What is the maximum point on the curve `x=e^(x)y` ?

Options

A. (1,e)
B. `(1,e^(-1))`
C. (e,1)
D. `(e^(-1),1)`

Explanation

To find the maximum point on the curve given by the equation \( x = e^{xy} \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x = e^{xy} \]
Taking the natural logarithm of both sides, we can rewrite it as:
\[ \ln x = xy \]
From this, we can express \( y \) in terms of \( x \):
\[ y = \frac{\ln x}{x} \]

### Step 2: Differentiate \( y \)
Next, we differentiate \( y \) with respect to \( x \):
\[ y = \frac{\ln x}{x} \]
Using the quotient rule, where \( u = \ln x \) and \( v = x \):
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Calculating \( \frac{du}{dx} = \frac{1}{x} \) and \( \frac{dv}{dx} = 1 \):
\[ \frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2} \]

### Step 3: Find critical points
To find the maximum point, we set the derivative equal to zero:
\[ \frac{1 - \ln x}{x^2} = 0 \]
This implies:
\[ 1 - \ln x = 0 \quad \Rightarrow \quad \ln x = 1 \quad \Rightarrow \quad x = e \]

### Step 4: Determine the nature of the critical point
To confirm that this is a maximum, we can check the second derivative or use the first derivative test. We will differentiate \( \frac{dy}{dx} \) again:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{1 - \ln x}{x^2} \right) \]
Using the quotient rule again:
\[ \frac{d^2y}{dx^2} = \frac{x^2 \cdot \left(-\frac{1}{x}\right) - (1 - \ln x) \cdot 2x}{x^4} \]
Simplifying this gives:
\[ \frac{d^2y}{dx^2} = \frac{-x - 2(1 - \ln x)}{x^3} \]
Evaluating at \( x = e \):
\[ \frac{d^2y}{dx^2} \bigg|_{x=e} = \frac{-e - 2(1 - 1)}{e^3} = \frac{-e}{e^3} < 0 \]
Since the second derivative is negative, \( x = e \) is indeed a maximum.

### Step 5: Find the corresponding \( y \) value
Now we substitute \( x = e \) back into the equation for \( y \):
\[ y = \frac{\ln e}{e} = \frac{1}{e} \]

### Conclusion
Thus, the maximum point on the curve is:
\[ \text{Maximum point} = (e, \frac{1}{e}) \]