$f_v = \dfrac{VQ}{Ib}$
Where
$Q = \frac{1}{2}xy \, [ \, \frac{2}{3}(h - y) \, ] = \frac{1}{3}xy \, (h - y)$
$I = \frac{1}{36}bh^3$
$b = x$
Thus,
$f_v = \dfrac{V \, [ \,\frac{1}{3}xy \, (h - y) \, ]}{(\frac{1}{36}bh^3)(x)}$
$f_v = \dfrac{12Vy(h - y)}{bh^3}$
$f_v = \dfrac{12V}{bh^3}(hy - y^2)$
$\dfrac{df_v}{dy} = \dfrac{12V}{bh^3}(h - 2y) = 0$
$h = 2y$
$y = \frac{1}{2} h$
Maximum shearing stress occurs at the mid-height of the altitude, h
$(\,f_v\,)_{max} = \dfrac{12V}{bh^3} [ \, h(\frac{1}{2} h) - (\frac{1}{2} h)^2 \, ]$
$(\,f_v\,)_{max} = \dfrac{12V}{bh^3} [ \, \frac{1}{2}h^2 - \frac{1}{4}h^2 \, ]$
$(\,f_v\,)_{max} = \dfrac{12V}{bh^3} (\frac{1}{4}h^2)$
$(\,f_v\,)_{max} = \dfrac{3V}{bh}$ answer
