Question

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

• A. \( \sqrt{\frac{x^3}{8}} \)
• B. \( \pi x^2 \)
• C. \( \frac{3}{2} x^2 \)
• D. \( \frac{1}{2} x^2 \)

Hint:

In optimization problems involving areas, use trigonometric identities and geometric properties to find the maximum or minimum values.

Answer 1

The Correct Option is D

We are given that a triangular park is enclosed on two sides by a fence, and the third side is along a straight river bank. The two sides with the fence are of length \( x \). We need to find the maximum area enclosed by the park. Step 1: Geometry of the Park The park forms a right-angled triangle because the river bank serves as one side of the triangle, and the two other sides are fenced with length \( x \). Let’s denote the angle between the two fenced sides as \( \theta \). The area \( A \) of the triangle can be given by the formula: \[ A = \frac{1}{2} \times x \times x \times \sin \theta = \frac{x^2}{2} \sin \theta \] Step 2: Maximize the Area The maximum area occurs when \( \sin \theta \) is maximum, which happens when \( \theta = 90^\circ \). At this point, \( \sin \theta = 1 \), so the maximum area is: \[ A_{\text{max}} = \frac{x^2}{2} \] Thus, the maximum area enclosed by the park is \( \frac{1}{2} x^2 \).