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

Answer 1

Given, the equation of parabola is. $x^{2} = 12 - y$

Area of the rectangle $ = \left(2t\right) \left(12 - t^{2}\right)$
$A =24t -2t^{3}$
$\frac{dA}{dt} 24 -6t^{2}$
Put $\frac{dA}{dt} = 0$
$\Rightarrow 24 -6t^{2} =0$
$\Rightarrow t=\pm 2$

At $t = 2$. area is maximum$= 24 \left(2\right) -2\left(2\right)^{3}$
$ =48 -16 =32$sq. units