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111 學年度台聯大微積分 A2 第 7 題

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111學年度 · 111台聯大微積分A2 · 第 7 題

題目

Problem

7. The production for a certain country in the early years following World War II is described by the function f(x,y)=30x2/3y1/3f(x, y) = 30x^{2/3}y^{1/3} units, when xx units of labor and yy units of capital were utilized. Find the approximate change in output if the amount expended on labor had been decreased from 125 units to 123 units and the amount expended on capital had been increased from 27 units to 29 units.

解答

解法一:全微分估計法

思路

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  1. 利用全微分(Total Differential)來估算函數值的變化量 Δfdf\Delta f \approx \mathrm{d}fdf=fx(x,y)dx+fy(x,y)dy\mathrm{d}f = f_x(x, y)\mathrm{d}x + f_y(x, y)\mathrm{d}y
  2. 題目中自變數的變化量為:
    • 勞動 xx 從 125 減少至 123: x0=125x_0 = 125dx=123125=2\mathrm{d}x = 123 - 125 = -2
    • 資本 yy 從 27 增加至 29: y0=27y_0 = 27dy=2927=2\mathrm{d}y = 29 - 27 = 2
  3. 計算在基準點 (125,27)(125, 27) 的偏導數 fxf_xfyf_y
  4. 代入公式求得產出的估計變動量。

答題過程

展開

給定 Cobb-Douglas 生產函數為:

f(x,y)=30x2/3y1/3f(x, y) = 30x^{2/3}y^{1/3}

設定基準點 (x0,y0)=(125,27)(x_0, y_0) = (125, 27)。自變數的微小變化量為:

dx=123125=2,dy=2927=2\mathrm{d}x = 123 - 125 = -2, \quad \mathrm{d}y = 29 - 27 = 2

1. 計算偏導數

  • xx 的偏導數

    fx(x,y)=3023x1/3y1/3=20x1/3y1/3f_x(x, y) = 30 \cdot \frac{2}{3} x^{-1/3} y^{1/3} = 20 x^{-1/3} y^{1/3}

    (125,27)(125, 27) 代入:

    fx(125,27)=20(125)1/3(27)1/3=20153=12f_x(125, 27) = 20 (125)^{-1/3} (27)^{1/3} = 20 \cdot \frac{1}{5} \cdot 3 = 12
  • yy 的偏導數

    fy(x,y)=3013x2/3y2/3=10x2/3y2/3f_y(x, y) = 30 \cdot \frac{1}{3} x^{2/3} y^{-2/3} = 10 x^{2/3} y^{-2/3}

    (125,27)(125, 27) 代入:

    fy(125,27)=10(125)2/3(27)2/3=102519=2509f_y(125, 27) = 10 (125)^{2/3} (27)^{-2/3} = 10 \cdot 25 \cdot \frac{1}{9} = \frac{250}{9}

2. 利用全微分計算變化量

Δfdf=fx(125,27)dx+fy(125,27)dy=12(2)+2509(2)=24+5009=216+5009=2849\Delta f \approx \mathrm{d}f = f_x(125, 27)\mathrm{d}x + f_y(125, 27)\mathrm{d}y = 12(-2) + \frac{250}{9}(2) = -24 + \frac{500}{9} = \frac{-216 + 500}{9} = \frac{284}{9}

因此,產出的近似變動量為增加 2849\boxed{\frac{284}{9}} 單位。


解法二:線性逼近(切平面逼近)法

思路

展開
  1. 利用函數在 (125,27)(125, 27) 的線性逼近(切平面)公式: L(x,y)=f(125,27)+fx(125,27)(x125)+fy(125,27)(y27)L(x, y) = f(125, 27) + f_x(125, 27)(x-125) + f_y(125, 27)(y-27)
  2. 近似變化量為 ΔfL(123,29)f(125,27)\Delta f \approx L(123, 29) - f(125, 27)
  3. 此方法在概念上與全微分相同,但在解題格式上以線性逼近函數的形態呈現,能使邏輯更加清晰。

答題過程

展開

在點 (125,27)(125, 27) 處,生產函數的值為:

f(125,27)=30(125)2/3(27)1/3=30253=2250f(125, 27) = 30(125)^{2/3}(27)^{1/3} = 30 \cdot 25 \cdot 3 = 2250

偏導數值分別為:

fx(125,27)=12,fy(125,27)=2509f_x(125, 27) = 12, \quad f_y(125, 27) = \frac{250}{9}

因此, f(x,y)f(x, y) 在點 (125,27)(125, 27) 的線性逼近函數 L(x,y)L(x, y) 為:

L(x,y)=2250+12(x125)+2509(y27)L(x, y) = 2250 + 12(x - 125) + \frac{250}{9}(y - 27)

(x,y)(x, y) 變為 (123,29)(123, 29) 時,產出的近似值為:

L(123,29)=2250+12(123125)+2509(2927)=2250+12(2)+2509(2)=2250+2849L(123, 29) = 2250 + 12(123 - 125) + \frac{250}{9}(29 - 27) = 2250 + 12(-2) + \frac{250}{9}(2) = 2250 + \frac{284}{9}

因此,產出的近似變化量為:

ΔfL(123,29)f(125,27)=2849\Delta f \approx L(123, 29) - f(125, 27) = \boxed{\frac{284}{9}}