(a) Find the sub-interval of (0, π) in which f (x)= tan^-1( sin x- cos x) is increasing and decreasing.

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

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

$\displaystyle \Rightarrow \cos x+\sin x=0$

getting $\displaystyle x=\dfrac{3\pi}{4} \in(0, \pi)$

$\displaystyle \mathrm{f}^{\prime}(x)>0$ when $\displaystyle \left(0, \dfrac{3\pi}{4}\right) \Rightarrow \mathrm{f}(x)$ is increasing when $\displaystyle x \in\left(0, \dfrac{3\pi}{4}\right)$.

$\displaystyle \mathrm{f}^{\prime}(x)<0$ when $\displaystyle \left(\dfrac{3\pi}{4}, \pi\right) \Rightarrow \mathrm{f}(x)$ is decreasing when $\displaystyle x \in\left(\dfrac{3\pi}{4}, \pi\right)$.

Let it be revolved around the side of length $\displaystyle (12-x)$, so that $\displaystyle x$ becomes the radius of the cylinder.

Volume of cylinder, $\displaystyle V=\pi x^{2}(12-x)=\pi\left(12 x^{2}-x^{3}\right)$

$\displaystyle \dfrac{dV}{dx}=\pi\left(24 x-3 x^{2}\right)$

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

$\displaystyle \Rightarrow x=8$ cm $\displaystyle (\because x \neq 0)$

Now $\displaystyle \dfrac{d^{2} V}{dx^{2}}=\pi(24-6 x)$ ; $\displaystyle \left.\dfrac{d^{2} V}{dx^{2}}\right]_{x=8\ \text{cm}}=-24\pi<0$

So, volume is maximum when $\displaystyle x=8$ cm.

Hence the dimensions of rectangle are $\displaystyle 8$ cm and $\displaystyle 4$ cm.