Question

Maximum value of the function \( f(r) = r^2 e^{-r} \), when \( 0<r<\infty \) is

• A. \( 4 e^{-2} \)
• B. \( e^{-1} \)
• C. \( 2 e^{-\sqrt{2}} \)
• D. \( 4 e^{-\sqrt{2}} \)

Hint:

To find the maximum of a function, first take the derivative, set it equal to zero, and solve for the critical points. Then, check the values of the function at these points.

Answer 1

The Correct Option is A

Step 1: Analyze the function.
The given function is \( f(r) = r^2 e^{-r} \). To find its maximum value, we need to take the first derivative, \( f'(r) \), and set it equal to zero.
Step 2: Differentiate the function.
Using the product rule of differentiation, we get: \[ f'(r) = \frac{d}{dr}(r^2) \cdot e^{-r} + r^2 \cdot \frac{d}{dr}(e^{-r}) \] \[ f'(r) = 2r e^{-r} - r^2 e^{-r} \] Simplify: \[ f'(r) = e^{-r}(2r - r^2) \] Step 3: Set the derivative equal to zero to find the critical points.
\[ f'(r) = 0 \Rightarrow e^{-r}(2r - r^2) = 0 \] Since \( e^{-r} \neq 0 \) for any \( r \), we have: \[ 2r - r^2 = 0 \] Factor: \[ r(2 - r) = 0 \] This gives us two critical points: \[ r = 0 \quad \text{or} \quad r = 2 \] Step 4: Verify which critical point gives the maximum value.
To determine whether \( r = 2 \) or \( r = 0 \) gives the maximum, we can check the second derivative or simply evaluate \( f(r) \) at both points.
- At \( r = 0 \), \( f(0) = 0^2 e^{0} = 0 \).
- At \( r = 2 \), \( f(2) = 2^2 e^{-2} = 4 e^{-2} \).
Thus, the maximum value of \( f(r) \) is \( 4 e^{-2} \).
Final Answer: \[ \boxed{4 e^{-2}} \]