## University Calculus: Early Transcendentals (3rd Edition)

$x=e$ and $y=\dfrac{1}{e^2}$ giving the maximum area $A=\frac{1}{e}$
Let the area of a rectangle be $a=xy$ Here, $a= x(\dfrac{\ln x}{x^2})=\dfrac{\ln x}{x}$ Then $a'=\dfrac{1-\ln x}{x^2}=0$ This implies that $x=e$ Now, $y=(\dfrac{\ln x}{x^2})=(\dfrac{\ln e}{e^2})=\dfrac{1}{e^2}$ Hence, $x=e$ and $y=\dfrac{1}{e^2}$ giving the maximum area $A=\frac{1}{e}$