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

1. 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,L0 where xSP, 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.

1. The domain of production is convex;

2. Q(0,0)=0 (if it is defined on (0,0));

3. The production function is of class C² on D_p that is it admits partial derivatives of order 2 and they are continuous;

4. The production function is monotonically increasing in each variable;

5. The production function is quasiconcave that is: Q(x+(1-)y)min(Q(x),Q(y)) [0,1] x,yR_p.

In a preceding paper ([5]), one of the authors define a rational production function with constant return to scale as:

Q:D_pR²R₊, (K,L)Q(K,L)R₊ (K,L)D_p

K,L0

where P and R are homogenous polynomials in K and L, deg P=n, deg R=n-1, n2.

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.

2. A Type of a Rational Production Function

Let now:

, a,d0

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: g0

From the third inequality, we have that . From the first: therefore a0. Also: and because we get: or .

But is equivalent with: therefore e0 and .

Analysing the inequality: we have first = .

If c0 then 0 therefore = = =-1. In consequence:

 R.

If c0 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 e0. In this case: .

Case 2: g0

From the third inequality, we have that . From the first: .

If now a0 the inequality holds for all R. From the relation: we have = .

If c0 we obtain 0 therefore if then the inequality holds for all R. If then: .

If ae=b then: g=c0 – contradiction.

If now c0 we have: 0 therefore, if : or

If e0 then 0 therefore . If e0 then .

If now : that is: .

If e0 then 0 therefore . If e0 then: .

If c=0 then =0 and we must have and or .

Let suppose now that a0. In this case, from the first: therefore and because we get: or .

For the inequality we have = .

If c0 we obtain 0 therefore if then the inequality holds for all R. If then: .

If ae=b then: g=c0 – contradiction.

If now c0 we have: 0 therefore, if : or