Transcript Lecture 11: Minimisation of Cost and Demand for Factors

Microeconomics
Course E
John Hey
This week: The Firm
• Tuesday
• Chapter 11: Cost minimisation and the
demand for factors.
• Wednesday
• Chapter 12: Cost curves.
• Thursday
• Exercise 4: A mathematical exercise on
profit maximisation.
Chapter 11
• In Chapter 10 we introduced the idea of an
isoquant – the locus of the points (in the
space (q1,q2) of the quantities of the
inputs) for which the output is constant.
• Also the production function:
• y = f (q1,q2) where y denotes the output.
• An isoquant is given by:
• y = f (q1,q2) = constant.
Particular cases
• Perfect substitutes 1 to a: isoquants are
straight lines with slope a.
• Perfect complements 1 with a: isoquants
are L-shaped and the line joining the
corners has slope a.
• Cobb-Douglas with parameter a:
isoquants are smoothly convex
everywhere.
Two dimensions
• The shape of the isoquants: depends on
the substitution between the two inputs.
(We call the slope of an isoquant the
marginal rate of substitution between the
inputs).
• The way in which the output changes from
one isoquant to another – depends on the
returns to scale.
Returns to scale with CobbDouglas technology : examples
•
•
•
•
•
•
•
Case 1: f(q1,q2) = q10.4 q20.6
Constant returns to scale.
Case 2: f(q1,q2) = q10.3 q20.45
Decreasing returns to scale.
Case 3: f(q1,q2) = q10.6 q20.9
Increasing returns to scale.
Note: the ratio of the exponents is the same:
hence the shape of the isoquants is the same –
but they have different returns to scale.
Chapters 11, 12 and 13
• We assume that a firm wants to maximise
its profits.
• We start with a small firm that has to take
the price of its output and those of its
inputs as given and fixed.
• Given these prices, the firm must choose
the optimal quantity of its output and the
optimum quantities of its inputs.
Chapters 11, 12 and 13
• We will do the analysis in two stages…
• …in Chapter 11 we find the optimal
quantities of the inputs – given a level of
output.
• …in Chapters 12 and 13 we will find the
optimal quantity of the output.
• (Recall that we are assuming that all
prices are given.)
Chapter 11
• So today we are finding the cheapest way of
producing a given level of output at given factor
(input) prices.
• This implies demands for the two factors...
• ... which are obviously dependent on the ‘givens’
– namely the level of output and the factor
prices.
• If we vary these ‘givens’ we are doing
comparative static exercises.
• The way that input demands vary depends upon
the technology.
Chapter 11
•
•
•
•
•
We use the following notation:
y for the level of the output.
p for the price of the output.
w1 and w2 for the prices of the inputs.
q1 and q2 for the quantities of the inputs.
• We define an isocost by
• w1q1 + w2q2 = constant
• …a line with slope –w1/w2
• Let’s go to Maple…
Chapter 11
• The optimal combination of the inputs is
given by the conditions:
• The slope of the isoquant at the optimal
point must be equal to to the relative
prices of the two inputs.
• (this assumes that the isoquants are
strictly convex)
• The output must be equal to the desired
output.
Factor demands with CD technology
q1y
 1


 ab
q2y




1



 ab
 a w2 




 b w1 




 b


 ab
 b w1 




 a w2 




a



 ab




Factor demands with CRS C-D
• The production function:
• y= q1a q2b where a + b =1
• The factor demands:
• q1 = y (aw2/bw1)b
• q2 = y (bw1/aw2 )a
Chapter 11
• What do we note?
• The demand curve for an input is a
function of the prices of the inputs and the
desired output.
• The shape of the function depends upon
the technology.
• From the demands we can infer the
technology of the firm.
Compito a casa/Homework
• CES technology with parameters c1=0.4, c2=0.5,
ρ=0.9 and s=1.0.
• The production function:
• y = ((0.4q1-0.9)+(0.5q2-0.9))-1/0.9
• I have inserted the isoquant for output = 40 (and
also that for output=60).
• I have inserted the lowest isocost at the prices
w1 = 1 and w2 = 1 for the inputs.
• The optimal combination: q1 = 33.38 q2 = 37.54
• and the cost = 33.58+37.54 = 70.92.
What you should do
• Find the optimal combination (either graphically
or otherwise) and the (minimum) cost to produce
the output for the following:
• w1 = 2 w2 = 1 y=40
• w1 = 3 w2 = 1 y=40
• w1 = 1 w2 = 1 y=60
• w1 = 2 w2 = 1 y=60
• w1 = 3 w2 = 1 y=60
• Put the results in a table.
Chapter 11
• Goodbye!