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113 台綜大微積分(B) 第 9 題

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113學年度 · 113微積分B · 第 9 題

題目

Problem

9. The Cobb-Douglas production function is P(x,y)=x1/3y2/3P(x, y) = x^{1/3} y^{2/3} (xx : capital, yy : labour) subject to budget constraint 3x1/2+5y1/2=453x^{1/2} + 5y^{1/2} = 45. Use the method of Lagrange multiplier to find the values of x,yx, y such that PP is maximized. (10%)

解答

解法一

思路

展開
  1. 本題要在條件約束 g(x,y)=3x1/2+5y1/245=0g(x,y) = 3x^{1/2} + 5y^{1/2} - 45 = 0 下,最大化目標函數 P(x,y)=x1/3y2/3P(x,y) = x^{1/3} y^{2/3}。其中 x>0x > 0y>0y > 0
  2. 第一步:建立拉格朗日乘子方程式系統P=λg\nabla P = \lambda \nabla g
    • 求偏導數: P=13x2/3y2/3, 23x1/3y1/3\nabla P = \left\langle \frac{1}{3}x^{-2/3}y^{2/3}, \ \frac{2}{3}x^{1/3}y^{-1/3} \right\rangle g=32x1/2, 52y1/2\nabla g = \left\langle \frac{3}{2}x^{-1/2}, \ \frac{5}{2}y^{-1/2} \right\rangle
    • 建立方程組:
      1. 13x2/3y2/3=λ32x1/2\frac{1}{3}x^{-2/3}y^{2/3} = \lambda \cdot \frac{3}{2}x^{-1/2}
      2. 23x1/3y1/3=λ52y1/2\frac{2}{3}x^{1/3}y^{-1/3} = \lambda \cdot \frac{5}{2}y^{-1/2}
  3. 第二步:兩式相除以消去 λ\lambda13x2/3y2/323x1/3y1/3=32λx1/252λy1/2    y2x=3y1/25x1/2\frac{\frac{1}{3}x^{-2/3}y^{2/3}}{\frac{2}{3}x^{1/3}y^{-1/3}} = \frac{\frac{3}{2}\lambda x^{-1/2}}{\frac{5}{2}\lambda y^{-1/2}} \implies \frac{y}{2x} = \frac{3 y^{1/2}}{5 x^{1/2}} 因為 x,y>0x, y > 0,可同除以 y1/2y^{1/2} 與同乘 x1/2x^{1/2}y1/22x1/2=35    5y1/2=6x1/2    x1/2=56y1/2\frac{y^{1/2}}{2 x^{1/2}} = \frac{3}{5} \implies 5 y^{1/2} = 6 x^{1/2} \implies x^{1/2} = \frac{5}{6} y^{1/2}
  4. 第三步:代入預算約束條件求解 xxyy
    • 代入 3x1/2+5y1/2=453 x^{1/2} + 5 y^{1/2} = 45
    • 求出 y=36y = 36x=25x = 25

答題過程

展開

我們設定拉格朗日乘子法系統。 目標函數:

P(x,y)=x1/3y2/3P(x, y) = x^{1/3} y^{2/3}

約束條件:

g(x,y)=3x1/2+5y1/245=0g(x, y) = 3x^{1/2} + 5y^{1/2} - 45 = 0

根據拉格朗日乘子法,最優解滿足偏導梯度平行:

P(x,y)=λg(x,y)\nabla P(x, y) = \lambda \nabla g(x, y)

我們對兩函數求偏導:

  • Px=13x2/3y2/3\displaystyle P_x = \frac{1}{3} x^{-2/3} y^{2/3}Py=23x1/3y1/3\displaystyle P_y = \frac{2}{3} x^{1/3} y^{-1/3}
  • gx=32x1/2\displaystyle g_x = \frac{3}{2} x^{-1/2}gy=52y1/2\displaystyle g_y = \frac{5}{2} y^{-1/2}

聯立方程組:

13x2/3y2/3=λ(32x1/2)— (1)\frac{1}{3} x^{-2/3} y^{2/3} = \lambda \left( \frac{3}{2} x^{-1/2} \right) \quad \text{--- (1)} 23x1/3y1/3=λ(52y1/2)— (2)\frac{2}{3} x^{1/3} y^{-1/3} = \lambda \left( \frac{5}{2} y^{-1/2} \right) \quad \text{--- (2)} 3x1/2+5y1/2=45— (3)3x^{1/2} + 5y^{1/2} = 45 \quad \text{--- (3)}

將式 (1) 除以式 (2) 以消去 λ\lambda

13x2/3y2/323x1/3y1/3=λ(32x1/2)λ(52y1/2)\frac{\frac{1}{3} x^{-2/3} y^{2/3}}{\frac{2}{3} x^{1/3} y^{-1/3}} = \frac{\lambda \left( \frac{3}{2} x^{-1/2} \right)}{\lambda \left( \frac{5}{2} y^{-1/2} \right)}

化簡左側與右側分式:

y2x=3y1/25x1/2\frac{y}{2x} = \frac{3 y^{1/2}}{5 x^{1/2}}

因為 x>0x > 0y>0y > 0,我們將兩側同除以 y1/2y^{1/2} 並同乘以 x1/2x^{1/2} 整理得:

y1/22x1/2=35    5y1/2=6x1/2    x1/2=56y1/2— (4)\frac{y^{1/2}}{2 x^{1/2}} = \frac{3}{5} \implies 5 y^{1/2} = 6 x^{1/2} \implies x^{1/2} = \frac{5}{6} y^{1/2} \quad \text{--- (4)}

將式 (4) 代回預算約束條件式 (3):

3(56y1/2)+5y1/2=453 \left( \frac{5}{6} y^{1/2} \right) + 5 y^{1/2} = 45 52y1/2+5y1/2=45    152y1/2=45\frac{5}{2} y^{1/2} + 5 y^{1/2} = 45 \implies \frac{15}{2} y^{1/2} = 45 y1/2=45215=6    y=36y^{1/2} = 45 \cdot \frac{2}{15} = 6 \implies y = 36

y1/2=6y^{1/2} = 6 代回式 (4) 求 xx

x1/2=56(6)=5    x=25x^{1/2} = \frac{5}{6} (6) = 5 \implies x = 25

因此,當資本投入 x=25x = 25 且勞動力投入 y=36y = 36 時,生產函數 P(x,y)P(x, y) 達到最大值。

結論:x=25,y=36x = 25, y = 36 時,生產函數 PP 取得最大值。