Answer to Question #187140 in Microeconomics for sad

Question

Given the cost function is

TC = 6L + 3K

Find out the optimal quantities of the two factor using Lagrangian method, if it is given that output is equal to 13.46 = L^3/4 . K^1/4.

Answer 1

Given,

$TC=6L+3K$

$13.46=L^{\frac{3}{4}}K^{\frac{1}{4}}$

Using Lagrangian method

L*= minimise cost subject to output constraint

$L^*=6L+3K-\lambda[L^{\frac{3}{4}}K^{\frac{1}{4}}-13.46]$

$\frac{\delta L^*}{\delta L}=6-\lambda[\frac{3}{4}L^{\frac{-1}{4}}K^{\frac{1}{4}}]=0$

$\frac{6(4)}{3L^{\frac{-1}{4}}K^{\frac{1}{4}}}=\lambda$

$8(\frac{L}{K})^{\frac{1}{4}}=\lambda............................................................(1)$

$\frac{\delta L}{\delta K}=3-\lambda[\frac{1}{4}L^{\frac{3}{4}}K^{\frac{1}{4}}]=0$

$\frac{3(4)}{L^{\frac{3}{4}}K^{\frac{-3}{4}}}=\lambda$

$12(\frac{K}{L})^{\frac{3}{4}}=\lambda$

equating

$\lambda'^s, (1) \& (2)$

$8(\frac{L}{K})^{\frac{1}{4}}=12(\frac{K}{L})^{\frac{3}{4}}$

$8L=12K$

$L=\frac{6}{4}K=\frac{3}{2}K$

$L=1.5K$