If a rectangle of maximum area is inscibed in the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`, then the dimensions are

Options

A. `asqrt(2), bsqrt(2)`
B. `(a)/(sqrt(2)), (b)/(sqrt(2))`
C. `(1)/(2sqrt(a)), (1)/(2sqrt(b))`
D. `(1)/(sqrt(2a)), (1)/(sqrt(2b))`

Answer

To find the dimensions of a rectangle of maximum area inscribed in the ellipse given by the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can follow these steps:

### Step 1: Parametrize the ellipse
We can represent a point on the ellipse using the parameter \(\theta\):
\[ x = a \sin \theta, \quad y = b \cos \theta \]

### Step 2: Determine the dimensions of the rectangle
The rectangle inscribed in the ellipse will have vertices at:
\[ (a \sin \theta, b \cos \theta), \quad (-a \sin \theta, b \cos \theta), \quad (a \sin \theta, -b \cos \theta), \quad (-a \sin \theta, -b \cos \theta) \]
The length \(L\) and breadth \(B\) of the rectangle can be expressed as:
\[ L = 2 |x| = 2a \sin \theta \]
\[ B = 2 |y| = 2b \cos \theta \]

### Step 3: Write the area of the rectangle
The area \(A\) of the rectangle can be calculated as:
\[ A = L \cdot B = (2a \sin \theta)(2b \cos \theta) = 4ab \sin \theta \cos \theta \]
Using the double angle identity, we can rewrite this as:
\[ A = 2ab \sin(2\theta) \]

### Step 4: Maximize the area
To maximize the area, we need to differentiate \(A\) with respect to \(\theta\):
\[ \frac{dA}{d\theta} = 2ab \cdot 2 \cos(2\theta) = 4ab \cos(2\theta) \]
Setting the derivative equal to zero for maximization:
\[ 4ab \cos(2\theta) = 0 \]
This gives us:
\[ \cos(2\theta) = 0 \]
Thus, \(2\theta = \frac{\pi}{2} + n\pi\) for \(n \in \mathbb{Z}\). The simplest solution is:
\[ 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4} \]

### Step 5: Calculate the dimensions
Substituting \(\theta = \frac{\pi}{4}\) back into the expressions for length and breadth:
\[ L = 2a \sin\left(\frac{\pi}{4}\right) = 2a \cdot \frac{1}{\sqrt{2}} = \frac{2a}{\sqrt{2}} = a\sqrt{2} \]
\[ B = 2b \cos\left(\frac{\pi}{4}\right) = 2b \cdot \frac{1}{\sqrt{2}} = \frac{2b}{\sqrt{2}} = b\sqrt{2} \]

### Final Dimensions
Thus, the dimensions of the rectangle of maximum area inscribed in the ellipse are:
\[ \text{Length} = a\sqrt{2}, \quad \text{Breadth} = b\sqrt{2} \]

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