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Unformatted text preview: ucsc econ/ams 11b winter 2008 Review Questions 8 Solutions 1. a. Q (20 , 15 , 5) ≈ 575 . 866 b. First compute the marginal products of labor and real estate at the given point: Q L = 15 K 2 / 5 L 1 / 2 R 1 / 4 = ⇒ Q L (20 , 15 , 5) ≈ 19 . 195 , Q R = 7 . 5 K 2 / 5 L 1 / 2 R 3 / 4 = ⇒ Q R (20 , 15 , 5) ≈ 28 . 793 . Now, use linear approximation: Δ Q ≈ Q L (20 , 15 , 5) · Δ L + Q R (20 , 15 , 5) · Δ R ≈ (19 . 195)(0 . 75) + (28 . 793)(0 . 5) ≈ 28 . 793 . 2. a. We want to maximize the output, Q = 30 K 2 / 5 L 1 / 2 R 1 / 4 , subject to the budget constraint K + L + R = 69, since the inputs are all being measured in $millions. The Lagrangian for this problem is F ( K,L,R,λ ) = 30 K 2 / 5 L 1 / 2 R 1 / 4 λ ( K + L + R 69) , and the first order conditions are F K = 0 = ⇒ 12 K 3 / 5 L 1 / 2 R 1 / 4 = λ, F L = 0 = ⇒ 15 K 2 / 5 L 1 / 2 R 1 / 4 = λ, F R = 0 = ⇒ 7 . 5 K 2 / 5 L 1 / 2 R 3 / 4 = λ, F λ = 0 = ⇒ K + L + R = 69 . The first two equations imply that 12 K 3 / 5 L 1 / 2 R 1 / 4 = 15 K 2 / 5 L 1 / 2 R 1 / 4 = ⇒ 12 L 1 / 2 K 3 / 5 = 15 K 2 / 5 L 1 / 2 , after canceling the common factor of R 1 / 4 . Clearing denominators gives 12 L = 15 K = ⇒ L = 1 . 25 K . Likewise, comparing the first and third equations implies that 12 K 3 / 5 L 1 / 2 R 1 / 4 = 7 . 5 K 2 / 5 L 1 / 2 R 3 / 4 = ⇒ 12 R 1 / 4 K 3 / 5 = 7 . 5 K 2 / 5 R 3 / 4 , and clearing denominators gives 12 R = 7 . 5 K = ⇒ R = 0 . 625 K . Substituting for R and L in the fourth equation (the constraint) gives K + 1 . 25 K + 0 . 625 K = 69 = ⇒ 2 . 875 K = 69 . Thus, the critical values for the inputs are K * = 24 , L * = 30 and R * = 15 , and the hotel chain’s maximum output is Q * = Q (24 , 30 , 15) ≈ 1152 . 894 . b. When output is maximized, the critical value of λ is λ * = 12( K * ) 3 / 5 ( L * ) 1 / 2 ( R * ) 1 / 4 ≈ 19 . 215 . c. By the envelope theorem, dQ * dB = λ * , where B is the budget. It follows that Δ Q * ≈ λ * · Δ B ≈ 9 . 607 , since we measure the budget in the same units as the inputs, so an increase of $500,000, means that Δ B = 0 . 5. 3. The equation η q/p = . 1 √ p gives a differential equation for the demand function, namely dq dp · p q = . 1 p 1 / 2 ....
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This note was uploaded on 03/15/2010 for the course ECON 11 taught by Professor Yk during the Spring '10 term at UCSC.
 Spring '10
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