Question

A company produces car engines with the following technology: Q=5KL where K denotes the number of conveyor belts and L denotes the number of working teams. The rent for one conveyor belt is 10000 per week and one team of workers costs 5000 per week. Additionally, each engine requires materials costing 2000. Imagine that the number of conveyor belts is fixed to five in the short-run What is the average short-run production cost of one engine, when total output equals 250 engines per week? Wählen Sie eine Antwort: o a. ACLQ-250)-52000 Ob. AC(Q-250) 2700 О с. АС(0-250) 2400 Od. AC(Q 250 7200

Answer 1

Given, Q = 5KL

Rent of capital, r = € 10,000 per week

Wage of 1 team of workers, w = € 5,000 per week

Additional material cost = € 2,000

Short run output, Q = 250 per week

First determine Marginal product of labor

Differentiating Q wrt L we get

$\LARGE \frac{dQ}{dL} = \frac{d(5KL)}{dL}$

$\large MP_L = 5K$

Similarly by differentiating Q wrt K we get

$\LARGE \frac{dQ}{dK} = \frac{d(5KL)}{dK}$

$\large MP_K= 5L$

Here we must satisfy the equimarginal criterion we get

$\large \frac{MP_L}{MP_K}= \frac{w}{r}$

Plug in the values in this we get

$\large \frac{5K}{5L}= \frac{5000}{10000}$

$\large \frac{K}{L}= \frac{1}{2}$

K = 0.5L

In short run Q = 250

Q = 5KL

250 = 5*0.5L*L

250 = 2.5L²

L² = 100

L = 10 teams of workers

K = 0.5L = 5 units

The total cost can be written as

TC = FC + VC

TC = 2,000*250 + (rK+wL)

TC =500,000+ 10,000*5 + 5,000*10

TC =500,000 + 50,000+ 50,000

TC = € 600,000

AC = TC /Q

AC = € 600,000/2,500

AC = € 2,400

Please contact if having any query thank you.