Acta Universitatis Danubius. Œconomica, Vol 10, No 5 (2014)
A Type of a Rational Production Function
Catalin Angelo Ioan1, Gina Ioan2
Abstract: The article deals with a particular rational production function of two factors with constant scale return. It were determined from the compatibility conditions with the axioms of production function all the cases for a such function.
Keywords: production function; marginal productivity; average productivity
JEL Classification: D00
Introduction
In what follows we shall presume there is a certain number of resources, supposedly indivisible needed for the proper functioning of the production process.
We define on R2 – the production space for two resources: K – capital and L - labor as SP=(K,L)K,L0 where xSP, x=(K,L) is an ordered set of resources and we restrict the production area to a subset DpSP called domain of production.
It is called production function an application Q:DpR+, (K,L)Q(K,L)R+ (K,L)Dp.
For an efficient and complex mathematical analysis of a production function, we impose a number of axioms both its definition and its scope.
The domain of production is convex;
Q(0,0)=0 (if it is defined on (0,0));
The production function is of class C2 on Dp that is it admits partial derivatives of order 2 and they are continuous;
The production function is monotonically increasing in each variable;
The production function is quasiconcave that is: Q(x+(1-)y)min(Q(x),Q(y)) [0,1] x,yRp.
In a preceding paper ([5]), one of the authors define a rational production function with constant return to scale as:
Q:DpR2R+, (K,L)Q(K,L)R+ (K,L)Dp
K,L0
where P and R are homogenous polynomials in K and L, deg P=n, deg R=n-1, n2.
The compatibility conditions for that function to be of production were (from theorem 2):
where
and
,
are the average productivity relative to L and K respectively.
A Type of a Rational Production Function
Let now:
,
a,d0
We shall suppose that d=1 with loss of generality, after a simplification of the ratio with d.
Therefore,
let:
=
.
The average productivity relative to K and L are:
=
=
,
=
=
The first and the second derivatives are:
=
=
=
The compatibility conditions become:
or after simplifying:
Let
now the transformation:
,
therefore
and also: g=
.
The conditions become:
Case 1: g0
From
the third inequality, we have that
.
From the first:
therefore a0.
Also:
and because
we get:
or
.
But
is equivalent with:
therefore e0
and
.
Analysing
the inequality:
we have first =
.
If
c0
then 0
therefore
=
=
=-1.
In consequence:
R.
If
c0
then 0.
The roots of the equation
are:
therefore if
then:
which is possible because
if
that is:
which is true. In this case:
or
.
If
now
then:
.
Because, in this case:
we finally find that:
or
.
If
ae=b then:
which is true because e0.
In this case:
.
Case 2: g0
From
the third inequality, we have that
.
From the first:
.
If
now a0
the inequality holds for all
R.
From the relation:
we have =
.
If
c0
we obtain 0
therefore if
then the inequality holds for all
R.
If
then:
.
If ae=b then: g=c0 – contradiction.
If
now c0
we have: 0
therefore, if
:
or
If
e0
then
0
therefore
.
If e0
then
.
If
now
:
that is:
.
If
e0
then
0
therefore
.
If e0
then:
.
If
c=0 then =0
and we must have
and
or
.
Let
suppose now that a0.
In this case, from the first:
therefore
and because
we get:
or
.
For
the inequality
we have =
.
If
c0
we obtain 0
therefore if
then the inequality holds for all
R.
If
then:
.
If ae=b then: g=c0 – contradiction.
If
now c0
we have: 0
therefore, if
:
or