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

Answer: 32

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