A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by

(i) 

Given : $M=\frac{W L}{2} x-\frac{W}{2} x^{2}$

$\Rightarrow \frac{d M}{d x}=\frac{W L}{2}-2 \times \frac{W x}{2}$

$\Rightarrow \frac{d M}{d x}=\frac{W L}{2}-W x$

For maximum or minimum values of $M$, we must have

$\frac{d M}{d x}=0$

$\Rightarrow \frac{W L}{2}-W x=0$

$\Rightarrow \frac{W L}{2}=W x$

$\Rightarrow x=\frac{L}{2}$

Now,

$\frac{d^{2} M}{d x^{2}}=-W<0$

So, $M$ is maximum at $x=\frac{L}{2}$.

(ii) 

Given : $M=\frac{W x}{3}-\frac{W x^{3}}{3 L^{2}}$

$\Rightarrow \frac{d M}{d x}=\frac{W}{3}-3 \times \frac{W x^{2}}{3 L^{2}}$

$\Rightarrow \frac{d M}{d x}=\frac{W}{3}-\frac{W x^{2}}{L^{2}}$

For maximum or minimum values of $M$, we must have

$\frac{d M}{d x}=0$

$\Rightarrow \frac{W}{3}-\frac{W x^{2}}{L^{2}}=0$

$\Rightarrow \frac{W}{3}=\frac{W x^{2}}{L^{2}}$

$\Rightarrow x=\frac{L}{\sqrt{3}}$

Now,

$\frac{d^{2} M}{d x^{2}}=-\frac{2 W x}{L^{2}}<0$

So, $M$ is maximum at $x=\frac{L}{\sqrt{3}}$.