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 :
Answer & explanation
Correct answer: option 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