Question

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

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

Answer 1

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

Two sides of triangle are on positive x and y axis.

So, one vertex of the rectangle is the origin.

Let $(h,k)$ be the co-ordinates of the upper right hand vertex of the rectange.

So, the co-ordinates of the other two vertices are $(h,0)$ and $(0,k)$

$(h,k)$ lies on the curve $y=\frac{\ln x}{x^2}$

$\therefore k=\frac{\ln h}{h^2}$ ...... (1)

Area of the rectangle = $h\times k$

$\therefore A=h\times \frac{\ln h}{h^2}$ ....[Using (1)]

$\therefore A=\frac{\ln h}{h}$

For area to be maximum, $\frac{dA}{dh}=0$

$\therefore \frac{d\left(\frac{\ln h}{h}\right)}{dh}=0$

$\therefore \frac{1-\ln h}{h^2}=0$

$\therefore h=e$

$\therefore A=\frac{\ln h}{h}=\frac{\ln e}{e}=e^{-1}$

Hence, the answer is $e^{-1}$.