By Kirchhoff

#i_1=i_2+i_3#

Along a loop

#V_b=R_1i_1+R_2i_2#

Along the other loop

#R_3i_3-V_b-R_2i_2=0#

Joining the equations

#{(i_1=i_2+i_3),(V_b=R_1i_1+R_2i_2),(R_3i_3-V_b-R_2i_2=0):}#

or also

#((1,-1,-1),(R_1,R_2,0),(0,-R_2,R_3))((i_1),(i_2),(i_3))=((0),(V_b),(V_b))#

Solving for #i_1,i_2,i_3# we obtain

#((i_1 = ((2 R_2 + R_3) V_b)/(R_2 R_3 + R_1 (R_2 + R_3))),(i_2 = ((R_3-R_1) V_b)/(R_2 R_3 + R_1 (R_2 + R_3))),(i_3 = ((R_1 + 2 R_2) V_b)/(R_2 R_3 + R_1 (R_2 + R_3))))#

NOTE:

Supposing that the figures for #R_1, R_2# are relative to maximum allowed dissipation, the problem would be formulated as:

solve for #i_1,i_2,R_1,R_2# the system of equations

#{
(i_1^2R_1=10),
(i_2^2R_2=15),
(i_1 = ((2 R_2 + R_3) V_b)/(R_2 R_3 + R_1 (R_2 + R_3))),
(i_2 = ((R_3-R_1) V_b)/(R_2 R_3 + R_1 (R_2 + R_3)))
:}#

obtaining

#i_1=2.8,i_2=1.78,R_1=1.27,R_2=4.74#