u(x, y) = x 1 3 y v(p, q, I) = u(x(p, q, I), y(p, q, I))

Transcription

1 December 4, 207 Let p, q, I > 0. We try to maximize the utility function u(x, y) = x y under the constraint I px qy = 0 () Find the statinary point of f(x, y subject to g(x, y = 0 to find (2) We define the indirect utility function by x(p, q, I) = I 2p, y(p, q, I) = I 2q, λ = 2 pqi v(p, q, I) = u(x(p, q, I), y(p, q, I)) and show () Show that the Roy s identy holds. I = λ(p, q, I) x + x(p, q, I) = 0 I Solution Assume that the function u(x, y) is maximal or minimal at (x, y) subject to g(x, y) = 0. Then there exists λ R satisfying It follows from () x and (2) y that Since () or (2) implies λ 0, we can derive x 2 y λp = () x y 2 λq = (2) I px qy = () λpx = x y, λqy = x y Moreover thanks to (), we find px = qy = λ x y Then it follows from () and (2) that px = qy = I 2 thus x = I 2p, y = I 2q 9 x y λ 2 pq = 0 2

2 and Accordingly Remark that () implies λ > 0. (2) The indirect utility function is λ 2 = 4pq 9pq I 2 = p 2 q 2 I 2 λ = 2 pqi We differentiate it in I to get Remark v(p, q, I) = I = 2 ( ) ( ) I I I 2 = 2p 2q 2 2 p q I 2 = 2 2 p q = λ p q I This is the reason why the function λ(p, q, I) is called the marginal utility of the income. () We partially differentiate v(p, q, I) を p to get p = I p 4 q = I 2p 2p I = I 2p 2 p q I I p 4 q This is Roy s identity. = I 2p λ = x I II Let p, q, I > 0. We try to maximize the utility function u(x, y) = log x + log y subject to the constraint I px qy = 0 Find the demand functions x(p, q, I) and y(p, q, I) and the marginal utility function of the budget λ(p, q, I). Solution Assume that the function u(x, y) is maximal or minimal at (x, y) subject to g(x, y) = 0. Then there exists λ R satisfying x λp = () y λq = (2) I px qy = () It folloes from () and (2) that { λp = x () λq = y (2) 22

3 and then from () /(2) that Moreover () implies p q = y x i.e. px = qy px = qy = I 2 Then the the marginal utility of the income is 従って x = I 2p, y = I 2q III Optimize under the constraint λ = p 2p I = 2 I z = f(x, y) = x + y g(x, y) := x 2 + 2y 2 24 = 0 Solution Assume that the function u(x, y) is maximal or minimal at (x, y) subject to g(x, y) = 0. Then there exists λ R satisfying + λ 2x = () + λ 4y = (2) x 2 + 2y 2 24 = () If λ = 0, it follows from () that = 0. Accordingly we find that λ 0 and from () and (2) that x = 2λ, y = 4λ We eliminate x and y in () by using these indentities to get Hence we find λ = ±8 and The statinary points of f under g(x, y) = 0 are where the double signs correspond. 4λ 2 + 8λ 2 = 24 i.e. 8λ 2 = 24 x = 4, y = mp2 (x, y, λ) = ( 4, 2, ± 8 ) IV Let p, q, I > 0. We try to maximize the utility function u(x, y) = x y 2 under the constraint g(x, y) := I px qy = 0 Find the demand functions x(p, q, I) and y(p, q, I) and the marginal utility function of the budget λ(p, q, I). 2

4 Solution If u is maximal or minimal at (x, y) subject to g(x, y) = 0, there exists λ R satisfying Then it follows from () and (2) that x 2 y 2 λp = () 2 x y λq = (2) I px qy = () { λp = x 2 y () λq = 2 x y (2) Moreover () /(2) leads to We substitute this to () to get p q = 2 y x thus px = 2 qy I 2 qy qy = I 2 qy = 0 Accordingly we get the demand functions Moreover the marginal utility of the budget is x = I p, y = 2I p λ = q ( I ) p = 2I q 2q2 p V We consider z = f(x, y) = x 2 y subject to the constraint g(x, y) := 2x 2 + y 2 = 0. Solution We find g x = 4x, g y = 2y, f x = 2xy, f y = x 2 If f is maximal or minimal subject to g(x, y) = 0 at (x, y), there exists λ R satisfying 2xy + λ 4x = 0 (i) x 2 + λ 2y = 0 (ii) 2x 2 + y 2 = 0 (iii) First we remark that (i) (x = 0 OR y + 2λ = 0) and divide the analysis into the following two cases. ()In case x = 0 (ii) implies λy = 0 which is equivallent to λ = 0 OR y = 0 If y = 0, we find (x, y) = (0, 0). It is, however, doesn t satisfies (iii). Accordingly we find y 0 hence λ = 0. We substitute x = 0 in (iii) to get y = ± 24

5 We find the stationary points in this case are (x, y, λ) = (0, ±, 0) (2)In case y + 2λ = 00 We substitute 2λ = y in (ii) to get x 2 y 2 = 0 i.e. (x = y OR x = y) (2a)In case x = y it follows from (iii) that (x, y) = (±, ± ) (Double Signs Correspond) and thus from λ = 2 ythat λ = 2 (DSC). The statinary points in this case are (2b)In case x = y it follows from (iii) that (x, y, λ) = (±, ±, 2 ) (DSC) (x, y) = (±, ) (DSC) and thus from λ = 2 y that λ = ± 2 (DSC) (x, y, λ) = (±,, ± 2 ) (DSC) VI A function in x, y = ϕ(x) satisfies the identity x 2 + ϕ(x) 2 xϕ(x) = 0 Express ϕ (x) and ϕ (x) by x and ϕ(x). Solution We diffferentiate the both sides of (#) to get 2x + 2ϕ(x)ϕ (x) ϕ(x) xϕ (x) = 0 ($) Namely from which it follows (2ϕ(x) x)ϕ (x) = 2x + ϕ(x) ϕ (x) = ϕ(x) 2x 2ϕ(x) x We differentiate the both sides of ($) to get 2 + 2(ϕ (x)) 2 + 2ϕ(x)ϕ (x) ϕ (x) ϕ (x) xϕ (x) = 0 25

6 Namely (2ϕ(x) x) = 2 2(ϕ (x)) 2 + 6ϕ (x) = 0ϕ(x)2 0xϕ(x) + 0x 2 (2ϕ(x) x) 2 0 = (2ϕ(x) x) 2 from which it follows ϕ (x) = 0 (2ϕ(x) x) 26