The task is to calculate the small-signal voltage gain based on \$R_s\$, \$R_c\$, \$I_c\$, \$V_A\$, and \$ \beta \$ for the given schematic diagram. Therefore, the small-signal model you have drawn does not correspond to the task because you have used small-signal parameters (\$r_\pi \$, \$g_m\$, and \$r_o\$) different from the required ones. Also, you do not specify anywhere what \$V_A\$ means, neither you use it and I must confess I do not have any idea as to its meaning here (any suggestion?). However, the designation \$ \beta \$ is normally used as a synonym for the hybrid small-signal parameter \$h_{21e}\$ (the suffix e indicates the common emitter topology), so it is obvious the small-signal model based on the \$h_e\$-parameters was assumed to be used for the calculations. The explored circuit including the mentioned full small-signal model then looks like this:
simulate this circuit – Schematic created using CircuitLab
The corresponding equations of that model are as follows:
\$ v_1 = h_{11e} \cdot i_1 + h_{12e} \cdot v_2 \$
\$ i_2 = h_{21e} \cdot i_1 + h_{22e} \cdot v_2 \$
and the equations for the whole explored circuit can obviously be written as:
\$ V_i = (R_s + h_{11e}) \cdot i_1 + h_{12e} \cdot V_o \$
\$ -\frac{V_o}{R_c} = h_{21e} \cdot i_1 + h_{22e} \cdot V_o \$
From the above equations there can be derived the small-signal voltage gain \$ A_v \$ of the circuit:
\$ A_v = \frac{V_o}{V_i} = \frac{1}{h_{12e} - \frac{(R_s + h_{11e}) \cdot (h_{22e} + \frac{1}{R_c})}{h_{21e}}} = \frac{-h_{21e} \cdot R_c}{(R_s + h_{11e}) \cdot (h_{22e} \cdot R_c + 1) - h_{12e} \cdot h_{21e}} = \frac{-\beta \cdot R_c}{(R_s + h_{11e}) \cdot (h_{22e} \cdot R_c + 1) - h_{12e} \cdot \beta} \$
In the first approximation, let's presume both \$ h_{12e} \rightarrow 0 \$ and \$ h_{22e} \rightarrow 0 \$, then the formula would simplify to:
\$ A_v = \frac{-h_{21e} \cdot R_c}{R_s + h_{11e}} = \frac{-\beta \cdot R_c}{R_s + h_{11e}} \$
To find the \$I_c \$ at which the voltage gain has its maximum, the h-parameters as functions of \$I_c \$ have to be known (they are nonlinear functions, of course; W. Shockley transistor equations). Then the first derivative \$ \frac{\partial A_v}{\partial I_c}\$ of the function \$ A_v = f(I_c) \$ has to be set to 0 and the corresponding \$I_c(\frac{\partial A_v}{\partial I_c}=0) \$ expressed out of it. The second condition for the gain to be a maximum is that the second derivative of the same function at that calculated \$I_c \$ has to be less than 0. In fact, if the function in charge is known, it can also be plotted and its maximum found this way.
Another view: the maximum theoretical accessible amplitude of the amplified signal is \$ \frac{V_{cc}}{2} \$, it means the operating point must be set to \$ I_c = \frac{V_{cc}}{2R_c} \$.
\$instead of just$to start and end inline math, since prices of things are often on-topic here. \$\endgroup\$ – The Photon Aug 4 '17 at 15:43