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Microeconomics I (Fall 2009) Problem Set 2 Sample Solutions

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Microeconomics I (Fall 2009) Problem Set 2 Sample Solutions Microeconomics I (Fall 2009) Problem Set 2 Sample Solutions 1. (a) True: The demand for an inferior good decreases as income increases. The Engel curve is a graph of the demand for the good as a function of income. Thus, it is downward sloping. (b) True: The demand for a Gi®en good declines as its price falls. A fall in price gives rise to two e®ects: the substitution e®ect and the income e®ect. The substitution e®ect of a price decrease is always non-negative. Therefore, the income e®ect of a Gi®en good must be negative. That is, a Gi®en good is always an inferior good. (c) False: If the demand for a good rises as its price falls, we call such a good an ordinary good. On the other hand, if an increase in income leads to a rise in the demand for a good, it is a normal good. As seen in (b), a fall in price brings about the substitution and income e®ects. If the former is greater than the latter, the demand for an inferior good increases with a decline in its price. Thus, ordinary goods are not always normal goods. 2. (a) The income elasticity of demand (´) is the percentage change in demand with the percentage change in income: ¢Qd µ ¶ Qd ¢Qd Y ´ = ¢Y = ; Y ¢Y Qd where Qd is demand and Y is income. In this case, (10 ¡ 5)=5 ´ = (110 ¡ 100)=100 = 10: Since ´ > 1 > 0, the good is a luxury good. (b) Suppose that the total income is Y . Then, the budget constraint is given by 2x1 + x2 = Y . Since utility is maximized at x1 = x2, the demand functions for goods 1 and 2 are x1 = Y=3 and x2 = Y=3, respectively. The Engel curve for good 1 is the graph of Y = 3x1, which is a straight line. (c) Given Y , utility is maximized at (x1; x2) = (0;Y ). This means that the demands for goods 1 and 2 are x1 = 0 and x2 = Y . The Engel curve for good 2 is the graph of Y = x2, which is the 45-degree line. 3. (a) By solving the following utility maximization problem, 1 1 2 2 max x1 x2 s:t: p1x1 + p2x2 = Y we have x1 = Y=2p1 and x2 = Y=2p2. 1 (b) The equations of the Engel curves for goods 1 and 2 are Y = 2p1x1 and Y = 2p2x2, respectively. 0 0 (c) By substituting p1 = p2 = 1 and Y = 32 into the demand functions for goods 1 0 0 and 2, the initialp p demands are x1 = x2 = 32=2 = 16. Therefore, the initial utility 0 0 0 level is u = 16 16 = 16. As the price of good 1 falls from p1 to p1 = 1=2, the 0 demand for good 1 becomes x1 = 32=(2 £ 1=2) = 32. Thus, the total e®ect is 0 0 x1 ¡ x1 = 16. In order to obtain the substitution e®ect, we need to ¯nd a point 0 1=2 1=2 at which the indi®erence curve for u = u , which is x1 x2 = 16, is tangent to a 0 1=2 1=2 2 line of slope ¡p1=p2 = ¡1=2. From x1 x2 = 16, x1x2 = 16 holds. The slope of 2 2 the indi®erence curve is thus dx2=dxp 1 = ¡16 =x1. From the tangency condition 2 2 ¡16 =x1 = ¡1=2, we havex ^1 = 16 2. Therefore, the substitution e®ect and the income e®ect are p 0 Substitution e®ect =x ^1 ¡ x1 = 16( 2 ¡ 1) and p 0 Income e®ect = x1 ¡ x^1 = (Total e®ect) ¡ (Substitution e®ect) = 32 ¡ 16 2: 1=2 1=2 2 (d) Solving the utility maximization problems for u(x1; x2) = (x1 x2 ) = x1x2 and 1=2 1=2 1 1 u(x1; x2) = ln(x1 x2 ) = 2 ln x1 + 2 ln x2 yields x1 = Y=2p1 and x2 = Y=2p2, as 1=2 1=2 in the case of u(x1; x2) = x1 x2 . In general, let U(x1; x2) denote a monotonic transformation of an arbitrary utility 0 function u(x1; x2): U(x1; x2) = f(u(x1; x2)), where f > 0. Let us ¯rst consider the utility maximization for u(x1; x2): Max u(x1; x2) s:t: p1x1 + p2x2 = Y: The Lagrangian function is of the form L = u(x1; x2) + ¸(Y ¡ p1x1 ¡ p2x2): Thus, the ¯rst-order conditions (for an interior solution) are given by @u(x1; x2) = ¸p1 @x1 and @u(x1; x2) = ¸p2: @x2 Dividing the ¯rst expression by the second one gives @u(x ; x )=@x p 1 2 1 = 1 (1) @u(x1; x2)=@x2 p2 2 The demand functions are obtained from (1) and the budget constraint. Next, the utility maximization problem for U(x1; x2) = f(u(x1; x2)) is Max f(u(x1; x2)) s:t: p1x1 + p2x2 = Y: Since the Lagrangian function is of the form L = f(u(x1; x2)) + ¸(Y ¡ p1x1 ¡ p2x2); the ¯rst-order conditions are given by df(u(x1; x2)) @u(x1; x2) = ¸p1 du @x1 and df(u(x1; x2)) @u(x1; x2) = ¸p2: du @x2 By dividing the ¯rst expression by the second one, we have @u(x ; x )=@x p 1 2 1 = 1 ; @u(x1; x2)=@x2 p2 which is identical to (1). Since the demand functions are derived from this con- dition and the budget constraint, they must be the same as those in the case of u(x1; x2). Therefore, the demand functions are invariant with respect to mono- tonic transformations of the utility function. 4. (a) Let us measure leisure on the horizontal axis and consumption on the vertical axis. If the individual enjoy N = N¹, she does not work at all. So her consumption is zero. This means that the budget line passes through (N; x) = (N;¹ 0). If she reduces leisure time by one hour, she works for one hour and earns w. Since the unit price of consumption is p, the amount of consumption available to her is w=p. Therefore, the budget line also passes through (N; x) = (N¹ ¡ 1; w=p). It can be seen that the slope of the budget line is ¡w=p. By combining these pieces of information, the equation of the budget line is x = ¡wN=p + wN=p¹ , or equivalently, px + wN = wN:¹ (b) The maximization problem is of the form Max N 1=2x1=2 s:t: px + wN = wN:¹ Solving, we have wN¹ x = : 2p 3 (c) The demand function for leisure is N = N=¹ 2. Therefore, the optimal number of work hours is N¹ H = N¹ ¡ N = ; 2 which is clearly independent of p and w. This is because the substitution e®ect on the demand for leisure of an increase in its price o®sets the income e®ect (due to a rise in total income). 5. (a) Yes/True: The derivative of the average product of, say, labor (APL) is @(AP ) @ ³q(L; K)´ L = @L @L L @q(L;K) L ¡ q(L; K) = @L ; L2 which is positive if and only if @q(L; K)=@L £ L > q(L; K). Rearranging this yields @q(L; K) q(L; K) > : @L L The left-hand side of the above inequality is the marginal product of labor, and the right-hand side is the average product of labor. Therefore, the average product of labor increases if and only if the marginal product of labor is greater than the average product of labor. (b) f(L; K) AP = = 3K¹ 2 + 2; L L @f(L; K) MP = = 3K¹ 2 + 2: L @L APL = MPL holds because the production function is linear with respect to L given K = K¹ : f(L; K¹ ) = (3K¹ 2 + 2)L. (c) The marginal product of labor is given by @f(L; K) MP = L @L = ®AL®¡1K¯: MPL decreases if the following condition holds: @MP L = ®(® ¡ 1)AL®¡2K¯ < 0; @L from which we have 0 < ® < 1 (because ® > 0 and AL®¡2K¯ > 0). 4.
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