Assume h_{FE3} = h_{FE4} =
100. This means that the total h_{FE} of each
darlington is 2000. I_{RL,max} = 1.4A =>
I_{E1,max} = 1.4A. I_{B1,max} =
I_{E1,max}/(h_{FE}+1) = 1.4A/2001
= 0.7mA = I_{R1}. V_{B1,max} =
11.3 + 2∙0.7 = 12.7V => V_{R1} = VS -
V_{B1} = 15 - 12.7 = 2.3V. R1 =
V_{R1}/I_{R1} = 2.3V/0.7mA =
3.3kΩ.

Let's now calculate the maximum power dissipation in T1 and T2.
P_{T1} =
V_{CE1}∙I_{E1} = (VS -
V_{RL})∙(V_{RL}/RL) =
(VS∙V_{RL} -
V^{2}_{RL})/RL.
P_{T1} reaches its maximum value if
dP_{T1}/dV_{RL} = VS -
2∙V_{RL} = 0 => V_{RL} = VS/2 =
7.5V. P_{T1,max} = (15∙7.5 -
7.5^{2})/8 = 7W. Since T1 works only during the
positive half periods, T1's (and T2's) maximum dissipation will be
3.5W.