Answer is: option1

Solution:

Compute the derivative of \( y \) with respect to \( x \):

Using the product rule:

\( \frac{dy}{dx} = \frac{d}{dx} (xe^{-kx}) = e^{-kx} + x(-ke^{-kx}) = e^{-kx}(1 - kx) \)

Set the first derivative equal to zero to find critical points:

\( e^{-kx}(1 - kx) = 0 \)

Since \( e^{-kx} \neq 0 \), solve:

\( 1 - kx = 0 \Rightarrow x = \frac{1}{k} \)

Compute the second derivative to determine the nature of the critical point:

\( \frac{d^2y}{dx^2} = \frac{d}{dx}(e^{-kx}(1 - kx)) = -ke^{-kx}(1 - kx) + e^{-kx}(-k) = e^{-kx}(k^2x - 2k) \)

Evaluate the second derivative at \( x = \frac{1}{k} \):

\( \frac{d^2y}{dx^2} \Big|_{x=\frac{1}{k}} = e^{-1}(k^2 \cdot \frac{1}{k} - 2k) = e^{-1}(k - 2k) = -ke^{-1} < 0 \)

Since the second derivative is negative, the function has a local maximum at \( x = \frac{1}{k} \).

Substitute \( x = \frac{1}{k} \) back into the original function to find \( y \):

\( y = \frac{1}{k}e^{-k \cdot \frac{1}{k}} = \frac{1}{k}e^{-1} = \frac{1}{ke} \)

The coordinates of the absolute maximum point are \( \left( \frac{1}{k}, \frac{1}{ke} \right) \), which corresponds to option (A).