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PS5_solutions

# PS5_solutions - Econ 300 Solutions to Problem Set 1 Julien...

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Econ 300 - Solutions to Problem Set 1 - Julien Bengui 10.1.8 First we write the firm’s profit function using the factor and output price information as Π( K, L ) = 4 * 9 L 1 / 3 K 1 / 3 - 12 L - 6 K = 36 L 1 / 3 K 1 / 3 - 12 L - 6 K. To find the optimal levels for each input, we write the first order conditions: Π K = 12 L 1 / 3 K - 2 / 3 - 6 = 0 , Π L = 12 L - 2 / 3 K 1 / 3 - 12 = 0 . Solving the second equation for K yields K = L 2 . This can be substituted into the first equation: 2 L 1 / 3 ( L 2 ) - 2 / 3 = 1 , which can be simplified to 2 L - 1 = 1, or L = 2. Plugging this back into K = L 2 yields K = 4. Thus, the stationary point of the profit function is ( K, L ) = (4 , 2). To check that this stationary point is a maximum, we compute the second order derivatives of the profit function and evaluate them at the stationary point: Π KK = - 8 L 1 / 3 K - 5 / 3 = - 8 * 2 1 / 3 4 - 5 / 3 = - 8 * 2 1 / 3 2 - 10 / 3 = - 8 * 2 - 9 / 3 = - 8 * 2 - 3 = - 1 < 0 , Π LL = - 8 L - 5 / 3 K 1 / 3 = - 8 * 2 - 5 / 3 4 1 / 3 = - 8 * 2 - 5 / 3 2 2 / 3 = - 8 * 2 - 3 / 3 = - 8 * 2 - 1 = - 4 < 0 . We also calculate the cross partial derivative Π KL = 4 L - 2 / 3 K - 2 / 3 = 4 * 2 - 2 / 3 4 - 2 / 3 = 4 * 2 - 2 / 3 2 - 4 / 3 = 4 * 2 - 6 / 3 = 4 * 2 - 2 = 4 / 4 = 1 , from which we see that Π KK Π LL = 4 > 1 = (Π KL ) 2 . We can thus conclude that the point ( K * , L * ) = (4 , 2) is a maximum. 10.1.10 (a) With linear total costs, the profit function of AWL is given by Π( Q T , Q M ) = ( α - βQ T ) Q T + ( γ - θQ M ) Q M - Ψ - c ( Q M + Q T ) = - βQ

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PS5_solutions - Econ 300 Solutions to Problem Set 1 Julien...

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