A Cobb-Douglas production function for tables is Q=2K^0.5 L^0.5 a) Can any table
ID: 1206955 • Letter: A
Question
A Cobb-Douglas production function for tables is Q=2K^0.5 L^0.5
a) Can any tables be built without tools?
b) Can the production of tables be mechanized?
c) There are currently 100 units of capital/tools, and in the short run no more tools can be either acquired or sold. What is the short-run production function?
d) If capital/tools rents for $100 per unit per day, and Labor can be hired for $200 per unit per day, what are the fixed costs and variable costs? What is the Total cost function?
e) What is the average cost function? To minimize average costs, how many tables should be produced?
f) What is the long term (meaning: capital/tool level is flexible) cost minimizing amount of labor and capital to produce 50 tables?
Explanation / Answer
a. No , tables cann't be built without tools because the production function of table is
Q=2K^0.5 L^0.5 , So when K or tools = 0 , then Q (Q =2L^0.5*0) is also 0.
b. No , tables cann't be Mechanized because the production function of table is
Q=2K^0.5 L^0.5 , So whenL or Labour = 0 , then Q =2L^0.5*0 is also 0.
c. Q=2K^0.5 L^0.5
when K = 100 , then
Q = 2(100)^0.5*L^o.5 = 20L^0.5
d. Fixed Cost = Cost of 100 Tools = rent of tools*number of tools = $100*100 = $10,000
Variabl Cost = Cost of Labour = Wage*Number of Labours = $200*L
Total Cost TC = $1000 + $200*L
e. Average Cost AC = TC/Q = [$1000 + $200*L]/20L^0.5
dAC/dL = -0.5*50L^-1.5 + 0.5*10L^-0.5
Putting dAC/aL = 0
25L^-1.5 = 5L^-0.5
L = 5
So, Q = 20*(5)^0.5 = 44.7
f.
For cost minimizing
(dQ/dL)/(dQ/dK) =MPL/MPK = wage/rent on tools
K/L = 200/100
K = 2L
50 = 2K^0.5 L^0.5
50 = 2(2L)^0.5 L^0.5
50 = 2.82L
L = 17.7
and k = 2L = 35.4
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