Maximum output and efficiency for a simple transistor based amplifier circuit
What if we start with:

$$U_{a,eff,max} = \frac{U_0}{2\sqrt{2}}\frac{R_L}{R_L+R_C}$$

Then we get: 

$$P_L = \frac{(U_{a,eff,max})^2}{R_L}=\frac{(\frac{U_0}{2\sqrt{2}}\frac{R_L}{R_L+R_C})^2}{R_L} = \frac{U^2_0R^2_L}{8R_L(R_L+R_C)^2}=\frac{U^2_0R_L}{8(R_L+R_C)^2}$$

And with \$R_C=R_L\$ we get for\$P_{L,max}\$:

$$P_{L,max}=\frac{U^2_0}{32\cdot R_L}$$

For the circuit power we have to consider the power loss of \$R_C\$ and the losses of the transistor itself.

$$P_{R_C} = P_{R_C,DC}+P_{R_C,AC}$$
$$P_{R_C,DC} = \frac{U_{R_C,DC}^2}{R_C}=\frac{(\frac{1}{2}(U_{R_C,max}+U_{R_C,min}))^2}{R_C}=\frac{(\frac{1}{2}(U_0+U_{a,max}))^2}{R_C}$$

If we set again \$R_C=R_L\$ then \$U_{a,max} = \frac{1}{2}U_0\$ and \$U_{R_C,DC}=\frac{3}{4}U_0\$ we get:

$$P_{R_C,DC} = \frac{9U^2_0}{16R_C}$$

For the AC-Part follows:

$$P_{R_C,AC}=\frac{(U_{R_C,eff})^2}{R_C}=\frac{(\frac{U_{R_C,max}-U_{R_C,DC}}{\sqrt{2}})^2}{R_C}=\frac{(\frac{U_{0}-\frac{3}{4}U_{0}}{\sqrt{2}})^2}{R_C}=\frac{U^2_0}{32R_C}$$

Finlay we get:

$$P_{R_C} = \frac{9U^2_0}{16R_C} + \frac{U^2_0}{32R_C} = \frac{19U^2_0}{32R_C}$$

Next how do we get \$P_{transistor}\$ ?