The maximum area of a rectangle whose two vertices lie on th x-axis and two on the curve $y = 3 - |x|, -3\leq x \leq 3$ is : 

\(\frac{9}{2}\)

The correct answer is Option (2) → \(\frac{9}{2}\)

Area: $A = 2x (3 - x)$

$A' = 2x (3 - x) - 2x$

$x = 3/2$

Maximum area of the rectangle occurs when $= 3/2$

maximum area = $2.[3/2][3-\frac{3}{2}]=9/2$ sq. units