Question

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

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

Answer 1

Answer: (B)

$\frac {1}{e}$

Answer 2

Let $y=x e^{-x}\,\,\,\,\,$
On differentiating w.r.t. ' $x$ ', we get
$\frac{d y}{d x}=x e^{-x}(-1)+e^{-x}$
$\frac{d y}{d x}=e^{-x}(1-x)$
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 E (ii), we get
$\frac{d^{2} y}{d x^{2}}=e^{-x}(-1)+(1-x) e^{-x}(-1)$
$=-e^{-x}-e^{-x}+x e^{-x}$
$=-2 e^{-x}+x e^{-x}$
$\frac{d^{2} y}{d x^{2}}=e^{-x}(x-2)$
$\left(\frac{d^{2} y}{d x^{2}}\right)_{\text {at } x=1} =e^{-1}(1-2)=\frac{1}{e}(-1)$
Negative value
$\therefore$ At $x=1, \,y$ is maximum and maximum value $=1 e^{-1}$
$=\frac{1}{e}$