The dimension of the rectangle of maximum area that can be inscribed in the ellipse ` (x//4)^(2) +(y//3)^(2) =1` are

Options

A. `sqrt(8),sqrt(2)`
B. `4,3`
C. `2sqrt(8),3sqrt(2)`
D. `sqrt(2),sqrt(6)`

Answer

To find the dimensions of the rectangle of maximum area that can be inscribed in the ellipse given by the equation \(\frac{x^2}{4^2} + \frac{y^2}{3^2} = 1\), we will follow these steps:

### Step 1: Identify Parameters of the Ellipse
The standard form of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
From the given equation, we can identify:
- \(a = 4\)
- \(b = 3\)

### Step 2: Parametric Representation of the Ellipse
A point \((x, y)\) on the ellipse can be represented parametrically as:
\[ x = a \cos \theta = 4 \cos \theta \]
\[ y = b \sin \theta = 3 \sin \theta \]

### Step 3: Dimensions of the Rectangle
The rectangle inscribed in the ellipse will have vertices at:
\[ (4 \cos \theta, 3 \sin \theta), (-4 \cos \theta, 3 \sin \theta), (4 \cos \theta, -3 \sin \theta), (-4 \cos \theta, -3 \sin \theta) \]
The length \(L\) and breadth \(B\) of the rectangle can be expressed as:
\[ L = 2 \times (4 \cos \theta) = 8 \cos \theta \]
\[ B = 2 \times (3 \sin \theta) = 6 \sin \theta \]

### Step 4: Area of the Rectangle
The area \(A\) of the rectangle is given by:
\[ A = L \times B = (8 \cos \theta)(6 \sin \theta) = 48 \cos \theta \sin \theta \]
Using the identity \(2 \sin \theta \cos \theta = \sin(2\theta)\), we can rewrite the area as:
\[ A = 24 \sin(2\theta) \]

### Step 5: Maximizing the Area
To find the maximum area, we need to maximize \(\sin(2\theta)\). The maximum value of \(\sin(2\theta)\) is 1, which occurs when:
\[ 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4} \]

### Step 6: Calculate Maximum Area
Substituting \(\theta = \frac{\pi}{4}\) into the area formula:
\[ A_{\text{max}} = 24 \times 1 = 24 \]

### Step 7: Dimensions at Maximum Area
Now we can find the dimensions of the rectangle at this maximum area:
- Length:
\[ L = 8 \cos\left(\frac{\pi}{4}\right) = 8 \times \frac{1}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2} \]
- Breadth:
\[ B = 6 \sin\left(\frac{\pi}{4}\right) = 6 \times \frac{1}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \]

### Final Answer
The dimensions of the rectangle of maximum area that can be inscribed in the ellipse are:
\[ \text{Length} = 4\sqrt{2}, \quad \text{Breadth} = 3\sqrt{2} \]