# Complement in production and the slope of factor demand curves

Considering a firm taking prices for granted and maximizing profits

$$pf(x_1,...,x_K) - \sum_{i=1}^K q_i x_i,$$

where $$f$$ is strictly concave. Furthermore, let the factor demand curves be the solutions $$x^*_i(p,q)$$ and the slopes the derivative with respect to factor prices $$\frac{\partial x^*_i(p,q)}{\partial q_j}$$.

Is it then possible to say anything about the slope of factor demand curves if it is assumed that factors are complements in the sense of all

$$(A) \ \ \frac{\partial^2 f(x_1,...,x_K)}{\partial x_i \partial x_j} > 0, \phantom{xxx}i\not = j$$

Motivation of the question

Intuitively, if some factor becomes more expensive less of the factor is used $$\frac{\partial x^*_j(p,q)}{\partial q_j}<0$$. Using less of the factor $$j$$ would decrease the marginal productivity of the factor $$i$$ if $$\frac{\partial^2 f(x_1,...,x_K)}{\partial x_i \partial x_j} > 0$$ which in itself would lead one to assume that $$\partial x_i^*(p,q)/\partial q_j <0$$. However, if I remember correctly this result does not hold in general for $$K>2$$.

However, it seems to me that condition (A) would rule out the existence of indirect effects resulting from a price increase in $$q_j$$ leading to lower demand of $$x_j$$ and with less use of $$x_j$$ less use of $$x_i$$ being counteracted by the fact that less use of some third factor $$x_s$$ having a positive effect on the marginal productivity of $$x_i$$. Hence, contrary to condition (A) requiring that

$$\frac{\partial^2 f(x_1,...,x_K)}{\partial x_i \partial x_s} < 0.$$

I think your intuition is correct and can be shown formally under your constraint.

When you say that:

$$\frac{\partial^2 f(x_1,...,x_K)}{\partial x_i \partial x_j} > 0, \phantom{xxx}i\not = j$$

You are forcing all the off-diagonal elements of the Hessian ($$H$$) to be positive. Now we know that $$H$$ is negative semi-definite. If we assume that it is negative definite (which will also force $$f(.)$$ to be strictly concave) then $$H$$ is invertible and its inverse is also symmetric negative definite. So we can write the Jacobian $$(\nabla_qx^*)$$ using SOC as:

$$\nabla_qx^* = (1/p)H^{-1}$$

Now use the result (it took me a while to find this but couldn't prove myself) that if off-diagonal elements of $$H$$ are non-negative then for $$H^{-1}$$ they will be non-positive (in the link it is for positive definite but should be directly extendible). This proves your intuition that the off-diagonal elements of the Jacobian will also be negative (or at least non-positive) which is what you are looking for.

• Thx. That makes it very clear. Commented Feb 28, 2022 at 13:04