Let P be the perimeter of rectangle of maximum area which can be inscribed inside ellipse `x^(2)/25+y^(2)/16=1` then evaluate `log_(3sqrt2)(3,P)`.

Answer

To solve the problem, we need to find the perimeter \( P \) of the rectangle of maximum area that can be inscribed in the ellipse given by the equation:

\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]

### Step 1: Understanding the ellipse
The ellipse is centered at the origin (0, 0) with a semi-major axis of 5 (along the x-axis) and a semi-minor axis of 4 (along the y-axis).

### Step 2: Setting up the rectangle
Let the vertices of the rectangle inscribed in the ellipse be at the points \( (x, y) \), \( (-x, y) \), \( (-x, -y) \), and \( (x, -y) \). The area \( A \) of the rectangle can be expressed as:

\[ A = 2x \cdot 2y = 4xy \]

### Step 3: Expressing \( y \) in terms of \( x \)
From the equation of the ellipse, we can express \( y \) in terms of \( x \):

\[ y = 4 \sqrt{1 - \frac{x^2}{25}} \]

### Step 4: Substituting \( y \) into the area formula
Substituting \( y \) into the area formula gives:

\[ A = 4x \cdot 4 \sqrt{1 - \frac{x^2}{25}} = 16x \sqrt{1 - \frac{x^2}{25}} \]

### Step 5: Maximizing the area
To maximize the area \( A \), we can differentiate it with respect to \( x \) and set the derivative equal to zero.

Let \( A = 16x \sqrt{1 - \frac{x^2}{25}} \).

Using the product rule and chain rule, we differentiate:

\[ \frac{dA}{dx} = 16 \left( \sqrt{1 - \frac{x^2}{25}} + x \cdot \frac{d}{dx} \left( \sqrt{1 - \frac{x^2}{25}} \right) \right) \]

Calculating \( \frac{d}{dx} \left( \sqrt{1 - \frac{x^2}{25}} \right) \):

\[ \frac{d}{dx} \left( \sqrt{1 - \frac{x^2}{25}} \right) = \frac{-\frac{2x}{25}}{2\sqrt{1 - \frac{x^2}{25}}} = -\frac{x}{25\sqrt{1 - \frac{x^2}{25}}} \]

Thus,

\[ \frac{dA}{dx} = 16 \left( \sqrt{1 - \frac{x^2}{25}} - \frac{x^2}{25\sqrt{1 - \frac{x^2}{25}}} \right) \]

Setting \( \frac{dA}{dx} = 0 \):

\[ \sqrt{1 - \frac{x^2}{25}} = \frac{x^2}{25\sqrt{1 - \frac{x^2}{25}}} \]

Squaring both sides and simplifying leads to:

\[ 25(1 - \frac{x^2}{25}) = x^2 \implies 25 - x^2 = x^2 \implies 25 = 2x^2 \implies x^2 = \frac{25}{2} \implies x = \frac{5}{\sqrt{2}} \]

### Step 6: Finding \( y \)
Substituting \( x \) back into the equation for \( y \):

\[ y = 4 \sqrt{1 - \frac{(\frac{5}{\sqrt{2}})^2}{25}} = 4 \sqrt{1 - \frac{25/2}{25}} = 4 \sqrt{1 - \frac{1}{2}} = 4 \cdot \frac{1}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \]

### Step 7: Finding the perimeter \( P \)
The perimeter \( P \) of the rectangle is given by:

\[ P = 2(2x + 2y) = 2(2 \cdot \frac{5}{\sqrt{2}} + 2 \cdot 2\sqrt{2}) = 2\left( \frac{10}{\sqrt{2}} + 4\sqrt{2} \right) = 2\left( 5\sqrt{2} + 4\sqrt{2} \right) = 2 \cdot 9\sqrt{2} = 18\sqrt{2} \]

### Step 8: Evaluating \( \log_{3\sqrt{2}}(3P) \)
Now we compute \( 3P \):

\[ 3P = 3 \cdot 18\sqrt{2} = 54\sqrt{2} \]

Now we need to evaluate:

\[ \log_{3\sqrt{2}}(54\sqrt{2}) \]

Using the change of base formula:

\[ \log_{3\sqrt{2}}(54\sqrt{2}) = \frac{\log(54\sqrt{2})}{\log(3\sqrt{2})} \]

Calculating \( \log(54\sqrt{2}) \):

\[ \log(54\sqrt{2}) = \log(54) + \log(\sqrt{2}) = \log(54) + \frac{1}{2}\log(2) \]

Calculating \( \log(3\sqrt{2}) \):

\[ \log(3\sqrt{2}) = \log(3) + \frac{1}{2}\log(2) \]

Thus, we have:

\[ \log_{3\sqrt{2}}(54\sqrt{2}) = \frac{\log(54) + \frac{1}{2}\log(2)}{\log(3) + \frac{1}{2}\log(2)} \]

### Final Result
After evaluating, we find that:

\[ \log_{3\sqrt{2}}(54\sqrt{2}) = 3 \]