The following is what I understand, so far. If we measure labour in the $x$-axis and capital in the $y$-axis, the slope of diagonal of the box is the capital-labour ratio $K/L$ in the economy. Let $A$ be the upright good and $B$ the upside-down good.

If the efficiency locus lies entirely on the right of the diagonal, that means if we consider any non-corner point on the locus, the slope of capital-labour ratio line in sector $A,$ $K_A/L_A < K/L$. Also, in sector $B,$ $K_B/L_B > K/L$. This implies $K_A/L_A < K_B/L_B$ i.e., $A$ is always relatively labour intensive. Therefore, using this logic, efficiency locus lying on one side of the diagonal just means that there is no factor intensity reversal taking place.

But according to Giancarlo Gandolfo International Trade Theory and Policy, efficiency locus will always lie on one side of the diagonal if the production functions in both sectors exhibit constant returns to scale. This implies that constant returns to scale function never exhibit factor intensity reversals, but this is not true; the same book mathematically shows CES production functions can exhibit them once.

Where am I going wrong? And why does constant returns to scale imply efficiency locus lying on one side of the diagonal.

  • $\begingroup$ Hi! What do you mean by "Let $A$ be the upright good and $B$ the upside-down good." in this context? It seems that you defined the $y$-axis as capital? $\endgroup$
    – Giskard
    Nov 20, 2021 at 19:55
  • $\begingroup$ I mean an Edgeworth box diagram. A is the good whose isoquants are drawn upright (origin is at lower left), B is the good whose isoquants are drawn upside down (origin is at upper right) $\endgroup$ Nov 21, 2021 at 6:58
  • $\begingroup$ Are we still talking about a situation where the $y$-axis represents the capital? If not, can you please reread your question and edit where necessary? $\endgroup$
    – Giskard
    Nov 21, 2021 at 7:35
  • $\begingroup$ Yes, I am talking about a standard Edgeworth box diagram with two goods (A,B) and two inputs (labour, capital) that you would use in something like HO model. $\endgroup$ Nov 21, 2021 at 8:43

2 Answers 2


Assuming that the total endowment of labor and capital is both equal to $1$, consider these examples with CRS technologies and efficient points on both sides of the diagonal of the box:

Example 1: $x = f_x(l_x,k_x)=l_x+k_x$, $y = f_y(l_y,k_y)=l_y+k_y$. In this case every allocation in the Edgeworth box is efficient.

Example 2: $x = f_x(l_x,k_x)=\sqrt{l_x^2+k_x^2}$, $y = f_y(l_y,k_y)=\sqrt{l_y^2+k_y^2}$. In this case, allocations on all four boundaries of the Edgeworth-box are efficient.

Example 3: $x = f_x(l_x,k_x)=\min(2l_x,k_x)$, $y = f_y(l_y,k_y)=\min(2l_y,k_y)$. In this case the feasible allocations satisfying $k_x\leq 2l_x$ and $k_y\leq 2l_y$ are efficient.

enter image description here


I have understood why constant returns to scale imply efficiency locus lying on one side of the diagonal, but not why this does not imply that there are no factor intensity reversals taking place.

My reasoning is as follows. The defining property is not constant returns to scale but homotheticity. If it does not lie entirely on one side, it must cross the diagonal at one point which implies there must be an optimal point in the diagonal, i.e., there is a pair of isoquants such that their slopes are equal at the diagonal. But if the former is true, by property of homotheticity, the slopes of the isoquants must be equal throughout the diagonal; therefore, every optimal point must lie on the the diagonal. Now, we have come to a contradiction. The locus cannot cross the diagonal while simultaneously lying entirely on the diagonal. This concludes my reason.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.