The largest area of a rectangle which has one side on the x-axis and the two vertices on the curve `y=e^(-x^(2)` is

Options

A. `sqrt2e^(-1//2)`
B. `2e^(-1//2)`
C. `e^(-1//2)`
D. None of these

Answer

To find the largest area of a rectangle that has one side on the x-axis and two vertices on the curve \( y = e^{-x^2} \), we can follow these steps:

### Step 1: Define the Area of the Rectangle
Let the rectangle have vertices at \( (a, 0) \) and \( (-a, 0) \) on the x-axis, and the top vertices at \( (a, e^{-a^2}) \) and \( (-a, e^{-a^2}) \) on the curve. The width of the rectangle is \( 2a \) and the height is \( e^{-a^2} \).

The area \( A \) of the rectangle can be expressed as:
\[ A = \text{width} \times \text{height} = 2a \cdot e^{-a^2} \]

### Step 2: Differentiate the Area Function
To find the maximum area, we need to differentiate \( A \) with respect to \( a \) and set the derivative equal to zero.

The area function is:
\[ A(a) = 2a e^{-a^2} \]

Using the product rule for differentiation:
\[ \frac{dA}{da} = 2e^{-a^2} + 2a \cdot \frac{d}{da}(e^{-a^2}) \]
Using the chain rule, we have:
\[ \frac{d}{da}(e^{-a^2}) = e^{-a^2} \cdot (-2a) \]
Thus,
\[ \frac{dA}{da} = 2e^{-a^2} - 4a^2 e^{-a^2} \]
Factoring out \( 2e^{-a^2} \):
\[ \frac{dA}{da} = 2e^{-a^2}(1 - 2a^2) \]

### Step 3: Set the Derivative Equal to Zero
Setting the derivative equal to zero to find critical points:
\[ 2e^{-a^2}(1 - 2a^2) = 0 \]
Since \( e^{-a^2} \) is never zero, we can set:
\[ 1 - 2a^2 = 0 \]
Solving for \( a^2 \):
\[ 2a^2 = 1 \implies a^2 = \frac{1}{2} \implies a = \frac{1}{\sqrt{2}} \text{ (taking the positive root)} \]

### Step 4: Calculate the Maximum Area
Now we substitute \( a = \frac{1}{\sqrt{2}} \) back into the area formula:
\[ A = 2a e^{-a^2} = 2 \cdot \frac{1}{\sqrt{2}} \cdot e^{-\left(\frac{1}{\sqrt{2}}\right)^2} \]
Calculating \( e^{-\left(\frac{1}{\sqrt{2}}\right)^2} = e^{-\frac{1}{2}} \):
\[ A = 2 \cdot \frac{1}{\sqrt{2}} \cdot e^{-\frac{1}{2}} = \sqrt{2} e^{-\frac{1}{2}} \]

### Final Answer
The largest area of the rectangle is:
\[ A = \sqrt{2} e^{-\frac{1}{2}} \]

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