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(A) Let the upper right vertex of the rectangle be $(x, y)$ on the curve $y = \frac{\ln x}{x^2}$.Since the rectangle has sides along the positive $x$

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$e^{-1}$

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(A) Let the upper right vertex of the rectangle be $(x, y)$ on the curve $y = \frac{\ln x}{x^2}$.Since the rectangle has sides along the positive $x$ and $y$ axes,its area $A$ is given by $A = x \cdot y$.Substituting the expression for $y$,we get $A = x \cdot \frac{\ln x}{x^2} = \frac{\ln x}{x}$.To find the maximum area,we differentiate $A$ with respect to $x$:$\frac{dA}{dx} = \frac{x \cdot (\frac{1}{x}) - \ln x \cdot (1)}{x^2} = \frac{1 - \ln x}{x^2}$.Setting $\frac{dA}{dx} = 0$,we get $1 - \ln x = 0$,which implies $\ln x = 1$,so $x = e$.To verify this is a maximum,we check the second derivative or the sign change of $\frac{dA}{dx}$. For $x  0$,and for $x > e$,$\frac{dA}{dx} The maximum area is $A_{max} = \frac{\ln e}{e} = \frac{1}{e} = e^{-1}$.

$A$ rectangle has one side on the positive $y-$ axis and one side on the positive $x-$ axis. The upper right hand vertex lies on the curve $y = \frac{\ln x}{x^2}$. The maximum area of the rectangle is

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