Question #199468

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.


1
Expert's answer
2021-05-31T14:47:05-0400

a. optimal capital/labor ratio

MPL=5KMPL =5 K

MPK=5LMPK = 5L

equating MPL=MPKMPL = MPK

K=LK = L

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

Max Q=5LKQ=5LK

subject to

64,000=20L+80K64,000=20L+80K

Form the Lagrangian function

G=5LK+λ(64,00020L80K)G=5LK+\lambda(64,000-20L-80K)

G=5LK+λ64,00020λL80λKG=5LK+\lambda64,000-20\lambda L-80\lambda K

First order conditions are:

  1. 5K20λ=05K-20\lambda=0
  2. 5L80λ=05L-80\lambda=0
  3. 64,00020L80K=064,000-20L-80K=0

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

5K20λ=05K-20\lambda=0

5L80λ=05L-80\lambda=0


20K80λ=020K-80\lambda=0

5L80λ=05L-80\lambda=0

20K5L=020K-5L=0


Solve

64,00020L80K=064,000-20L-80K=0

5L+20K=0-5L+20K=0


64,00020L80K=064,000-20L-80K=0

20L+80K=0-20L+80K=0

64,00040L=064,000-40L=0

64,000=40L64,000=40L

L=1600L=1600

5L+20K=0-5L+20K=0

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

8000+20K=0-8000+20K=0

20K=800020K=8000

K=400K=400


L=1600,K=400L=1600, K=400

Q=0.5(1600)(400)Q=0.5(1600)(400)

Q=3,200,000Q=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.

MinimizeC=20L+80KMinimize C=20L+80K

subjecttosubject to

3,200,000=5LK3,200,000=5LK

Form Lagrangian function:

G=20L+80K+λ(3,200,0005LK)G=20L+80K+\lambda(3,200,000-5LK)

G=20L+80K+λ3,200,0005λLKG=20L+80K+\lambda3,200,000-5\lambda LK


First order conditions are:

  1. 205λK=020-5\lambda K=0
  2. 805λL=080-5\lambda L=0
  3. 3,200,0005LK=03,200,000-5LK=0

Solve 1 and 2

205λK=020-5\lambda K=0

805λK=080-5\lambda K=0


20K5λ=0\frac {20}{K} -5\lambda=0


80L5λ=0\frac {80}{L}-5\lambda=0


20K80L=0\frac {20}{K}-\frac {80}{L}=0


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

20K80L=0\frac {20}{K}-\frac {80}{L}=0

3,200,0005LK=03,200,000-5LK=0

Multiply top equation by L2K.

20L20L 280LK=0-80LK=0

3,200,0005LK=03,200,000-5LK=0


20L20L280LK=0-80LK=0

51,200,00080LK=051,200,000-80LK=0

20L20L251,200,000=0-51,200,000=0

LL2=2,560,000=2,560,000

L=1600L=1600


3,200,0005(1600)K=03,200,000-5(1600)K=0

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

K=400K=400



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS