Answer to Question #199468 in Microeconomics for nasreen

Question

Murray Manufacturing Company produces caps. The firm’s production function is given as: Q = 5LK, where Q = quantity of caps, L = labour measured in person hours and K = capital measured in machine hours. Murray’s labour cost, including fringe benefits, is R20 per hour, while the firm uses R80 per hour as an implicit machine rental charge per hour. Murray’s current budget is R64, 000 per month to pay labour and capital.

Using the Lagrangian technique, determine the quantities of labour and capital that will allow the firm to maximize output given their budgeted input expenditure. What is the firm’s output?

Using the Lagrangian technique, demonstrate the duality in production and cost theory.

Answer 1

a. optimal capital/labor ratio

$MPL =5 K$

$MPK = 5L$

equating $MPL = MPK$

$K = L$

b. We use the Lagrangian to maximize Q subject to cost constraint.

Max $Q=5LK$

subject to

$64,000=20L+80K$

Form the Lagrangian function

$G=5LK+\lambda(64,000-20L-80K)$

$G=5LK+\lambda64,000-20\lambda L-80\lambda K$

First order conditions are:

1. $5K-20\lambda=0$
2. $5L-80\lambda=0$
3. $64,000-20L-80K=0$

Solve (1) and (2) to eliminate $\lambda$.

$5K-20\lambda=0$

$5L-80\lambda=0$

$20K-80\lambda=0$

$5L-80\lambda=0$

$20K-5L=0$

Solve

$64,000-20L-80K=0$

$-5L+20K=0$

$64,000-20L-80K=0$

$-20L+80K=0$

$64,000-40L=0$

$64,000=40L$

$L=1600$

$-5L+20K=0$

$-5(1600)+20K=0$

$-8000+20K=0$

$20K=8000$

$K=400$

$L=1600, K=400$

$Q=0.5(1600)(400)$

$Q=3,200,000$

c.

For duality to be shown, we are supposed to show that the cost minimization approach leads to the same answer as the maximizing quantity.

$Minimize C=20L+80K$

$subject to$

$3,200,000=5LK$

Form Lagrangian function:

$G=20L+80K+\lambda(3,200,000-5LK)$

$G=20L+80K+\lambda3,200,000-5\lambda LK$

First order conditions are:

1. $20-5\lambda K=0$
2. $80-5\lambda L=0$
3. $3,200,000-5LK=0$

Solve 1 and 2

$20-5\lambda K=0$

$80-5\lambda K=0$

$\frac {20}{K} -5\lambda=0$

$\frac {80}{L}-5\lambda=0$

$\frac {20}{K}-\frac {80}{L}=0$

Combine with equation (3) to solve for L and K

$\frac {20}{K}-\frac {80}{L}=0$

$3,200,000-5LK=0$

Multiply top equation by L²K.

$20L$ ²$-80LK=0$

$3,200,000-5LK=0$

$20L$²$-80LK=0$

$51,200,000-80LK=0$

$20L$²$-51,200,000=0$

$L$²$=2,560,000$

$L=1600$

$3,200,000-5(1600)K=0$

$-8000K=-3,200,000$

$K=400$