Let $f(x)=\frac{\log x}{x}$
$f'(x)=\frac{1-\log x}{x^2}$
Set $f'(x)=0 \Rightarrow \log x=1 \Rightarrow x=e$
Maximum value $= \frac{\log e}{e}=\frac{1}{e}$
🎓 AMU MCA📅 Year: 2021📚 Mathematics🏷 Maxima and Minima
1
The greatest value of $f(x,y)=xy$ on ellipse
$ \frac{x^2}{8}+\frac{y^2}{2}=1 $