(a) Find the sub intervals in which f (x)= cot^-1( sin x+ cos x), x ∈(0, π) is increasing and decreasing.

$\displaystyle f^{\prime}(x)=\dfrac{-(\cos x-\sin x)}{1+(\sin x+\cos x)^{2}}=\dfrac{\sin x-\cos x}{1+(\sin x+\cos x)^{2}}$

For critical points of $\displaystyle f(x)$, put $\displaystyle f^{\prime}(x)=0$

$\displaystyle \Rightarrow \sin x=\cos x \Rightarrow x=\dfrac{\pi}{4}$

$\displaystyle f^{\prime}(x)>0$ when $\displaystyle x \in\left(\dfrac{\pi}{4}, \pi\right) \Rightarrow f(x)$ is increasing in $\displaystyle \left(\dfrac{\pi}{4}, \pi\right)$.

$\displaystyle f^{\prime}(x)<0$ when $\displaystyle x \in\left(0, \dfrac{\pi}{4}\right) \Rightarrow f(x)$ is decreasing in $\displaystyle \left(0, \dfrac{\pi}{4}\right)$.

Let it be revolved around the side of length $\displaystyle (18-x)$, so that $\displaystyle x$ becomes the radius of the cylinder.

Volume of cylinder, $\displaystyle V=\pi x^{2}(18-x)=\pi\left(18 x^{2}-x^{3}\right)$

$\displaystyle \dfrac{dV}{dx}=\pi\left(36 x-3 x^{2}\right)$

For critical points, put $\displaystyle \dfrac{dV}{dx}=0$

$\displaystyle \Rightarrow x=12$ cm $\displaystyle (\because x \neq 0)$

Now $\displaystyle \dfrac{d^{2} V}{dx^{2}}=\pi(36-6 x)$ ; $\displaystyle \left.\dfrac{d^{2} V}{dx^{2}}\right|_{x=12\ \text{cm}}=-36\pi<0$

So, volume is maximum when $\displaystyle x=12$ cm.

Hence the dimensions of rectangle are $\displaystyle 12$ cm and $\displaystyle 6$ cm.