The sides of the rectangle of greatest area that can be inscribed in the ellipse `x^(2)+4y^(2)=64` are

Options

A. `6sqrt(2),4sqrt(2)`
B. `8sqrt(2),4sqrt(2)`
C. `8sqrt(2),8sqrt(2)`
D. `16sqrt(2),4sqrt(2)`

Answer

To solve the problem of finding the sides of the rectangle of greatest area that can be inscribed in the ellipse given by the equation \( x^2 + 4y^2 = 64 \), we can follow these steps:

### Step 1: Write the equation of the ellipse in standard form
The given equation is:
\[ x^2 + 4y^2 = 64 \]
To convert this into standard form, we divide the entire equation by 64:
\[ \frac{x^2}{64} + \frac{4y^2}{64} = 1 \]
This simplifies to:
\[ \frac{x^2}{64} + \frac{y^2}{16} = 1 \]
Thus, the semi-major axis \( a = 8 \) and the semi-minor axis \( b = 4 \).

### Step 2: Express the area of the rectangle
Let the rectangle be inscribed in the ellipse with vertices at \( (x, y) \), \( (-x, y) \), \( (-x, -y) \), and \( (x, -y) \). The area \( A \) of the rectangle can be expressed as:
\[ A = \text{length} \times \text{breadth} = (2x)(2y) = 4xy \]

### Step 3: Use the constraint from the ellipse
From the ellipse equation, we can express \( y \) in terms of \( x \):
\[ 4y^2 = 64 - x^2 \implies y^2 = \frac{64 - x^2}{4} \implies y = \sqrt{\frac{64 - x^2}{4}} = \frac{\sqrt{64 - x^2}}{2} \]

### Step 4: Substitute \( y \) in the area formula
Substituting \( y \) into the area formula gives:
\[ A = 4x \left(\frac{\sqrt{64 - x^2}}{2}\right) = 2x\sqrt{64 - x^2} \]

### Step 5: Differentiate the area function
To find the maximum area, we differentiate \( A \) with respect to \( x \):
\[ \frac{dA}{dx} = 2\sqrt{64 - x^2} + 2x \cdot \frac{-x}{\sqrt{64 - x^2}} = 2\sqrt{64 - x^2} - \frac{2x^2}{\sqrt{64 - x^2}} \]
Setting the derivative equal to zero to find critical points:
\[ 2\sqrt{64 - x^2} - \frac{2x^2}{\sqrt{64 - x^2}} = 0 \]
Multiplying through by \( \sqrt{64 - x^2} \) (assuming \( \sqrt{64 - x^2} \neq 0 \)):
\[ 2(64 - x^2) - 2x^2 = 0 \implies 128 - 4x^2 = 0 \implies 4x^2 = 128 \implies x^2 = 32 \implies x = 4\sqrt{2} \]

### Step 6: Find \( y \) using the value of \( x \)
Substituting \( x = 4\sqrt{2} \) back into the equation for \( y \):
\[ y = \frac{\sqrt{64 - (4\sqrt{2})^2}}{2} = \frac{\sqrt{64 - 32}}{2} = \frac{\sqrt{32}}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \]

### Step 7: Calculate the dimensions of the rectangle
The sides of the rectangle are:
- Length: \( 2x = 2 \times 4\sqrt{2} = 8\sqrt{2} \)
- Breadth: \( 2y = 2 \times 2\sqrt{2} = 4\sqrt{2} \)

### Final Answer
Thus, the sides of the rectangle of greatest area that can be inscribed in the ellipse are:
\[ \text{Length} = 8\sqrt{2}, \quad \text{Breadth} = 4\sqrt{2} \]