Question

The maximum value of $x{e}^{-x}$ is

• A. $e$
• B. $\frac{1}{e}$
• C. $-e$
• D. $-\frac{1}{e}$

Answer 1

Answer: (B)

Let $y=x{e}^{-x}$
On differentiating w.r.t. ' $x$ ', we get
$\frac{dy}{dx}=x{e}^{-x}(-1)+e^{-x}$
$\frac{dy}{dx}=e^{-x}(1-x)$
For maximum or minimum,
$\frac{dy}{dx}=0$
$\Rightarrow e^{-x}(1-x)=0$
$\Rightarrow 1-x=0$
$\because e^{-x}\ne 0)$
$\Rightarrow x=1$
From Eq. (ii), we get
$\frac{d^{2}y}{d{x}^2}=e^{-x}(-1)+(1-x)e^{-x}(-1)$
$=-e^{-x}-e^{-x}+x{e}^{-x}$
$=-2e^{-x}+x{e}^{-x}$
$\frac{d^{2}y}{d{x}^2}=e^{-x}(x-2)$
${\left(\frac{d^{2}y}{d{x}^2}\right)}_{\text{at }x=1}=e^{-1}(1-2)=\frac{1}{e}(-1)$
Negative value
$\therefore$ At $x=1,y$ is maximum and maximum value $=1e^{-1}$
$=\frac{1}{e}$