Key Concepts and Formulas

• Optimization using Calculus: Finding the maximum or minimum of a function by finding critical points using derivatives.
• Area of a Rectangle: Area = width × height
• First and Second Derivative Tests: Using the first derivative to find critical points and the second derivative to determine if a critical point is a local maximum or minimum.

Step-by-Step Solution

Step 1: Set up the Coordinate System and Define Variables

We are given a rectangle ABCD inscribed under the parabola $y = x^2 - 1$, with A and B on the x-axis. The parabola is symmetric about the y-axis, so we can assume the rectangle is also symmetric about the y-axis to maximize area.

• Let the coordinates of point B be $(t, 0)$, where $t > 0$. Since A lies on the x-axis and the rectangle is symmetric, the coordinates of A are $(-t, 0)$.
• Since C and D lie on the parabola, their y-coordinates are given by $y = x^2 - 1$. Therefore, the coordinates of C are $(t, t^2 - 1)$ and the coordinates of D are $(-t, t^2 - 1)$.
• Since the vertices C and D lie below the x-axis, we must have $t^2 - 1 < 0$, which implies $t^2 < 1$, and since $t>0$ this means $0 < t < 1$.

Step 2: Formulate the Area Function

The area of the rectangle ABCD can be expressed in terms of $t$.

• The width of the rectangle is the distance between A and B, which is $2t$.
• The height of the rectangle is the absolute value of the y-coordinate of C (or D), which is $-(t^2 - 1) = 1 - t^2$ (since $t^2 - 1$ is negative).
• Therefore, the area $A(t)$ of the rectangle is given by: $A(t) = (2t)(1 - t^2) = 2t - 2t^3$ We want to maximize $A(t)$ on the interval $(0, 1)$.

Step 3: Find Critical Points by Taking the First Derivative

To find the critical points, we take the derivative of $A(t)$ with respect to $t$ and set it equal to zero.

• The first derivative of $A(t)$ is: $\frac{dA}{dt} = \frac{d}{dt}(2t - 2t^3) = 2 - 6t^2$
• Setting the derivative to zero: $2 - 6t^2 = 0$ $6t^2 = 2$ $t^2 = \frac{1}{3}$ $t = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}}$
• Since $0 < t < 1$, we choose the positive root: $t = \frac{1}{\sqrt{3}}$

Step 4: Confirm Maximum Using the Second Derivative Test

To confirm that $t = \frac{1}{\sqrt{3}}$ corresponds to a maximum, we use the second derivative test.

• The second derivative of $A(t)$ is: $\frac{d^2A}{dt^2} = \frac{d}{dt}(2 - 6t^2) = -12t$
• Evaluating the second derivative at $t = \frac{1}{\sqrt{3}}$: $\frac{d^2A}{dt^2}\Big|_{t=\frac{1}{\sqrt{3}}} = -12\left(\frac{1}{\sqrt{3}}\right) = -\frac{12}{\sqrt{3}}$
• Since the second derivative is negative at $t = \frac{1}{\sqrt{3}}$, this indicates that we have a local maximum at this point.

Step 5: Calculate the Maximum Area

Substitute $t = \frac{1}{\sqrt{3}}$ back into the area function $A(t) = 2t - 2t^3$ to find the maximum area.

• $A_{\text{max}} = 2\left(\frac{1}{\sqrt{3}}\right) - 2\left(\frac{1}{\sqrt{3}}\right)^3$ $A_{\text{max}} = \frac{2}{\sqrt{3}} - 2\left(\frac{1}{3\sqrt{3}}\right)$ $A_{\text{max}} = \frac{2}{\sqrt{3}} - \frac{2}{3\sqrt{3}}$ $A_{\text{max}} = \frac{6}{3\sqrt{3}} - \frac{2}{3\sqrt{3}}$ $A_{\text{max}} = \frac{4}{3\sqrt{3}}$

Common Mistakes & Tips

• Remember to consider the domain of the variable $t$. The condition $0 < t < 1$ is crucial.
• Always check for symmetry to simplify the problem.
• Don't forget to confirm whether the critical point corresponds to a maximum or a minimum.

Summary

We found the maximum area of the rectangle by expressing its dimensions in terms of a variable $t$ and using calculus to maximize the area function $A(t) = 2t - 2t^3$ subject to the constraint $0 < t < 1$. The critical point $t = \frac{1}{\sqrt{3}}$ was found using the first derivative, and the second derivative test confirmed it corresponded to a maximum. The maximum area is $\frac{4}{3\sqrt{3}}$ square units.

The final answer is $\boxed{\frac{4}{3\sqrt{3}}}$, which corresponds to option (C).