Question #289029

Two duopolists produce and quantities of a homogenous product . The market demand of the product is given by Q= 240-2p where Q= Qa +Qb and the price of the product. The total functions of the duopolists are given by: C(Qa)= 60+4Qa and C(Qb)= 50+0.625Qb^2



a) Find the level of output that maximizes the profit of each firm and the corresponding profit



and price.



b) Find the level of output that maximizes their profit if the two firms corporate and the



corresponding profit and price


1
Expert's answer
2022-01-24T03:16:19-0500

a)

Q=2402pQ=240−2p

c(Qa)=60+4Qac(Q_a​)=60+4Q_a

c(Qb)=50+0.625Qb​c(Q_b​)=50+0.625Q_b

where

Q=Qa+QbQ=Q_a​+Q_b

from Q=2402p,p=1200.5QQ=240−2p, p=120−0.5Q :

πa=pQc(Qa)=(1200.5Q)Qa(60+4Qa)=π_a=pQ−c(Q_a​)=(120−0.5Q)Q_a​−(60+4Q_a​)=

=120Qa0.5(Qa)20.5QaQb604Qa=120Q_a​−0.5(Q_a​)^2−0.5Q_a​Q_b​−60−4Q_a


πa=120Qa0.5Qb4=0​π_a​'=120−Q_a​−0.5Q_b​−4=0

Qa=1160.5QbQ_a​=116−0.5Q_b (1)


πb=pQc(Qb)=(1200.5Q)Qb(50+0.625Qb2)=π_b​=pQ−c(Q_b​)=(120−0.5Q)Q_b​−(50+0.625Q_b^2​)=

== 120Qb0.5QaQb501.125Qb2120Q_b​−0.5Q_a​Q_b​−50−1.125Qb^2

πb=1200.5Qa2.25Qb=0π_b'​=120−0.5Q_a​−2.25Q_b=0

Qb=53.30.22Qa​Q_b​=53.3−0.22Q_a (2)


from equation 1 and 2:

Qb=53.30.22(1160.5Qb)Q_b​=53.3−0.22(116−0.5Q_b​)

Qb=31.21Q_b​=31.21

Qa=1160.5Qb=1160.5(31.21)=100.4Q_a​=116−0.5Q_b=116−0.5(31.21)=100.4

Then we have:

πa=120Qa0.5(Qa)20.5QaQb604Qaπ_a​=120Q_a​−0.5(Q_a​)^2−0.5Q_a​Q_b​−60−4Qa

πa=120(100.4)0.5(100.4)20.5(100.4)(31.21)604(100.4)=4979.58​π_a​=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)501.125(31.21)2=1032.64π_b​=120(31.21)−0.5(100.4)(31.21)−50−1.125(31.21)^2=1032.64

p=1200.5(Qa+Qb)=1200.5(100.4+31.21)=54.195p=120−0.5(Q_a​+Q_b​)=120−0.5(100.4+31.21)=54.195


b)

Q=240+2pQ=240+2p

c(Q)=c(Qa)+c(Qb)=60+4Qa+50+0.0625Qb2=110+4Qa+0.625Qb2c(Q)=c(Q_a​)+c(Q_b​)=60+4Q_a​+50+0.0625Q_b^2=110+4Q_a​+0.625Q_b^2 ​

p=1200.5Qp=120−0.5Q

π=pQc(Q)π=pQ−c(Q)

π=(1200.5Q)Q1104Q0.625Q2=120Q0.5Q21104Q0.625Q2π=(120−0.5Q)Q−110−4Q−0.625Q^2=120Q−0.5Q^2−110−4Q−0.625Q^2

π=120Q41.25Q=0π'=120−Q−4−1.25Q=0

Q=51.56Q=51.56

p=1200.5Q=p=1200.5Q=94.22p=120−0.5Q=p=120−0.5Q=94.22

π=116Q1.125Q2110=2880.22π=116Q−1.125Q ^ 2 −110=2880.22


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