Inverse demand functions are PA = 10 - QA and PB = 5 - 0.5QB. Under a two part tariff, fixed fee consumer surplus of smaller demand which is PB. Hence the profit is given by Profit function = 2 x F + (P-MC)*(QA + QB) = 2 x 0.5*(5 - P)*(10 - 2P) + (P - 2)*(20 - 3P) = 50 + 2P^2 - 20P + 20P - 40 - 3P^2 + 6P
Inverse demand functions are PA = 10 - QA and PB = 5 - 0.5QB. Under a two part tariff, fixed fee consumer surplus of smaller demand which is PB. Hence the profit is given by Profit function = 2 x F + (P-MC)*(QA + QB) = 2 x 0.5*(5 - P)*(10 - 2P) + (P - 2)*(20 - 3P) = 50 + 2P^2 - 20P + 20P - 40 - 3P^2 + 6P
Chapter1: Making Economics Decisions
Section: Chapter Questions
Problem 1QTC
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given the following solution (see attached), how would you calculate the average
![Inverse demand functions are \( PA = 10 - QA \) and \( PB = 5 - 0.5QB \). Under a two-part tariff, the fixed fee is the consumer surplus of smaller demand, which is \( PB \). Hence the profit is given by
\[
\text{Profit function} = 2 \times F + (P - MC) \times (QA + QB)
\]
\[
= 2 \times 0.5 \times (5 - P) \times (10 - 2P) + (P - 2) \times (20 - 3P)
\]
\[
= 50 + 2P^2 - 20P + 20P - 40 - 3P^2 + 6P
\]
\[
= 10 - P^2 + 6P
\]
Profit is maximum when
\[
2P = 6
\]
\[
P^* = 3
\]
Hence the price per unit is 3 and is the same for A and B. The fees amount is
\[
F = 0.5 \times (5 - 3) \times (10 - 2 \times 3) = \$4
\]
Profit:
\[
= 4 \times 2 + (3 - 2) \times (20 - 3 \times 3) = \$19
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9ce17ee7-739c-4e84-bb58-c48dfc141b37%2F0e5507ea-38e2-4e75-be97-119fafe762ab%2Fa5myhso_processed.png&w=3840&q=75)
Transcribed Image Text:Inverse demand functions are \( PA = 10 - QA \) and \( PB = 5 - 0.5QB \). Under a two-part tariff, the fixed fee is the consumer surplus of smaller demand, which is \( PB \). Hence the profit is given by
\[
\text{Profit function} = 2 \times F + (P - MC) \times (QA + QB)
\]
\[
= 2 \times 0.5 \times (5 - P) \times (10 - 2P) + (P - 2) \times (20 - 3P)
\]
\[
= 50 + 2P^2 - 20P + 20P - 40 - 3P^2 + 6P
\]
\[
= 10 - P^2 + 6P
\]
Profit is maximum when
\[
2P = 6
\]
\[
P^* = 3
\]
Hence the price per unit is 3 and is the same for A and B. The fees amount is
\[
F = 0.5 \times (5 - 3) \times (10 - 2 \times 3) = \$4
\]
Profit:
\[
= 4 \times 2 + (3 - 2) \times (20 - 3 \times 3) = \$19
\]
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