressed in terms of \(x\) and \(\theta\):
\[ DC = x \cos(\theta) \]

5. **Calculating the Area of Triangle \(ABC\):**
- The area of triangle \(ABC\) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
- Thus, the area becomes:
\[ \text{Area} = \frac{1}{2} \times BC \times AD \]
- Since \(BC = 2 \times DC = 2x \cos(\theta)\), we have:
\[ \text{Area} = \frac{1}{2} \times (2x \cos(\theta)) \times (x \sin(\theta)) = x^2 \sin(\theta) \cos(\theta) \]

6. **Using the Double Angle Identity:**
- We can simplify \( \sin(\theta) \cos(\theta) \) using the double angle identity:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \Rightarrow \sin(\theta) \cos(\theta) = \frac{1}{2} \sin(2\theta) \]
- Therefore, the area can be rewritten as:
\[ \text{Area} = \frac{1}{2} x^2 \sin(2\theta) \]

7. **Maximizing the Area:**
- The maximum value of \( \sin(2\theta) \) is \(1\). Thus, the maximum area becomes:
\[ \text{Maximum Area} = \frac{1}{2} x^2 \times 1 = \frac{1}{2} x^2 \]

8. **Conclusion:**
- The maximum area enclosed by the park is:
\[ \boxed{\frac{1}{2} x^2} \]