Key Concepts and Formulas

• Parametric Representation of an Ellipse: The point $(x, y)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ can be represented as $(a \cos \theta, b \sin \theta)$.
• Area of a Rectangle: The area of a rectangle with length $l$ and width $w$ is given by $A = lw$.
• Optimization using Derivatives: To find the maximum or minimum of a function $f(x)$, find the critical points by setting $f'(x) = 0$, and use the first or second derivative test to confirm the nature of the extremum.

Step-by-Step Solution

Step 1: Visualize and Define the Rectangle

Consider the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. We want to inscribe a rectangle with maximum area. Due to the symmetry of the ellipse about both axes, we can assume the rectangle is also symmetric about both axes. Let $(x, y)$ be the vertex of the rectangle in the first quadrant. Then the vertices of the rectangle are $(x, y)$, $(-x, y)$, $(-x, -y)$, and $(x, -y)$.

The length of the rectangle is $2x$, and the width is $2y$. Therefore, the area of the rectangle is: $A = (2x)(2y) = 4xy$

Step 2: Express the Area in Parametric Form

We use the parametric representation of a point on the ellipse: $x = a \cos \theta$ and $y = b \sin \theta$, where $\theta$ is the eccentric angle. Substituting these into the area formula: $A(\theta) = 4(a \cos \theta)(b \sin \theta) = 4ab \cos \theta \sin \theta$ Using the trigonometric identity $2 \sin \theta \cos \theta = \sin(2\theta)$: $A(\theta) = 2ab (2 \sin \theta \cos \theta) = 2ab \sin(2\theta)$

Step 3: Optimize the Area Function

To find the maximum area, we need to maximize $A(\theta) = 2ab \sin(2\theta)$. Since $a$ and $b$ are constants, we only need to maximize $\sin(2\theta)$. The maximum value of the sine function is 1, which occurs when its argument is $\frac{\pi}{2}$. Therefore, we want: $2\theta = \frac{\pi}{2}$ $\theta = \frac{\pi}{4}$

Alternatively, we can use calculus. Differentiating $A(\theta)$ with respect to $\theta$: $\frac{dA}{d\theta} = \frac{d}{d\theta} (2ab \sin(2\theta)) = 2ab \cdot 2 \cos(2\theta) = 4ab \cos(2\theta)$ Setting the derivative to zero to find critical points: $4ab \cos(2\theta) = 0$ Since $a$ and $b$ are non-zero, we must have $\cos(2\theta) = 0$. In the interval $0 < \theta < \frac{\pi}{2}$, $2\theta = \frac{\pi}{2}$, which gives $\theta = \frac{\pi}{4}$.

Now, we check the second derivative to confirm that this is a maximum: $\frac{d^2A}{d\theta^2} = \frac{d}{d\theta} (4ab \cos(2\theta)) = -8ab \sin(2\theta)$ At $\theta = \frac{\pi}{4}$, $\frac{d^2A}{d\theta^2} = -8ab \sin\left(\frac{\pi}{2}\right) = -8ab < 0$, since $a > 0$ and $b > 0$. This confirms that $\theta = \frac{\pi}{4}$ gives a maximum area.

Step 4: Calculate the Maximum Area

Substitute $\theta = \frac{\pi}{4}$ back into the area function: $A_{max} = 2ab \sin\left(2 \cdot \frac{\pi}{4}\right) = 2ab \sin\left(\frac{\pi}{2}\right) = 2ab(1) = 2ab$

Common Mistakes & Tips

• Forgetting Symmetry: Always consider symmetry to simplify the problem setup. The rectangle should be centered on the ellipse's axes.
• Incorrect Area Formula: Ensure the area is $4xy$, not $xy$. Remember that $x$ and $y$ represent only one vertex in the first quadrant.
• Using Calculus Unnecessarily: Recognizing that the maximum of $\sin(2\theta)$ is 1 can save time compared to using derivatives.

Summary

We found the maximum area of a rectangle inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ by expressing the area in terms of the eccentric angle $\theta$, and then maximizing the resulting trigonometric function. The maximum area is $2ab$.

The final answer is $\boxed{2ab}$, which corresponds to option (A).