If points A and B lie on x-axis and points C and D lie on the curve `y=x^2-1` below the x-axis then maximum area of rectangle ABCD is

Options

A. `(4sqrt(3))/3`
B. `(4sqrt(3))/9`
C. `(4sqrt(3))/27`
D. `(8sqrt(3))/9`

Answer

To find the maximum area of rectangle ABCD, where points A and B lie on the x-axis and points C and D lie on the curve \( y = x^2 - 1 \) below the x-axis, we can follow these steps:

### Step 1: Define the points
Let the coordinates of points A and B on the x-axis be:
- \( A(t, 0) \)
- \( B(-t, 0) \)

Points C and D will be on the curve \( y = x^2 - 1 \):
- \( C(t, t^2 - 1) \)
- \( D(-t, t^2 - 1) \)

### Step 2: Determine the dimensions of the rectangle
The width of the rectangle (distance between A and B) is:
\[ \text{Width} = |t - (-t)| = 2t \]
The height of the rectangle (the y-coordinate of points C and D) is:
\[ \text{Height} = t^2 - 1 \]

### Step 3: Write the area of the rectangle
The area \( A \) of rectangle ABCD can be expressed as:
\[ A = \text{Width} \times \text{Height} = (2t)(t^2 - 1) = 2t(t^2 - 1) = 2t^3 - 2t \]

### Step 4: Find the critical points
To find the maximum area, we need to differentiate \( A \) with respect to \( t \):
\[ \frac{dA}{dt} = 6t^2 - 2 \]
Setting the derivative equal to zero to find critical points:
\[ 6t^2 - 2 = 0 \implies 6t^2 = 2 \implies t^2 = \frac{1}{3} \implies t = \pm \frac{1}{\sqrt{3}} \]

### Step 5: Determine if the critical point is a maximum
To determine whether this critical point corresponds to a maximum, we take the second derivative:
\[ \frac{d^2A}{dt^2} = 12t \]
Evaluating the second derivative at \( t = -\frac{1}{\sqrt{3}} \):
\[ \frac{d^2A}{dt^2} = 12\left(-\frac{1}{\sqrt{3}}\right) < 0 \]
Since the second derivative is negative, this indicates that we have a maximum at \( t = -\frac{1}{\sqrt{3}} \).

### Step 6: Calculate the maximum area
Substituting \( t = \frac{1}{\sqrt{3}} \) into the area formula:
\[ A = 2\left(-\frac{1}{\sqrt{3}}\right)^3 - 2\left(-\frac{1}{\sqrt{3}}\right) \]
Calculating:
\[ A = 2\left(-\frac{1}{3\sqrt{3}}\right) + \frac{2}{\sqrt{3}} = -\frac{2}{3\sqrt{3}} + \frac{2\sqrt{3}}{3} \]
Combining the terms:
\[ A = \frac{2\sqrt{3}}{3} - \frac{2}{3\sqrt{3}} = \frac{2}{3}\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = \frac{2}{3}\left(\frac{3 - 1}{\sqrt{3}}\right) = \frac{2}{3}\left(\frac{2}{\sqrt{3}}\right) = \frac{4}{3\sqrt{3}} \]
Rationalizing the denominator:
\[ A = \frac{4\sqrt{3}}{9} \]

### Final Answer
The maximum area of rectangle ABCD is:
\[ \boxed{\frac{4\sqrt{3}}{9}} \]