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Answers to Problem Set 3 Spring 2007 John Rust Economics 425 University of Maryland Answers to Problem Set 3 0. Jack has a utility function for two perfectly divisible goods, x and y. Jack's utility function is 2 u(x;y) = (x + y) . Derive Jack's demand function for the two goods as a function of px (the price of good x), py (the price of good y), and I, (Jack's total income to be allocated to the 2 goods). Answer As I showed in class the answer to this is easy if you perform a montonic transformation, f (u) = pu to get v(x;y) = f (u(x;y)) = u(x;y) = (x + y)2 = x + y. The transformed utility function has constant marginal utility for both goods, with marginal utility equal to 1 for both goods. Thus, p p the indifference curves for this consumer are straight lines with a slope of 1. You can see this − since the slope of the indifference curve is the negative of the ratio of the marginal utilities of the two good, which are both 1 in this case. Or you can solve for an indifference curve directly, treating the consumption of good y as an implicit function of the consumption of good x, i.e. to solve for y(x) in the equation u(x;y(x)) = u = x + y(x) (1) so we see that y(x) = u x (2) − which is a straight line with a slope of 1, as claimed. Then, with such indifference curves, it − is easy to see that the demand for the two goods has a “bang-bang” type of solution, i.e. Jack is almost always at a “corner solution” where he/she consumes either entirely x or entirely y depending on which one is cheaper. Thus, if py > px, then the slope of the budget line I = pxxpyy is px=py > 1 (i.e. less steep than the indifference curves), so that Jack will get the highest − − utility by spending his entire budget on x and consume none of y as in figure 1 below. We can see from figure 1 that when py > px, it is better for Jack to only buy good x, and so with an income of 10, Jack can afford 10 units of x. If py < px, then we have the opposite situation where the budget line is steeper than Jack's indifference curves and so Jack only buys good y and none of good x. The “”knife-edge” case occurs when px = py. Then the indifference curve and the budget line both have the same slope, 1. In this knife-edge case, any combination of x and y − on the budget line is utility maximizing, and Jack is indifferent about consuming from any point on the budget line. Writing all of this down mathematically we have I=px if py > px x(p ; p ;I) = 0 if p < p x y 8 y x < [0;I=px] if py = px p > p : 0 if y x y(p ; p ;I) = I=p if p < p (3) x y 8 y y x < [0;I=py] if py = px : 1 Figure 1: Jack's utility maximization solution when I = 10, px = 1 and py = 2 Jacks demand for x and y when I=10 and p =1 and p =2 x y 10 Indifference curve 9 Indifference curve Budget line 8 7 6 5 4 Quantity of Good y 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 Quantity of Good x In the demand equations above, the notation y(px; py;I) = [0;I=px] denotes the knife-edge case where y could be any value between 0 and the number I=py, which happens when Jack spends his whole budget on good y. In the knife edge case, if Jack consumes y units in this interval, then to satisfy his budget constraint, his consumption of x is given by x = (I pyy)=px, i.e. he spends the − rest of his budget on x. It is possible to arrive at the same conclusion by writing down the Lagrangian for this problem as I did in class. We have (x;y;λ) = (x + y) + λ(I pxx pyy) (4) L − − Taking the first order conditions (for maximization) with respect to x and y we have ∂L (x∗;y∗;λ∗) = 1 λ∗ px 0 ∂x − ≤ ∂L (x∗;y∗;λ∗) = 1 λ∗ py (5) ∂y − The inequalties reflect the possibility of corner solutions: if x = 0, (a corner solution for x), ∂ λ ∗ then we have L(x∗;y∗; ∗) 0. However if x∗ > 0 (and interior solution for x∗), then we have ∂ λ ≤ L(x∗;y∗; ∗) = 0. Suppose that px < py. Then we conjecture that x∗ > 0, and solving the first ∂ λ λ first order condition (for an interior solution for x∗), L(x∗;y∗; ∗) = 0, we deduce that ∗ = 1=px, i.e. the marginal utility of income in this case is 1=px. From the other first order condition, we have ∂ L py (x∗;y∗;λ∗) = 1 λ∗ py = 1 < 0 (6) ∂y − − px 2 which implies that we have a corner solution for y∗, i.e. y∗ = 0. Using the budget equation, it follows that when y∗ = 0 we have x∗ = I=px (i.e. Jack spends his entire budget on good x, nothing on y), and this is the same solution as we got above. Thus, by following through the various cases, we can see that solving the Lagrangian problem gives us the same solution, i.e. the same demand functions for goods x and y as we derived above by intuitive means. 1. Provide an example of a utility function that leads to at least one good being an inferior good. Provide a general proof that all goods cannot be inferior goods. 1=2 3=2 Answer Consider the utility function u(x1;x2) = x1 + x2 . The Lagrangian for the utility maximization problem is 1=2 3=2 (x ;x ;λ) = x + x + λ(y p xx p x ) (7) L 1 2 1 2 − 1 − 2 2 The first order conditions for the maximization of the Lagrangian with respect to x1 and x2 are ∂ L λ 1 1=2 λ (x1∗;x2∗; ∗) = [x1∗]− ∗ p1 = 0 ∂x1 2 − ∂ L λ 3 1=2 λ (x1∗;x2∗; ∗) = [x2∗] ∗ p2 = 0 (8) ∂x2 2 − Soving these two equations we get 1 x = 1∗ λ2 2 4 p1 4 2 2 x∗ = λ p (9) 2 9 2 λ Substituting the above solutions into the budget constraint, y = p1x1∗ + p2x2∗ and solving for ∗ we get 1 4 λ2 p3 y 2 + 2 = (10) 4λ p1 9 Multiply both sides of this equation by λ2 and rearrange terms to get 4 3λ4 λ2 1 p2 y + 2 = 0 (11) 9 − 4p2 Let γ λ2. Then we can rewrite the equation above as a quadratic equation for γ ≡ aγ2 + bγ + c = 0 (12) a 4 p3 b y c 1 quadratic formula: γ where = 9 2, = and = p2 . Recall the there are two solutions for − 4 2 from the equation above. They are b pb2 4ac γ = − − (13) 2a Plugging in the formulas for a, b and c above we get p3 y y2 4 2 λ γ − 9 p1 p = v (14) ≡ u q8 p3 u 9 2 t 3 Regardless of whether we take the + or root in the formula above, λ∗ is an increasing function − 1 of y, as we would expect. But since x1∗ = λ2 2 , it follows that x1∗ is a decreasing function of y, i.e. 4 p1 x1 is an inferior good. Another example of a utility function that leads to an inferior good is x2 u(x1;x2) = log(x1) + e (15) Guess which of the two goods is inferior. Yes, you're right, it is x1, the good with the diminishing marginal utility. The other good, x2 has increasing marginal utility, just like the previous example. Setting up the Lagrangian as in the previous case and taking first order conditions, you should be able to show that 1 x1∗ = λp1 x∗ e 2 = λp2 (16) So we just need to show that λ is an increasing function of y in this case and we will have another example of a utility function for which one of the goods is an inferior good. Plugging into the budget constraint to solve for λ∗ we get 1 λ y = p1x1∗ + p2x2∗ = λ + p2 log( p2) (17) Rearranging this equation, we get 1 y p2 log(p2) λ λ = exp − − (18) ( p2 ) Now, by totally differentiating the above equation (more exactly, using the implicit function the- orem), we get ∂λ A = > 0 (19) ∂y 1 + A where 1 1 y p2 log(p2) λ A = exp − − (20) p2 ( p2 ) 1 Thus, we see that λ is an increasing function of y, so that x∗ = is a decreasing function of y 1 λp1 and is thus and inferior good. 2. Define mathematically what a homothetic utility function is. a. Show that homothetic utility functions lead to demand functions that are linear in income, y. Answer A homothetic function is defined to be a monotonic transformation of a homogeneous of degree 1 function. Thus, a utility function is homothetic if it can be represented as u(x) = f (l(x)) (21) where f : R R is a monotonic increasing function (i.e. f (u) > 0 for all u R) and l : Rn R ! 0 2 ! is a linearly homogenous function of the vector x (i.e.
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