Someone help please. Microeconomics problem. QUESTION 03. Suppose AT&T; and Veri

Question

Someone help please. Microeconomics problem.

QUESTION 03. Suppose AT&T; and Verizon both have new cellular plans coming out in Santa Barbara, CA this winter. AT&T; will announce next week, and Verizon will announce the week after. Assume that these are the only two carriers in town and they face the following average revenue (demand) curve P = 400-0.05Q Where Q = Q Qv. Q is the total quantity of new cellular plans, QA is the quantity sold by carrier). The cost function for AT&T; is given by: The cost function for Verizon is given by: A aniy so C(Qa)= 0.5Q, Assume that both carriers maximize profit, these plans are the only available in town, and they choose quantity of plans to sell sequentially, with AT&T; moving first. A. What are the number of plans sold (Q*, Qa". Qv*), price (P*), and profits (.u", ')?

Explanation / Answer

In a sequential game, it is solved by backward induction. Because AT&T moves first followed by Verizon, hence we first solve profit maximisation for Verizon and then use it’s reaction function to solve for AT&T.

So the profit maximization problem of Verizon stands as,

Max: v = PQv – C(Qv)

              = (400 – 0.05(Qa+ Qv)) Qv – 0.5Qv

                    = 400Qv - 0.05Qa Qv - Qv2 – 0.5Qv

             = 399.5 Qv- 0.05Qa Qv - Qv2

Differentiating with respect to Qv,

d v/d Qv = 399.5 – 0.05 Qa - 2 Qv

Putting d v/d Qv = 0,

399.5 – 0.05 Qa - 2 Qv = 0

Qv = (399.5 – 0.05 Qa)/2......................................(1) which is the reaction function of Verizon

Similarly profit maximization problem of AT&T stands as

Max: A = PQA – C(QA)

              = (400 – 0.05(QA+ Qv)) QA – 0.5QA

                    = [400 – 0.05(QA+(399.5 – 0.05 Qa)/2)] QA – 0.5QA

                    = [400 – 0.05 QA – 0.05(399.5 – 0.05 Qa)/2] QA – 0.5QA

             = 400 QA – 0.05 QA2 – 9.99QA +0.00125 QA2 – 0.5QA

              = 389.51 QA – 0.049 QA2

Differentiating with respect to QA,

d A/d QA = 389.51 – 0.09 QA

Putting d v/d Qv = 0,

389.51 – 0.09 QA =0

QA* = 4327.88

Qv* = (399.5 – 0.05*4327.88)/2

      = 91.55

Q* = QA* + Qv*

= 4419.43

P* = 400 – 0.05 Q*

      = 179.03

A* = 389.51 QA* – 0.049 QA2*

      = 767955.819

v* = 399.5 Qv*- 0.05Qa* Qv* - Qv2*

        = 8381.9518

* = v* + A*

      = 776337.7708

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