If the area of the ellipse `((x^2)/(a^2))+((y^2)/(b^2))=1` is `4pi` , then find the maximum area of rectangle inscribed in the ellipse.

Explanation

To solve the problem of finding the maximum area of a rectangle inscribed in the ellipse given by the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) with an area of \(4\pi\), we can follow these steps:

### Step 1: Find the values of \(a\) and \(b\)
The area \(A\) of the ellipse is given by the formula:
\[ A = \pi a b \]
According to the problem, the area of the ellipse is \(4\pi\). Therefore, we can set up the equation:
\[ \pi a b = 4\pi \]
Dividing both sides by \(\pi\), we get:
\[ a b = 4 \]

### Step 2: Set up the rectangle's area
Let’s denote the vertices of the rectangle inscribed in the ellipse as \((a \cos \theta, b \sin \theta)\), \((-a \cos \theta, b \sin \theta)\), \((-a \cos \theta, -b \sin \theta)\), and \((a \cos \theta, -b \sin \theta)\). The width of the rectangle is:
\[ 2a \cos \theta \]
and the height is:
\[ 2b \sin \theta \]
Thus, the area \(A_r\) of the rectangle can be expressed as:
\[ A_r = \text{width} \times \text{height} = (2a \cos \theta)(2b \sin \theta) = 4ab \sin \theta \cos \theta \]

### Step 3: Use the double angle identity
Using the double angle identity for sine, we can rewrite the area of the rectangle:
\[ A_r = 4ab \sin \theta \cos \theta = 2ab \sin(2\theta) \]

### Step 4: Maximize the area
To find the maximum area of the rectangle, we need to maximize \(A_r = 2ab \sin(2\theta)\). The maximum value of \(\sin(2\theta)\) is \(1\). Therefore, the maximum area occurs when:
\[ \sin(2\theta) = 1 \]
Thus, the maximum area of the rectangle is:
\[ A_{r,\text{max}} = 2ab \]

### Step 5: Substitute \(ab\)
From Step 1, we know that \(ab = 4\). Therefore, substituting this value into the equation gives:
\[ A_{r,\text{max}} = 2 \times 4 = 8 \]

### Final Answer
The maximum area of the rectangle inscribed in the ellipse is:
\[ \boxed{8} \]