Question

Maximum area of rectangle whose two sides are $x=x_{0}, x=\pi-x_{0}$ and which is inscribed in a region bounded by $y=\sin x$ and $x$ -axis is obtained when $x_{0} \in$

• A. $\left(\frac{\pi}{4}, \frac{\pi}{3}\right)$
• B. $\left(\frac{\pi}{6}, \frac{\pi}{4}\right)$
• C. $\left(0, \frac{\pi}{6}\right)$
• D. None of these

Answer 1

Answer: (B)

$\quad A =$ Area $=\sin x(\pi-2 x)$
$\Rightarrow \frac{d A}{d x}=(\pi-2 x) \cos x-2 \sin x=0 $
$\Rightarrow \tan x=\frac{\pi}{2}-x$

Let $f(x)=\tan x+x-\frac{\pi}{2}$
$\Rightarrow f\left(\frac{\pi}{6}\right)$ is negative; $f\left(\frac{\pi}{4}\right)$ is positive
So, one root lies between $\left(\frac{\pi}{6}, \frac{\pi}{4}\right)$.
[Note that in this interval $\frac{d^{2} A}{d x^{2}}$ is negative]