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

Answer 1

The correct option is B $32$

$y=12-x^2$

$\therefore AB=2tAD=12-t^2\therefore$
area of rectangle $ABCD$
$\left(A_r\right)=2t(12-t^2)\Rightarrow A_r=24t-2t^3$
To find maximum area -
$\frac{d{A}_r}{dt}=24-6t^2=0\Rightarrow 24-6t^2=0\Rightarrow t=\pm 2$
$\frac{d^2{A}_r}{d{t}^2}=-12t$
at $t=2,\frac{d^2{A}_r}{d{t}^2}<0$
$\therefore A_r=|24\left(2\right)-2(2{)}^3|=|48-16|=|32|\Rightarrow A_r=32\text{sq. units}$
So, maximum area = $32$ sq. units