a)
Q=240−2p
c(Qa)=60+4Qa
c(Qb)=50+0.625Qb
where
Q=Qa+Qb
from Q=240−2p,p=120−0.5Q :
πa=pQ−c(Qa)=(120−0.5Q)Qa−(60+4Qa)=
=120Qa−0.5(Qa)2−0.5QaQb−60−4Qa
πa′=120−Qa−0.5Qb−4=0
Qa=116−0.5Qb (1)
πb=pQ−c(Qb)=(120−0.5Q)Qb−(50+0.625Qb2)=
= 120Qb−0.5QaQb−50−1.125Qb2
πb′=120−0.5Qa−2.25Qb=0
Qb=53.3−0.22Qa (2)
from equation 1 and 2:
Qb=53.3−0.22(116−0.5Qb)
Qb=31.21
Qa=116−0.5Qb=116−0.5(31.21)=100.4
Then we have:
πa=120Qa−0.5(Qa)2−0.5QaQb−60−4Qa
πa=120(100.4)−0.5(100.4)2−0.5(100.4)(31.21)−60−4(100.4)=4979.58
πb=120(31.21)−0.5(100.4)(31.21)−50−1.125(31.21)2=1032.64
p=120−0.5(Qa+Qb)=120−0.5(100.4+31.21)=54.195
b)
Q=240+2p
c(Q)=c(Qa)+c(Qb)=60+4Qa+50+0.0625Qb2=110+4Qa+0.625Qb2
p=120−0.5Q
π=pQ−c(Q)
π=(120−0.5Q)Q−110−4Q−0.625Q2=120Q−0.5Q2−110−4Q−0.625Q2
π′=120−Q−4−1.25Q=0
Q=51.56
p=120−0.5Q=p=120−0.5Q=94.22
π=116Q−1.125Q2−110=2880.22
Comments