均衡底下的"離散方式"或"連續方式"? - 普考
By Olive
at 2013-05-26T14:52
at 2013-05-26T14:52
Table of Contents
背景
a物20元,b物10元;a物的總效用函數為10x-x^2.若買b物的MU為2,問a物應該買多少
出處
某年普考 經濟學
解答
U(x) = 10x-x^2 則 MUa = (d/dx)U(x) = 10-2x. MU ≠ U(x)/x
Pa = 20, Pb = 10, MUb=2.
若邊際效用遞減法則成立, 則均衡狀態 MUa/Pa = MUb/Pb
所以 MUa/20 = 2/10. 則 MUa = 4.
10-2x = 4 則 x=3, a物買3個單位.
想法
我有上奇x知x+問過
因為我完全不懂為什麼 (10x-x^2)/x 可以變成 10-2x 消去x後不是10-x嗎?
版友的解釋是
MUa 是 a 物的邊際效用...什麼叫 "邊際效用"?
離散方式處理, 邊際效用是 "多消費一單位能增加多少效用", 就是 U(x+1)-U(x).
連續方式處理, 就是 lim_{△x→0} (U(x+△x)-U(x))/△x = U(x) 對 x 的微分.
就算用離散方式, MUa 也不會是 U(x)/x, 那是 "平均效用" 而不是 "邊際效用".
如果用離散方式, MUa = [10(x+1)-(x+1)^2]-(10x-x^2) = 10-(2x+1) = 9-2x.
嗄?什麼離散方式或連續方式的... 有強者可以用比較簡單的方式說明嗎= ="
--
a物20元,b物10元;a物的總效用函數為10x-x^2.若買b物的MU為2,問a物應該買多少
出處
某年普考 經濟學
解答
U(x) = 10x-x^2 則 MUa = (d/dx)U(x) = 10-2x. MU ≠ U(x)/x
Pa = 20, Pb = 10, MUb=2.
若邊際效用遞減法則成立, 則均衡狀態 MUa/Pa = MUb/Pb
所以 MUa/20 = 2/10. 則 MUa = 4.
10-2x = 4 則 x=3, a物買3個單位.
想法
我有上奇x知x+問過
因為我完全不懂為什麼 (10x-x^2)/x 可以變成 10-2x 消去x後不是10-x嗎?
版友的解釋是
MUa 是 a 物的邊際效用...什麼叫 "邊際效用"?
離散方式處理, 邊際效用是 "多消費一單位能增加多少效用", 就是 U(x+1)-U(x).
連續方式處理, 就是 lim_{△x→0} (U(x+△x)-U(x))/△x = U(x) 對 x 的微分.
就算用離散方式, MUa 也不會是 U(x)/x, 那是 "平均效用" 而不是 "邊際效用".
如果用離散方式, MUa = [10(x+1)-(x+1)^2]-(10x-x^2) = 10-(2x+1) = 9-2x.
嗄?什麼離散方式或連續方式的... 有強者可以用比較簡單的方式說明嗎= ="
--
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普考
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