Acta Universitatis Danubius. Œconomica, Vol 10, No 5 (2014)
A Type of a Rational Production Function
Catalin Angelo Ioan^{1}, Gina Ioan^{2}
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 R^{2} – 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 D_{p}SP called domain of production.
It is called production function an application Q:D_{p}R_{+}, (K,L)Q(K,L)R_{+} (K,L)D_{p}.
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 C^{2} on D_{p} 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,yR_{p}.
In a preceding paper ([5]), one of the authors define a rational production function with constant return to scale as:
Q:D_{p}R^{2}R_{+}, (K,L)Q(K,L)R_{+} (K,L)D_{p}
K,L0
where P and R are homogenous polynomials in K and L, deg P=n, deg R=n1, 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