The sides of the rectangle o.f greatest area which can be inscribed in the· ellipse `x^2/8 + y^2/4 = 1` are given by

Options

A. `(4,2sqrt2)`
B. `(2sqrt2,2)`
C. `(2, sqrt2)`
D. `(4sqrt2,4)`

Answer

To find the sides of the rectangle of greatest area that can be inscribed in the ellipse given by the equation \( \frac{x^2}{8} + \frac{y^2}{4} = 1 \), we can follow these steps:

### Step 1: Understand the Equation of the Ellipse
The equation of the ellipse can be rewritten as:
\[ x^2 = 8(1 - \frac{y^2}{4}) \quad \Rightarrow \quad x^2 = 8 - 2y^2 \]
This tells us that the semi-major axis is along the x-axis with length \( \sqrt{8} = 2\sqrt{2} \) and the semi-minor axis is along the y-axis with length \( \sqrt{4} = 2 \).

### Step 2: Define the Area of the Rectangle
Let the rectangle be inscribed in the ellipse with vertices at \( (h, k), (-h, k), (-h, -k), (h, -k) \). The area \( A \) of the rectangle is given by:
\[ A = 2h \cdot 2k = 4hk \]

### Step 3: Express \( k \) in Terms of \( h \)
From the ellipse equation, we have:
\[ \frac{h^2}{8} + \frac{k^2}{4} = 1 \]
Rearranging gives:
\[ k^2 = 4(1 - \frac{h^2}{8}) \quad \Rightarrow \quad k = \sqrt{4 - \frac{h^2}{2}} \]

### Step 4: Substitute \( k \) into the Area Formula
Substituting \( k \) into the area formula gives:
\[ A = 4h \sqrt{4 - \frac{h^2}{2}} \]

### Step 5: Differentiate the Area with Respect to \( h \)
To find the maximum area, we differentiate \( A \) with respect to \( h \):
\[ \frac{dA}{dh} = 4\sqrt{4 - \frac{h^2}{2}} + 4h \cdot \frac{d}{dh}\left(\sqrt{4 - \frac{h^2}{2}}\right) \]
Using the chain rule:
\[ \frac{d}{dh}\left(\sqrt{4 - \frac{h^2}{2}}\right) = \frac{-h}{\sqrt{4 - \frac{h^2}{2}}} \]
Thus,
\[ \frac{dA}{dh} = 4\sqrt{4 - \frac{h^2}{2}} - \frac{2h^2}{\sqrt{4 - \frac{h^2}{2}}} \]
Setting \( \frac{dA}{dh} = 0 \) for maximum area gives:
\[ 4\sqrt{4 - \frac{h^2}{2}} - \frac{2h^2}{\sqrt{4 - \frac{h^2}{2}}} = 0 \]
Multiplying through by \( \sqrt{4 - \frac{h^2}{2}} \) yields:
\[ 4(4 - \frac{h^2}{2}) = 2h^2 \]
Simplifying this results in:
\[ 16 - 2h^2 = 2h^2 \quad \Rightarrow \quad 16 = 4h^2 \quad \Rightarrow \quad h^2 = 4 \quad \Rightarrow \quad h = 2 \]

### Step 6: Find \( k \) when \( h = 2 \)
Substituting \( h = 2 \) back into the equation for \( k \):
\[ k = \sqrt{4 - \frac{2^2}{2}} = \sqrt{4 - 2} = \sqrt{2} \]

### Step 7: Calculate the Dimensions of the Rectangle
The dimensions of the rectangle are:
\[ \text{Length} = 2h = 2 \times 2 = 4 \]
\[ \text{Width} = 2k = 2 \times \sqrt{2} = 2\sqrt{2} \]

### Final Answer
The sides of the rectangle of greatest area that can be inscribed in the ellipse are \( 4 \) and \( 2\sqrt{2} \).