Answer
$10e$ is maximum
$0$ is minimum
Work Step by Step
$\frac{dy}{dx}$ = $10x(\frac{-1}{x})+(2-\ln{x})(10)$
$\frac{dy}{dx}$ = $10-10\ln{x}$
$\frac{dy}{dx}$ = $0$
$10-10\ln{x}$ = $0$
$x$ = $e$
if $x$ = $e$ then
$y$ = $10e(2-\ln{e})$ = $10e$ is maximum
if $x$ = $e^{2}$ then
$y$ = $10e^{2}(2-\ln{e^{2}})$ = $0$ is minimum
