Question

The maximum area (in sq. units) of a rectangle having its base on the $x$-axis and its other two vertices on the parabola, $y=12-x^2$ such that the rectangle lies inside the parabola, is :-

• A. $20\sqrt{2}$
• B. $18\sqrt{3}$
• C. $32$
• D. $36$

Answer 1

The correct option is C $32$

$f\left(a\right)=2a(12-a^2)$
$f^{\prime }\left(a\right)=2(12-3a^2)$
maximum at $a=2$
maximum area $=f\left(2\right)=32$