Question

Find the maximum area $(insq.units)$ 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.

Answer 1

Step 1: Finding the rectangle formed using the given conditions

Equation of parabola is $y=12-x^2$

A rectangle has its base on the x-axis and its upper two vertices on the parabola.

Now, the adjacent figure depicts the rectangle formed.

Length $=2xunits$

Breadth $=y=12-x^{2}units$

Step 2: Finding the maximum area of the rectangle:

Area, $A=length\times beadth$

$=2x\times (12-x^2)$

$=24x-2x^{3}sq.units$

On taking its derivative we get

$\Rightarrow A\text{'}=24-6x^2$

The area is large when $A\text{'}=0$

$\Rightarrow 24-6x^2=0$

$$\begin{gathered} \Rightarrow x^2=4 \\ \Rightarrow x=2units \end{gathered}$$

Substituting the value of $x$ in $y=12-x^2$

$$\begin{gathered} \Rightarrow y=12-4 \\ \Rightarrow y=8units \end{gathered}$$

Hence, the area of rectangle$=2\times 2\times 8=32sq.units$.