Suggested Solution to Problem Set 1 Seihui Kim September 30, 2021 Question 1 (a) The utility maximization problem for Mr. Stark is 1−α max xα s.t px1 + x2 ≤ m 1 x2 x1 ,x2 ∈R2+ where α ∈ (0, 1) and p > 0. The Lagrangian function to solve the problem is 1−α L (x1 , x2 , λ) =xα + λ(m − px1 − x2 ) 1 x2 The first-order conditions for optimality are: ∂L ∂L = α(xα−1 x1−α ) − λp ≤ 0, x1 = (αxα−1 x1−α ) − λp)x1 = 0, x1 ≥ 0 1 2 1 2 ∂x1 ∂x1 (1) ∂L ∂L −α −α = {(1 − α)xα x2 = {(1 − α)xα 1 x2 − λ ≤ 0, 1 x2 ) − λ}x2 = 0, x2 ≥ 0 ∂x2 ∂x2 (2) ∂L ∂L = m − px1 − x2 ≥ 0, λ = (m − px1 − x2 )λ = 0, λ ≥ 0 ∂λ ∂λ (3) • Case 1) x∗1 > 0 and x∗2 > 0 Since x∗1 > 0 and x∗2 > 0, by complementary slackness condtion in (1) and (2), −α αxα−1 x1−α − λp = 0 and {(1 − α)xα 1 x2 ) − λ} = 0 1 2 Each equations is equivalent to λ= αx1α−1 x1−α 2 p (4) and −α λ = (1 − α)xα 1 x2 (5) Note that λ > 0 for all x1 > 0 and x2 > 0, so m = px1 + x2 by complementary slackness condition in (3). Let (4) and (5) be equal. Then it yields x2 = p(1 − α) x1 α (6) Substituting (6) into the budget constraint, then we get x∗ = (x∗1 , x∗2 ) = ( αm , m(1 − α)) p • Case 2) x∗1 > 0 and x∗2 = 0 Since x∗2 = 0, by (3), −α (1 − α)xα 1 x2 − λ ≤ 0. 1 (7) −α 1−α However, (1 − α)xα → ∞ as x2 → 0+ , so (1 − α)xα − λp > 0 at x∗2 = 0, which is a contradiction. 1 x2 1 x2 ∗ Thus, x2 = 0 can not be a solution of the problem. • Case 3) x∗1 = 0 and x∗2 > 0 Since x∗1 = 0, by (1), αxα−1 x1−α − λp ≤ 0. 1 2 x∗1 However, αx1α−1 x1−α → ∞ as x1 → 0+ ,so αx1α−1 x1−α − λp > 0 at x∗1 = 0, which is a contradiction. Thus, 2 2 = 0 can not be a solution of the problem. Thus, x∗ = (x∗1 , x∗2 ) = ( αm p , (1 − α)m) is the solution of the maximization problem, considering three cases. (b) By (7), m= px∗ 1 α = x∗ 2 1−α . Expressing x∗2 in terms of x∗1 yields x∗2 = p(1−α) ∗ x1 α This is a straight line passing through the origin and having a slope of Its graph is the following: (1−α)p . α (c) The income elasticity of the demand for batteries is ∂x∗ 1 m ∂m x∗ 1 = α p p α =1 Whenever m changes a unit percent, both x∗1 and x∗2 change the same unit percent with m, because the income offer curve is a straight line passing through the origin, which implies homotheticity. (d),(e) By (a), only interior points can be a solution of the problem. Thus, using similar arguments with αm ∗∗ (a), x∗∗ = (x∗∗ 1 , x2 ) = ( p0 , (1 − α)m). (f) By definition of the normal good, batteries are a normal good because, ∂x∗∗ m 1 ∂m x∗∗ 1 = 0 α p p0 α >0 for all p0 > 0 and m > 0. (g) For removing the income effect by the Slutsky’s way, let ms be the amount of money that Mr. Stark can consume the original bundle, x∗ , at the changed price, p0 . That is, mS = p0 αm + (1 − α)m. p (8) When the budget is mS and the price p0 , the optimal choice of good 1, xS1 , is xS1 = α2 m m + α(1 − α) 0 , p p by (7) and (8). Note that we remove the income effect by using mS . Thus, xS1 − x∗1 = mα(1 − α)( p10 − p1 ) < 0 2 (9) is the Slutsky substitution effect. Also, the Slutsky income effect is 1 S 2 1 x∗∗ 1 − x1 = mα ( p0 − p ) < 0. (h) Let mCV be the amount of money that Mr. Stark can consume the optimal bundle that is indifferent to his original bundle, x∗ , at the changed price, p0 . Also, xCV is the optimal bundle, when the budget is mCV and the price is p0 . That is, CV αm CV xCV = (xCV 1 , x2 ) = ( p0 , (1 − α)mCV ), by (7). The utility level at the x∗ is U = αα (1 − α)(1−α) p−α m. (10) By definition of xCV , (10) is the utility constraint when Mr. Stark consumes the bundle, xCV , that is, α αmCV ((1 − α)mCV )1−α = αα (1 − α)(1−α) p−α m. (11) p0 (11) is equivalent to 0 α p CV m = m. (12) p Thus, 0 CV = mCV − m = {( pp )α − 1}m > 0, by definition of the CV and (12). Let mEV be the amount of money that Mr. Stark can consume the optimal bundle that is indifferent to his new bundle, x∗∗ , at the original price, p. Also, xEV is the optimal bundle, when the budget is mEV and the price is p. That is, EV αm EV xEV = (xEV 1 , x2 ) = ( p , (1 − α)mEV ), by (7). The utility level at the x∗∗ is U = αα (1 − α)(1−α) (p0 )−α m (13) By definition of xEV , (13) is the utility constraint when Mr. Stark consumes the bundle, xEV , that is, α αmEV ((1 − α)mEV )1−α = αα (1 − α)(1−α) (p0 )−α m. (14) p (14) is equivalent to α p EV m = m. (15) p0 Thus, EV = m − mEV = {1 − ( pp0 )α }m > 0, by definition of the EV and (15). Note that these CV and EV are ’welfare change’, not ’welfare loss’. If you want to see welfare loss, then take the absolute values for them. Their graphs are as follows: 3 Question 2 The answer to the Review Questions from the textbook is at the back of the book. 4