A rectangle has one side on the positive y-axis and one side on the positive x -axis. The upper right hand vertex of the rectangle lies on the curve `y=(lnx)/x^2`. The maximum area of the rectangle is

Options

A. `e^(-1)`
B. `e^(-1//2)`
C. 1
D. `e^(1//2)`

Answer

To solve the problem of finding the maximum area of a rectangle with one side on the positive x-axis and one side on the positive y-axis, where the upper right vertex lies on the curve \( y = \frac{\ln x}{x^2} \), we can follow these steps:

### Step 1: Define the Area of the Rectangle
Let the coordinates of the upper right vertex of the rectangle be \( (x, y) \). The area \( A \) of the rectangle can be expressed as:
\[ A = x \cdot y \]
Since the vertex lies on the curve \( y = \frac{\ln x}{x^2} \), we can substitute this into the area formula:
\[ A = x \cdot \frac{\ln x}{x^2} = \frac{\ln x}{x} \]

### Step 2: Differentiate the Area Function
To find the maximum area, we need to differentiate \( A \) with respect to \( x \):
\[ A = \frac{\ln x}{x} \]
Using the quotient rule for differentiation, where \( u = \ln x \) and \( v = x \):
\[ \frac{dA}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Calculating the derivatives:
- \( \frac{du}{dx} = \frac{1}{x} \)
- \( \frac{dv}{dx} = 1 \)

Substituting these into the quotient rule gives:
\[ \frac{dA}{dx} = \frac{x \cdot \frac{1}{x} - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2} \]

### Step 3: Set the Derivative to Zero
To find the critical points, set the derivative equal to zero:
\[ \frac{1 - \ln x}{x^2} = 0 \]
This implies:
\[ 1 - \ln x = 0 \quad \Rightarrow \quad \ln x = 1 \]
Thus, solving for \( x \):
\[ x = e \]

### Step 4: Find the Maximum Area
Now we substitute \( x = e \) back into the area formula to find the maximum area:
\[ A = \frac{\ln e}{e} = \frac{1}{e} \]

### Conclusion
The maximum area of the rectangle is:
\[ \boxed{\frac{1}{e}} \]