Point A lies on the curve `y = e^(x^2)` and has the coordinate `(x,e^(-x^2))` where `x>0.` Point B has the coordinates `(x, 0).` If `'O'` is the origin, then the maximum area of the `DeltaAOB` is

Options

A. `(1)/(sqrt8e)`
B. `(1)/(sqrt4e)`
C. `(1)/(sqrt2e)`
D. `(1)/(sqrte)`

Answer

To find the maximum area of triangle AOB, where point A lies on the curve \( y = e^{x^2} \) at coordinates \( (x, e^{-x^2}) \) and point B has coordinates \( (x, 0) \), we will follow these steps:

### Step 1: Identify the Area of Triangle AOB
The area \( A \) of triangle AOB can be calculated using the formula for the area of a triangle:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In this case, the base is the distance from the origin O to point B, which is \( x \), and the height is the y-coordinate of point A, which is \( e^{-x^2} \). Therefore, the area can be expressed as:
\[ A = \frac{1}{2} \times x \times e^{-x^2} \]

### Step 2: Differentiate the Area with Respect to x
To find the maximum area, we need to differentiate \( A \) with respect to \( x \):
\[ A = \frac{1}{2} x e^{-x^2} \]
Using the product rule, we differentiate:
\[ \frac{dA}{dx} = \frac{1}{2} \left( e^{-x^2} + x \frac{d}{dx}(e^{-x^2}) \right) \]
The derivative of \( e^{-x^2} \) is \( -2x e^{-x^2} \), so:
\[ \frac{dA}{dx} = \frac{1}{2} \left( e^{-x^2} - 2x^2 e^{-x^2} \right) \]
This simplifies to:
\[ \frac{dA}{dx} = \frac{e^{-x^2}}{2} (1 - 2x^2) \]

### Step 3: Set the Derivative to Zero
To find the critical points, we set \( \frac{dA}{dx} = 0 \):
\[ \frac{e^{-x^2}}{2} (1 - 2x^2) = 0 \]
Since \( e^{-x^2} \) is never zero, we solve:
\[ 1 - 2x^2 = 0 \implies 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \frac{1}{\sqrt{2}} \]

### Step 4: Find the Maximum Area
Now, we substitute \( x = \frac{1}{\sqrt{2}} \) back into the area formula to find the maximum area:
\[ A = \frac{1}{2} \times \frac{1}{\sqrt{2}} \times e^{-\left(\frac{1}{\sqrt{2}}\right)^2} = \frac{1}{2} \times \frac{1}{\sqrt{2}} \times e^{-\frac{1}{2}} \]
This simplifies to:
\[ A = \frac{1}{2\sqrt{2}} \times \frac{1}{\sqrt{e}} = \frac{1}{2\sqrt{2e}} \]

### Final Result
Thus, the maximum area of triangle AOB is:
\[ \boxed{\frac{1}{2\sqrt{2e}}} \]