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Step-by-Step Solution
Step 1: Identify the coordinates of the rectangle
Let the rectangle ABCD be positioned such that points A and B lie on the x-axis and points C and D lie on the parabola
$y = x^2 - 1$ below the x-axis. We assume symmetry about the y-axis to simplify the process:
• A has coordinates $(-a, 0)$
• B has coordinates $(a, 0)$
• D has coordinates $(-a, a^2 - 1)$
• C has coordinates $(a, a^2 - 1)$
Step 2: Express the area of the rectangle as a function of a
The base AB of the rectangle is the distance along the x-axis from $-a$ to $a$, which is $2a$.
The height is the (positive) distance from the x-axis down to the parabola $y = x^2 - 1$,
which is $|\,0 - (a^2 - 1)| = 1 - a^2$ for $|a| < 1$.
Thus, the area $A$ as a function of $a$ is:
$$
A(a) = (\text{base}) \times (\text{height}) = 2a \times (1 - a^2).
$$
Step 3: Differentiate and find the critical point
To maximize the area, we take the derivative of $A(a)$ with respect to $a$ and set it to zero:
$$
A(a) = 2a - 2a^3,
$$
$$
\frac{dA}{da} = 2 - 6a^2.
$$
We set the derivative equal to zero to find the critical points:
$$
2 - 6a^2 = 0 \quad \Longrightarrow \quad 6a^2 = 2 \quad \Longrightarrow \quad a^2 = \frac{1}{3} \quad \Longrightarrow \quad a = \frac{1}{\sqrt{3}}.
$$
(We take $a>0$ because we can rely on symmetry.)
Step 4: Determine the maximum area
Substitute $a = \frac{1}{\sqrt{3}}$ back into $A(a)$:
$$
A\!\Bigl(\frac{1}{\sqrt{3}}\Bigr) = 2 \cdot \frac{1}{\sqrt{3}} \times \Bigl(1 - \frac{1}{3}\Bigr)
= \frac{2}{\sqrt{3}} \times \frac{2}{3}
= \frac{4}{3\sqrt{3}}.
$$
Thus, the area of the largest rectangle under the given conditions is
$$
\frac{4}{3\sqrt{3}}.
$$
Step 5: Final Answer
The area (in square units) of the largest rectangle is
$$
\frac{4}{3\sqrt{3}}.
$$
