Question

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

• A. $e^{-1}$
• B. $e^{-\frac{1}{2}}$
• C. $1$
• D. $e^{\frac{1}{2}}$

Answer 1

The correct option is A $e^{-1}$

Area of rectangle is,$A=\frac{lnx}{x}$
$\Rightarrow A^{\prime }=\frac{1-\ln x}{x^2}=0$
For minimum or maximum area $A^{\prime }=0\Rightarrow x=e$
Now $A"=\frac{-x-2x(2-\ln x)}{x^2}$
Clearly $A"\left(e\right)<0$
Thus at $x=e$ area will be maximum
$\Rightarrow A_{max}=\frac{1}{e}$