(a) Find the sub-interval of (0, (π)/2) in which f (x)= log ( sin x+ cos x) is increasing and decreasing.

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

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

$\displaystyle \Rightarrow \dfrac{\cos x-\sin x}{\sin x+\cos x}=0 \Rightarrow \cos x=\sin x$

$\displaystyle \Rightarrow \tan x=1$

$\displaystyle \Rightarrow x=\dfrac{\pi}{4} \in\left(0, \dfrac{\pi}{2}\right)$

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

$\displaystyle \mathrm{f}^{\prime}(x)<0$ when $\displaystyle x \in\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right) \Rightarrow \mathrm{f}(x)$ is decreasing in $\displaystyle \left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)$.

Let it be revolved around the side of length $\displaystyle (15-x)$, so that $\displaystyle x$ becomes the radius of the cylinder.

Volume of cylinder, $\displaystyle V=\pi x^{2}(15-x)=\pi\left(15 x^{2}-x^{3}\right)$

$\displaystyle \dfrac{dV}{dx}=\pi\left(30 x-3 x^{2}\right)$

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

$\displaystyle \Rightarrow x=10$ cm $\displaystyle (\because x \neq 0)$

Now $\displaystyle \dfrac{d^{2} V}{dx^{2}}=\pi(30-6 x)$ ; $\displaystyle \left.\dfrac{d^{2} V}{dx^{2}}\right]_{x=10\ \text{cm}}=-30\pi<0$

So, volume is maximum when $\displaystyle x=10$ cm.

Hence the dimensions of rectangle are $\displaystyle 10$ cm and $\displaystyle 5$ cm.