Water stands at a depth H in a tank whose side walls are vertical. A hole is made in one of the walls at a height h below the water surface. The stream of water emerging from the hole strikes the floor at a distance R from the tank (given $R=2\sqrt{h(H-h)}$. R is maximum if

R will be maximum at value of h for which $\frac{dR}{dh}=0$ and $\frac{d^{2}R}{d{h}^2}<0$
Now $R=2{\left(hH-h^2\right)}^{1/2}$
Differentiating with respect to h, we have
$\frac{dR}{dh}=2\times \frac{1}{2}{\left(hH-h^2\right)}^{-1/2}$
$\frac{dR}{dh}=\frac{(H-2h)}{{\left(hH-h^2\right)}^{1/2}}$                (i)
It is clear that $\frac{dR}{dh}$ will be zero at a value of h given by 
$H-2h=0$
i.e. $h=\frac{H}{2}$
To find out whether $\frac{d^{2}R}{d{h}^2}$ is negative at this value of h, we differentiate Eq. (i) with respect to h to get
$\frac{d^{2}R}{d{h}^2}=-\frac{2h}{(H-h{)}^{1/2}}\left\{1+\frac{1}{4}\frac{(H-2h)}{h(H-h)}\right\}$
Putting h=H/2, we have
${\left(\frac{d^{2}R}{d{h}^2}\right)}_{h=H/2}=-\sqrt{2H}$
which is negative. Hènce R wili be maximum at h=H/2. Thus for range R to be maximum the hole must be exactly in the middle of the tank.