STRUCTURED HOMEWORK ASSIGNMENT 3
STRUCTURED HOMEWORK ASSIGNMENT 3
PART 1
Question 1
(A) First, invert the demand function QD= 8,300 - 2.1P into the price function, so that price is on the left hand side on its own.
QD= 8,300 - 2.1P → 1QD/2.1 = 8,300/2.1 – 2.1P/2.1
0.5QD = 3,952.4 – P → P = 3,952.40 – 0.5QD
TR = P*Q → TR = (3,952.40 – 0.5Q) *Q → TR = 3,952.40Q – 0.5Q^2
MR = 3,952.40 – Q
(B) Profit = TR – TC
Profit = 3,952.40Q – 0.5Q^2 – (2,200 + 480Q + 20Q^2)
Profit = -2,200 + 3,472.40Q – 20.5Q^2
Marginal Profit = 3,472.40 – 41Q → MP = 0
3,472.40 – 41Q = 0 → 41Q = 3,472.40
→ Optimal Quantity = 85 (rounded off)
Therefore, Coolidge Corporation should produce and sell 85 lasers each month to maximise profit.
(C) Substitute Q = 85 into Profit = -2,200 + 3,472.40Q – 20.5Q^2 to determine Optimal Profit
Profit = -2,200 + (3,472.40 * 85) – (20.5 * (85)^2)
→ Profit = $144,841.50
Question 2
C = 100 + 2Q^2 P = 90 – 2Q
(A) TR = P*Q → TR = (90 – 2Q) * Q → TR = 90Q – 2Q^2
Profit = TR – TC → Profit = 90Q – 2Q^2 – (100 + 2Q^2)
Profit = -100 + 90Q – 4Q^2
Marginal Profit = 90 – 8Q → MP = 0
90 – 8Q = 0 → 8Q = 90
→ Monopoly Quantity = 11.25
Substitute Q = 11.25 into P = 90 – 2Q to determine Monopoly Price
P = 90 – (2 * 11.25)
→ P = $67.50
Substitute Q = 11.25 into Profit = -100 + 90Q – 4Q^2 to determine Monopoly Profit
Profit = -100 + (90 * 11.25) – (4 * (11.25)^2)
→ Profit = $406.25
(B) Price = Marginal Cost in a competitive industry, therefore, set P = MC to determine the Optimal Quantity at which profit is maximised.
Cost = 100 + 2Q^2 → Marginal Cost = 4Q
P = MC → 90 – 2Q = 4Q → 6Q = 90
→ Optimal Quantity = 15
Substitute Q = 15 into P = 90 – 2Q to determine the Optimal Price
P = 90 – (2 * 15)
→ P = $60
Substitute Q = 15 into Profit = -100 + 90Q – 4Q^2 to determine Optimal Profit
Profit = -100 + (90 * 15) – (4 * (15)^2)
→ Profit = $350
(C) The profit that is lost by having the firm produce at the competitive industry as compared to the monopoly is the difference of the two
PART 1
Question 1
(A) First, invert the demand function QD= 8,300 - 2.1P into the price function, so that price is on the left hand side on its own.
QD= 8,300 - 2.1P → 1QD/2.1 = 8,300/2.1 – 2.1P/2.1
0.5QD = 3,952.4 – P → P = 3,952.40 – 0.5QD
TR = P*Q → TR = (3,952.40 – 0.5Q) *Q → TR = 3,952.40Q – 0.5Q^2
MR = 3,952.40 – Q
(B) Profit = TR – TC
Profit = 3,952.40Q – 0.5Q^2 – (2,200 + 480Q + 20Q^2)
Profit = -2,200 + 3,472.40Q – 20.5Q^2
Marginal Profit = 3,472.40 – 41Q → MP = 0
3,472.40 – 41Q = 0 → 41Q = 3,472.40
→ Optimal Quantity = 85 (rounded off)
Therefore, Coolidge Corporation should produce and sell 85 lasers each month to maximise profit.
(C) Substitute Q = 85 into Profit = -2,200 + 3,472.40Q – 20.5Q^2 to determine Optimal Profit
Profit = -2,200 + (3,472.40 * 85) – (20.5 * (85)^2)
→ Profit = $144,841.50
Question 2
C = 100 + 2Q^2 P = 90 – 2Q
(A) TR = P*Q → TR = (90 – 2Q) * Q → TR = 90Q – 2Q^2
Profit = TR – TC → Profit = 90Q – 2Q^2 – (100 + 2Q^2)
Profit = -100 + 90Q – 4Q^2
Marginal Profit = 90 – 8Q → MP = 0
90 – 8Q = 0 → 8Q = 90
→ Monopoly Quantity = 11.25
Substitute Q = 11.25 into P = 90 – 2Q to determine Monopoly Price
P = 90 – (2 * 11.25)
→ P = $67.50
Substitute Q = 11.25 into Profit = -100 + 90Q – 4Q^2 to determine Monopoly Profit
Profit = -100 + (90 * 11.25) – (4 * (11.25)^2)
→ Profit = $406.25
(B) Price = Marginal Cost in a competitive industry, therefore, set P = MC to determine the Optimal Quantity at which profit is maximised.
Cost = 100 + 2Q^2 → Marginal Cost = 4Q
P = MC → 90 – 2Q = 4Q → 6Q = 90
→ Optimal Quantity = 15
Substitute Q = 15 into P = 90 – 2Q to determine the Optimal Price
P = 90 – (2 * 15)
→ P = $60
Substitute Q = 15 into Profit = -100 + 90Q – 4Q^2 to determine Optimal Profit
Profit = -100 + (90 * 15) – (4 * (15)^2)
→ Profit = $350
(C) The profit that is lost by having the firm produce at the competitive industry as compared to the monopoly is the difference of the two
Bibliography: (8th edition), Competitive Firms vs Monopoly, D. L. Robert S. Pindyck, Microeconomics, Pearson