11Bfinrevsol

# 11Bfinrevsol - ucsc econ/ams 11b winter 2008 Review...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

11Bfinrevsol - ucsc econ/ams 11b winter 2008 Review...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online