Answer
$V_{max}=54.7V$
Work Step by Step
$R_{eq}=\frac{1}{\frac{1}{2.5k\Omega}+\frac{1}{3.7k\Omega}}+1.4k\Omega=2.89k\Omega$
$P=I^2R$
$I_{max}=\sqrt{\frac{P_{max}}{R}}$
$I_{max}=I_1=\sqrt{\frac{0.5W}{1.4k\Omega}}=0.0189A$
$I_{max}=I_2+I_3$
$I_2=I_{max}-I_3$
$V_2=V_3$
$(I_{max}-I_3)R_2=I_3R_3$
$I_3=\frac{I_{max}R_2}{R_2+R_3}=\frac{(0.0189A)(2500\Omega)}{6200\Omega}=0.00762A$
$I_2=0.0113A$
To check if resistors 2 and 3 will exceed their maximum power output when resistor 1 is working at maximum power output, we need to find their power output.
$P_2=I_2^2R_2=(0.0113A)^2(2500\Omega)=0.318V$
$P_3=I_3^2R_3=(0.00762A)^2(3700\Omega)=0.215V$
Both resistors are working below their maximum power output, so the maximum voltage that can be applied across the whole network is
$V_{max}=I_{max}R_1+I_{max}R_{2,3}=(0.0189A)(1400\Omega)+(0.0189A)\Big(\frac{1}{\frac{1}{R_2}+\frac{1}{R_3}}\Big)=54.7V$
