Anonymous
Anonymous asked in 科學數學 · 6 years ago

# 獨佔市場經濟學問題求解答!!!!!不懂如何計算

based on market research, a film production company in Ectenia obtains the

following information about the demand and production cost of its new DVD

P=1000-10Q

TR=1000Q-10Q(Q二次方)

MR=1000-20Q

MC=100+10Q

suppose, in addition to the cost above, the director of the film has to be paid.

the company is considering four options

a. A flat fee of 2000 Ectenian dollars

b. 50 percent of the profiles

c. 150 Ectenian dollars per unit sold

d. 50 percent of the revenue

for each option, calculate the profit maximizing price and quantity.

which, if any of these compensation schemes would alter the deadweight loss

from monopoly ?

C小題...為什麼新的MC=100+160Q???

Rating
• 慈信
Lv 4
6 years ago

P = 1000 - 10Q

TR = 1000Q - 10Q²

MR = 1000 - 20Q

MC = 100 + 10Q

TC = 100Q + 5Q² + C

原均衡

MR = MC

1000 - 20Q = 100 + 10Q , Q = 30

P = 1000 - 10Q = 1000 - 10*30 = 700

(P,Q) = ( 700 , 30 )

π = TR - TC = 700*30 - (100*30 + 5*30² + C )

= 7500 - C

a. A flat fee of 2000 Ectenian dollars (定額費用)

TC1 = TC + F = 100Q + 5Q² + C + 2000

MC1 = 100+ 10Q = MC

MR = MC1 = MC

所以 (P1,Q1) = (P , Q) = (700,30)

π1 = TR - ( TC + 2000 )

= 700*30 - (100*30 + 5*30² + C + 2000 )

= 5500 - C

消費者剩餘沒有影響 , 但獨占廠商超額利潤減少

b. 50 percent of the profiles

πT = π/2 = ( TR - TC ) / 2

dπT/dQ = 1/2 * dπ/dQ = (MR-MC)/2 = 0

MR=MC

所以 (P2,Q2) = (P , Q) = (700,30)

πT= π/2 = (7500 - C)/2 = 3750 - C/2

c. 150 Ectenian dollars per unit sold ( 從量費用)

每單位產量 , 支付 t 元從量費用

TC3 = TC + tQ = 100Q + 5Q² + C + tQ

MC3 = 100 + 10Q + t = 100 + 10Q + 150 = 160 + 10Q

MR = MC3

1000 - 20Q3 = 160 + 10Q3 , Q3 = 28

P3 = 1000 - 10Q3 = 1000 - 10*28 = 720

所以 (P3,Q3) = ( 720 , 28 )

π3 = TR - TC3 = P*Q - ( 100Q + 5Q² + C + tQ )

= 720*28 - ( 100*28 + 5*28² + C + 150*28 )

= 9240 - C

d. 50 percent of the revenue

π4 = TR/2 - TC

dπ4/dQ = MR/2 - MC = 0

MR/2 = MC

(1000 - 20Q4)/2 = 100 + 10Q4 , Q4=20

P4 = 1000 - 10Q4 = 1000 - 10*20 = 800

所以 (P4,Q4) = ( 800 , 20 )

π4 = TR/2 - TC = 800*20/2 - (100*20 + 5*20² + C )

= 4000 - C