a)Q=240−2pc(Qa)=60+4Qac(Qb)=50+0.625Qb,whereQ=Qa+Qb,from Q=240−2p,p=120−0.5Qπa=P.Q−c(Qa)=(120−0.5Q)Qa−(60+4Qa)=120Qa−0.5(Qa)2−0.5QaQb−60−4Qaπa′=120−Qa−0.5Qb−4πa′=0⟹120−Qa−0.5Qb−4=0Qa=116−0.5Qb......equ(1)πb=P.Q−c(Qb)=(120−0.5Q)Qb−(50+0.625Qb2)=120Qb−0.5QaQb−0.5Qb2−50−0.625Qb2=120Qb−0.5QaQb−50−1.125Qb2πb′=120−0.5Qa−2.25Qbπb′=0⟹120−0.5Qa−2.25Qb=0Qb=53.3−0.22Qa......equ(2)from equation 1 and 2Qb=53.3−0.22(116−0.5Qb)Qb=53.3−25.52+0.11QbQb=31.21. Also,Qa=116−0.5Qb=116−0.5(31.21)=100.4Then we haveπa=120Qa−0.5(Qa)2−0.5QaQb−60−4Qaπa=120(100.4)−0.5(100.4)2−0.5(100.4)(31.21)−60−4(100.4)πa=4979.578πb=120(31.21)−0.5(100.4)(31.21)−50−1.125(31.21)2πb=1032.64p=120−0.5(Qa+Qb)p=120−0.5(100.4+31.21)p=54.195b)Q=240+2pc(Q)=c(Qa)+c(Qb)=60+4Qa+50+0.0625Qb2=110+4Qa+0.625Qb2p=120−0.5Qπ=P.Q−c(Q)π=(120−0.5Q)Q−110−4Q−0.625Q2π=120Q−0.5Q2−110−4Q−0.625Q2π′=120−Q−4−1.25Q=02.25Q=116, Q=51.56p=120−0.5Qp=120−0.5(51.56)p=94.22π=116Q−1.125Q2−110π=2880.22
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