Answer is: option3

Solution:

First, let's write out what \(xy\) is:

\[ xy = x \cdot y = x \left( \frac{1}{\sqrt{x}} - \sqrt{x} \right) \]

Simplify the expression:

\[ xy = x \cdot \frac{1}{\sqrt{x}} - x \cdot \sqrt{x} = \sqrt{x} - x^{3/2} \]

So, the product \(xy\) is:

\[ xy = \sqrt{x} - x^{3/2} \]

To find the maximum value, we'll take the derivative of \(xy\) with respect to \(x\) and set it to zero.

First, rewrite \(xy\) using exponents:

\[ xy = x^{1/2} - x^{3/2} \]

Now, take the derivative:

\[ \frac{d}{dx}(xy) = \frac{d}{dx}\left(x^{1/2} - x^{3/2}\right) = \frac{1}{2}x^{-1/2} - \frac{3}{2}x^{1/2} \]

Set \( \frac{d}{dx}(xy) = 0 \):

\[ \frac{1}{2}x^{-1/2} - \frac{3}{2}x^{1/2} = 0 \]

Let's solve for \(x\):

\[ \frac{1}{2}x^{-1/2} = \frac{3}{2}x^{1/2} \]

Divide both sides by \( \frac{1}{2} \):

\[ x^{-1/2} = 3x^{1/2} \]

Recall that \( x^{-1/2} = \frac{1}{x^{1/2}} \), so:

\[ \frac{1}{x^{1/2}} = 3x^{1/2} \]

Multiply both sides by \(x^{1/2}\):

\[ 1 = 3x \Rightarrow x = \frac{1}{3} \]

Now, evaluate the second derivative to ensure it is a maximum:

\[ \left.\frac{d^2}{dx^2}(xy)\right|_{x = 1/3} = -\frac{1}{4} \cdot 3\sqrt{3} - \frac{3}{4} \cdot \sqrt{3} = \frac{3\sqrt{3}}{4} - \frac{6\sqrt{3}}{4} = -\frac{3\sqrt{3}}{4} \]

Since the second derivative is negative at \( x = \frac{1}{3} \), this confirms that the function \( xy \) has a maximum at this point.

Now, substitute \( x = \frac{1}{3} \) back into the expression for \( xy \):

\[ xy = \sqrt{\frac{1}{3}} - \left(\frac{1}{3}\right)^{3/2} = \frac{1}{\sqrt{3}} - \frac{1}{3\sqrt{3}} = \frac{2}{3\sqrt{3}} \]

Rationalize the denominator:

\[ xy = \frac{2}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{9} \]

The maximum value of \( xy \) is \( \frac{2\sqrt{3}}{9} \), which corresponds to option (C).