Question

The area (in sq. units) of the largest rectangle $ABCD$ whose vertices $A$ and $B$ lie on the x-axis and vertices $C$ and $D$ lie on the parabola, $y=x^2-1$ below the x-axis, is

• A. $\frac{2}{3\sqrt{3}}$
• B. $\frac{4}{3}$
• C. $\frac{1}{3\sqrt{3}}$
• D. $\frac{4}{3\sqrt{3}}$

Answer 1

The correct option is D

$\frac{4}{3\sqrt{3}}$

Explanation for correct answer Option(s)

Step 1: Find the coordinates of A and B

Equation of parabola is given as:

$y=x^2-1$

The diagram is as shown:

Let the coordinates of $A$ and $B$ are $\left(-a,0\right)$ and $\left(a,0\right)$ respectively as the parabola is symmetric.

Hence, the distance between $A$ and $B$ is equal to $2a$

Substitute $x=a$ in, the given parabola we get

$y=a^2-1$

Hence y-coordinate is equal to $a^2-1$

Now the area of a rectangle

$\text{ABCD=length × breadth}$

$$\begin{gathered} A=2a\left(a^2-1\right)......\left(1\right) \\ A=2a^3-2a \end{gathered}$$

Differentiate $A$ with respect to $a$ to find the maximum area of rectangle.

$$\begin{gathered} \Rightarrow \frac{dA}{da}=\frac{d\left(2a^3-2a\right)}{da} \\ \Rightarrow \frac{dA}{da}=6a^2-2 \\ \Rightarrow 6a^2=2\left[\because \frac{dA}{da}=0\right] \\ \Rightarrow a^2=\frac{2}{6} \\ \Rightarrow a^2=\frac{1}{3} \\ \Rightarrow a=\pm \frac{1}{\sqrt{3}} \\ \Rightarrow a=\frac{1}{\sqrt{3}} \end{gathered}$$

Step 2: Find the area (in sq. units) of the largest rectangle $ABCD$.

substitute the value of $a$ in the Equation $\left(1\right)$ we get, maximum area

$$\begin{gathered} A_{max}=2a\left(a^2-1\right) \\ {A}_{max}=2\times \frac{1}{\sqrt{3}}\left({\left(\frac{1}{\sqrt{3}}\right)}^2-1\right) \\ {A}_{max}=\frac{2}{\sqrt{3}}\left(\frac{1}{3}-1\right) \\ {A}_{max}=\frac{2}{\sqrt{3}}\left(-\frac{2}{3}\right) \\ {A}_{max}=-\frac{4}{3\sqrt{3}} \end{gathered}$$

Since the area can not be negative

Hence, the required area is $A_{max}=\frac{4}{3\sqrt{3}}$

Therefore, the correct answer is Option D.