To determine the maximum area enclosed by the triangular park, we need to consider that the park is bound by two sides of equal length,
x, and the third side by the river. One optimal configuration for maximum area in such cases is an isosceles triangle where the third side (formed by the river) is also a variable.
Let's denote the length of the third side by river as
y. The angle between the two sides of length
x will be
θ. The area of a triangle formed by two sides and the included angle is given by:
A=‌ab‌sin‌(θ)where
a=x and
b=x are sides of length
x, yielding:
A=‌x2‌sin‌(θ)To find the angle that maximizes
sin‌(θ), and thus
A, we note that
sin‌(θ) reaches its maximum value of 1 when
θ=90∘. This configuration describes a right-angled triangle.
In this specific optimally configured triangle:
The base is
x.
The height is
x.
The angle between
x and
x is
90∘ (making it a right triangle).
Thus, the maximum area of the park will be:
Hence, the correct answer to the maximum area enclosed by the park, given the side lengths and constraints, is:
Option D:
‌x2