If the function `f(x)=ax e^(-bx)` has a local maximum at the point (2,10), then

Options

A. a = 5e
B. a = 5
C. b = 1
D. b = 1/2

Explanation

To solve the problem, we need to find the values of \( a \) and \( b \) for the function \( f(x) = ax e^{-bx} \) given that it has a local maximum at the point \( (2, 10) \).

### Step-by-Step Solution:

1. **Use the given point to set up an equation:**
Since \( f(2) = 10 \), we can substitute \( x = 2 \) into the function:
\[ f(2) = a(2)e^{-2b} = 10 \]
Simplifying this gives:
\[ 2a e^{-2b} = 10 \]
Dividing both sides by 2:
\[ a e^{-2b} = 5 \quad \text{(Equation 1)} \]

2. **Find the derivative of the function:**
To find the local maximum, we need to take the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(ax e^{-bx}) = a e^{-bx} + ax \frac{d}{dx}(e^{-bx}) \]
Using the product rule and the chain rule, we get:
\[ f'(x) = a e^{-bx} - abx e^{-bx} = e^{-bx}(a - abx) \]

3. **Set the derivative equal to zero at the maximum point:**
Since there is a local maximum at \( x = 2 \), we set \( f'(2) = 0 \):
\[ e^{-2b}(a - 2ab) = 0 \]
Since \( e^{-2b} \neq 0 \), we can simplify:
\[ a - 2ab = 0 \]
Factoring out \( a \):
\[ a(1 - 2b) = 0 \]
This gives us two cases:
- \( a = 0 \) (not valid since \( f(x) \) would be zero everywhere)
- \( 1 - 2b = 0 \) which leads to:
\[ b = \frac{1}{2} \quad \text{(Equation 2)} \]

4. **Substitute \( b \) back into Equation 1:**
Now we substitute \( b = \frac{1}{2} \) into Equation 1:
\[ a e^{-2(\frac{1}{2})} = 5 \]
This simplifies to:
\[ a e^{-1} = 5 \]
Therefore:
\[ a = 5e \quad \text{(Equation 3)} \]

5. **Final values:**
From Equations 2 and 3, we find:
\[ a = 5e, \quad b = \frac{1}{2} \]

### Summary of Results:
The values of \( a \) and \( b \) are:
\[ a = 5e, \quad b = \frac{1}{2} \]