Question

If $y=\frac{2lnx}{3x},$ then y has a maximum at x equal to

• A. 0
• B. 1
• C. $\frac{1}{e}$
• D. e

Answer 1

The correct option is D e

$y=\frac{2}{3}\frac{lnx}{x}$

$\frac{dy}{dx}=\frac{2}{3x}\frac{1}{x}+\frac{2}{3}lnx\left(\frac{-1}{x^2}\right)$

$=\frac{2}{3x^2}(1-lnx)$

For maxima, $\frac{dy}{dx}=0$

$lnx=1\Rightarrow e$ is a stationary point

$\frac{d^{2}y}{d{x}^2}=\frac{-2}{3x^3}(3-2lnx)$

At x = e

${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=e}=\frac{-2}{3e^3}$

Hence maxima at x = e