Question

The maximum value of area of rectangle $OABC$, where $O$ is origin, $A$ lies on positive $y-$axis and $B$ lies on the curve $y=\frac{\ln x}{x^2}$, is

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

Answer 1

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

Let the vertices of rectangle be $O(0,0),A(0,y),B(x,y),C(x,0)$ therefore, Area$A=xy=x\left(\frac{\ln x}{x^2}\right)=\frac{\ln x}{x}$
For maximum area ,$\frac{dA}{dx}=0$
$\Rightarrow \frac{1-\ln x}{x^2}=0$
$\Rightarrow x=e$
Now,$\frac{d^{2}A}{d{x}^2}=\frac{-(3-2\ln x)}{x^3}$
At $x=e$, $\frac{d^{2}A}{d{x}^2}<0$
therefore, $A$ is maximum at $x=e$
Area $=\frac{\ln x}{x}=\frac{\ln e}{e}=\frac{1}{e}$