The largest area of a rectangle which has one side on the $x$-axis and the two vertices on the curve $\mathrm{y}={\mathrm{e}}^{-{\mathrm{x}}^2}$ is

It is Given that two vertices of the rectangle lie on $y=e^{-x^2}$and other two vertices of the rectangle lie on the $x-$axis.

Let the Vertices of the rectangle on the $x$-axis be  $(\mathrm{a},0) \mathrm{and}(-\mathrm{a},0)$

Thus, the Vertices of the rectangle on the curve $y=e^{-x^2}$ will be  $\left(a,e^{-a^2}\right)\text{ and }\left(-a,e^{-a^2}\right)$

So, all the Vertices of the rectangle are $(a,0),(-a,0),\left(-a,e^{-a^2}\right),\left(a,e^{-a^2}\right)$

Length of the rectangle $=\mathrm{a}-\left(-\mathrm{a}\right)=2\mathrm{a}$

Breadth of the rectangle will be ${\mathrm{e}}^{-{\mathrm{a}}^2}$

As we know the area of the rectangle is given as $A=L\times B$

$\Rightarrow \mathrm{A}=2\mathrm{a}\times {\mathrm{e}}^{-{\mathrm{a}}^2}$

Differentiating area with respect to $a$ we get, $\frac{\mathrm{dA}}{\mathrm{da}}=2{\mathrm{e}}^{-{\mathrm{a}}^2}-4{\mathrm{a}}^2{\mathrm{e}}^{-{\mathrm{a}}^2}$

$\Rightarrow \frac{\mathrm{dA}}{\mathrm{da}}=2{\mathrm{e}}^{-{\mathrm{a}}^2}\left(1-2{\mathrm{a}}^2\right)$

To maximize the area put , $\frac{dA}{da}=0$

So, $2{\mathrm{e}}^{-{\mathrm{a}}^2}\left(1-2{\mathrm{a}}^2\right)=0$

Since ${\mathrm{e}}^{-{\mathrm{a}}^2}\ne -0 \forall \mathrm{a}\in \mathrm{R}$

$\Rightarrow a=\frac{\pm 1}{\sqrt{2}}$

Replacing the value of $a$ in the equation of Area of the rectangle we get $A=2a\times {\mathrm{e}}^{-{\mathrm{a}}^2}$

$\Rightarrow \mathrm{A}=2\times \frac{1}{\sqrt{2}}\times {\mathrm{e}}^{-{\left(\frac{1}{\sqrt{2}}\right)}^2}$

$\Rightarrow A=\sqrt{2}\times {\mathrm{e}}^{-\frac{1}{2}}$

Hence option-$1$ is the correct answer.