Find the maximum value of:
$\displaystyle f(x) = x^2 ln(1/x)$
$\displaystyle \frac{d}{dx} x^{2}\ln \Big(\frac{1}{x}\Big) = 2x \ln \Big(\frac{1}{x}\Big) +x^{2}*\frac{1}{1/x}\Big(\frac{-1}{x^{2}}\Big) = 2x \ln \Big(\frac{1}{x} \Big) -x$
$\displaystyle 2x \ln \Big(\frac{1}{x} \Big) -x = 0 $
$\displaystyle \ln \Big(\frac{1}{x} \Big) = \frac{1}{2} $
$\displaystyle x = \frac{1}{e^{1/2}} $