Step-by-step Solution:

Given: The dimensions of the rectangle inscribed in the ellipse are: \[ \text{Length} = 2a \cos \theta \] \[ \text{Breadth} = 2b \sin \theta \] Step 1: Compute the Area of the Rectangle \[ \text{Area} = (\text{Length}) \times (\text{Breadth}) \] \[ = (2a \cos \theta) \times (2b \sin \theta) \] \[ = 4ab \cos \theta \sin \theta \] Using the identity: \[ 2 \sin \theta \cos \theta = \sin 2\theta \] \[ \Rightarrow \text{Area} = 2ab \sin 2\theta \] Step 2: Maximizing the Area To maximize the area, \( \sin 2\theta \) should be maximum. \[ \sin 2\theta \leq 1 \] \[ \Rightarrow \sin 2\theta = 1 \text{ at } 2\theta = \frac{\pi}{2}, \frac{5\pi}{2}, \dots \] \[ \Rightarrow \text{Maximum Area} = 2ab \] Thus, the maximum area of the rectangle inscribed in the ellipse is: \[ \boxed{2ab} \]