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)

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

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]