If the function $$f(x) = \left(\frac{1}{x}\right)^{2x}; x \gt 0$$ attains the maximum value at $$x = \frac{1}{e}$$ then :

We are given that $$f(x)$$ is maximized at $$x = 1/e$$. This implies that for any $$x \neq 1/e$$, $$f(1/e) > f(x)$$.

Let's look at a simpler related function $$g(x) = x^{1/x}$$. It is well known (via $$g'(x)$$) that $$x^{1/x}$$ is maximized at $$x = e$$.

Step 2: Compare the values.

Since $$e < \pi$$, and the function $$g(x) = x^{1/x}$$ is strictly decreasing for $$x > e$$:

$$e^{1/e} > \pi^{1/\pi}$$

Raise both sides to the power of $$e\pi$$:

$$(e^{1/e})^{e\pi} > (\pi^{1/\pi})^{e\pi} \implies e^\pi > \pi^e$$

Correct Option: B ($$e^\pi > \pi^e$$)