Question

Determine the largest area of the rectangle whose base is on the x-axis and two of its vertices lie on the curve $y=e^{-x^2}$.

• A. $\sqrt{2}e$
• B. $\sqrt{\frac{e}{2}}$
• C. $\sqrt{\frac{2}{e}}$
• D. $\sqrt{e}$

Answer 1

Answer: (C)

$\sqrt{\frac{2}{e}}$

The curve of $y=e^{-x^2}$ will be symmetric about the origin.
Therefore let the vertices of the rectangle lying on the x axis be $(a,0)$ and $(-a,0)$.
Then the other two vertices lying on the curve will be $(a,e^{-a^2})$ and $(-a,e^{-a^2})$.
Hence the vertices of the rectangle in clockwise sense will be
$(a,0),(-a,0),(-a,e^{-a^2}),(a,e^{-a^2})$.
Therefore the length of the base of the rectangle will be
$=a-(-a)=2a$
And height will be equal to
$\sqrt{(a-a{)}^2+(0-e^{-a^2}{)}^2}$
$=e^{-a^2}$.
Hence the total area enclosed by the rectangle will be
$A=2a.e^{-a^2}$.
Differentiating the above expression with respect to '$a$'.
$\frac{dA}{da}=2e^{-a^2}-4a^2{e}^{-a^2}$.
$=2e^{-a^2}(1-2a^2)$
$=0$
Hence
$a=\frac{\pm 1}{\sqrt{2}}$.
Hence the required area will be
$=2a.e^{-a^2}$.
$=\frac{2}{\sqrt{2}}.e^{-\frac{1}{2}}$
$=\sqrt{2}.\frac{1}{\sqrt{e}}$
$=\sqrt{\frac{2}{e}}$.