A triangular park is enclosed on two sides of 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

Options

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

Explanation

Let PQR be the triangular park such that PQ =PR =x and let `angleQPR =theta`.Then,
Area of the park is given by `Delta=1/2(PQxxPRxxsintheta)`
`rArr=1/2x^(2)sintheta`
Clearly, `Delta` is maximum when sin `theta` is maximum i.e. `theta` = 90°.
The maximum value of `Delta` is `1/2x^(2)` .