Answer to Question #117719 in Microeconomics for Suzanne

Question

. The market for biscuit consist of two firms. Competition in the market is such that each. Of the tow firms independently produces a quantity of output taking the output of the other firm as given and these quantities are then sold in the market at a price determined by the total amount produced by the two firms. The short-run total cost functions of the firms are given by:

STC 1 = 0.1q 1 2 + 20q 1 + 100,000

STC 2 = 0.4q 2 2 + 32q 2 + 20,000

The two firms produce a homogenous product, the market demand is:

Q = 4,000 − 10P

If a Cournot equilibrium is achieved, calculate:

i. The equilibrium price.

ii. The equilibrium output of firm 1.

iii. The equilibrium output of firm 2.

iv. The pure profit of firm 1.

v. The pure profit of firm 2.

Answer 1

Market demand:

Q = 4,000 – 10P

Cournot equilibrium

Determine:

Profit = TR – TC

Market demand:

Q = $Q_{1} + Q_{2}$

$Q_{1} + Q_{2} = 4,000 - 10P$

Inverse demand function: $10P = 4,000 - (Q_{1} + Q_{2})$

10P = 4,000 – $Q_{1}-Q_{2}$

P = 400 – $0.1Q_{1}-0.1Q_{1}$

Firm 1

$TR = P \times Q$

$TR_{1} = P \times Q_{1} = (400 – 0.1Q_{1} – 0.1Q_{2}) \times Q_{1}$

TR₁ = P*Q₁ = (400 – 0.1Q₁ – 0.1Q₂)*Q₁

$TR_{1} = 400 – 0.1Q_{1}^2 – 0.1Q_{1}\times Q_{2}$

Profit

Profit ($\varPi$ ) = $TR_{1} – TC_{1}$

Profit ($\varPi$ ) = $400Q_{1} – 0.1Q_{1}^2 – 0.1 Q_{1}Q_{2} – TC_{1} (0.1Q_{1}^2 + 20Q_{1} + 100,000)$

Profit ($\varPi$ ) = $400 Q_{1} – 0.1Q_{1}^2 – 0.1 Q_{1}Q_{2} – 0.1Q_{1}^2 - 20Q_{1} - 100,000$

Profit ($\varPi$ ) = $380 Q_{1} – 0.2Q_{1}^2 – 0.1 Q_{1}Q_{2} - 100,000$

Getting the first derivative of the profit function and equating it to 0

$\varDelta Profit (\varPi)/\varDelta Q_{1} = 380 – 0.4Q_{1} – 0.1Q_{2} = 0$

$\varDelta Profit (\varPi)/\varDelta Q_{1} = 0.4Q_{1} + 0.1Q_{2} = 380$

Firm 2

$TR = P \times Q$

$TR_{2} = P \times Q_{2} = (400 – 0.1Q_{1} – 0.1Q_{2}) \times Q_{2}$

$TR_{2} = 400 Q_{2} – 0.1Q_{1} Q_{2}– 0.1Q_{2}^2$

$Profit (\varPi) = TR_{2} – TC_{2}$

$Profit (\varPi) = 400 Q_{2} – 0.1Q_{1} Q_{2}– 0.1Q_{2}^2– TC_{2} (0.4Q_{2}^2 + 32Q_{2} +20,000)$

$Profit (\varPi) = 400 Q_{2} – 0.1Q_{1} Q_{2}– 0.1Q_{2}^2– 0.4Q_{2}^2 - 32Q_{2} -20,000$

$Profit (\varPi) = 368Q_{2} – 0.1Q_{1} Q_{2} – 0.5Q_{2}^2 -20,000$

Getting the first derivative of the profit function and equating it to 0

$\varDelta Profit (\varPi)/ \varDelta Q_{2} = 368 – 0.1Q_{1} – Q_{2} = 0$

$\varDelta Profit (\varPi)/ \varDelta Q_{2} = 0.1Q_{1} + Q_{2} = 368$

Solving the simultaneous equation

Equilibrium output

$\varDelta Profit (\varPi)/ \varDelta Q_{1} = 0.4Q_{1} + 0.1Q_{2} = 380$

$\varDelta Profit (\varPi)/ \varDelta Q_{2} = 0.1Q_{1} + Q_{2} = 368$

$Q_{2} = 368 - 0.1Q_{1}$

$0.4Q_{1} + 0.1\times(368 - 0.1Q_{1}) = 380$

$0.4Q_{1} + 36.8 - 0.01Q_{1} = 380$

$0.39Q_{1} = 380 - 36.8$

$Q_{1} = 343.2/0.39$

$Q_{1} = 880$

Equilibrium output of firm 1 = 880

$Q_{2} = 368 - 0.1Q_{1}$

$Q_{2} = 368 - 0.1\times880$

$Q_{2} = 368 - 88$

$Q_{2} = 280$

Equilibrium output of firm 2 = 280

Equilibrium price

$P = 400 – 0.1Q_{1} – 0.1Q_{2}$

$P = 400 – 0.1\times880 – 0.1\times280$

$P = 400 – 88 – 28$

$P = 284$

Pure profit:

Firm 1

$Profit (\varPi_{1}) = 380 Q_{1} – 0.2Q_{1}^2 – 0.1 Q_{1}Q_{2} - 100,000$

$Profit (\varPi_{1}) = 380 \times 880 – 0.2 \times (880^2) – 0.1 (880 \times 280) - 100,000$

$Profit (\varPi_{1}) = 334400 – 154880 – 24640 - 100,000$

$Profit (\varPi_{1}) = 54,880$

Firm 2

$Profit (\varPi_{2}) = 368\times280 – 0.1 \times 880 \times 280 – 0.5 \times (280^2) -20,000$

$Profit (\varPi_{2}) = 103040 – 24640 – 39200 -20,000$

$Profit (\varPi_{2}) = 19,200$