<p>Assuming that the total endowment of labor and capital is both equal to <span class="math-container">$1$</span>, consider these examples with CRS technologies and efficient points on both sides of the diagonal of the box:</p>
<p>Example 1: <span class="math-container">$x = f_x(l_x,k_x)=l_x+k_x$</span>, <span class="math-container">$y = f_y(l_y,k_y)=l_y+k_y$</span>. In this case every allocation in the Edgeworth box is efficient.</p>
<p>Example 2: <span class="math-container">$x = f_x(l_x,k_x)=\sqrt{l_x^2+k_x^2}$</span>, <span class="math-container">$y = f_y(l_y,k_y)=\sqrt{l_y^2+k_y^2}$</span>. In this case, allocations on all four boundaries of the Edgeworth-box are efficient.</p>
<p>Example 3: <span class="math-container">$x = f_x(l_x,k_x)=\min(2l_x,k_x)$</span>, <span class="math-container">$y = f_y(l_y,k_y)=\min(2l_y,k_y)$</span>. In this case the feasible allocations satisfying <span class="math-container">$k_x\leq 2l_x$</span> and <span class="math-container">$k_y\leq 2l_y$</span> are efficient.</p>
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CC BY-SA 4.0
2023-08-14T14:19:37.443Z
2023-08-14T14:32:55.837Z
2023-08-14T14:32:55.837Z
Amit
<p>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.</p>
<p>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.</p>

CC BY-SA 4.0
2021-11-22T10:50:22.390Z
2021-11-22T10:50:22.390Z
Shaikh Ammar
<p>The following is what I understand, so far.
If we measure labour in the <span class="math-container">$x$</span>-axis and capital in the <span class="math-container">$y$</span>-axis, the slope of diagonal of the box is the capital-labour ratio <span class="math-container">$K/L$</span> in the economy. Let <span class="math-container">$A$</span> be the upright good and <span class="math-container">$B$</span> the upside-down good.</p>
<p>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 <span class="math-container">$A,$</span> <span class="math-container">$K_A/L_A &lt; K/L$</span>. Also, in sector <span class="math-container">$B,$</span> <span class="math-container">$K_B/L_B &gt; K/L$</span>. This implies <span class="math-container">$K_A/L_A &lt; K_B/L_B$</span> i.e., <span class="math-container">$A$</span> 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.</p>
<p>But according to Giancarlo Gandolfo <em>International Trade Theory and Policy</em>, 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.</p>
<p>Where am I going wrong? And why does constant returns to scale imply efficiency locus lying on one side of the diagonal.</p>

CC BY-SA 4.0
2021-11-20T19:55:44.763Z
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?
Giskard
CC BY-SA 4.0
2021-11-21T06:58:23.267Z
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)
Shaikh Ammar
CC BY-SA 4.0
2021-11-21T07:35:44.120Z
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?
Giskard
CC BY-SA 4.0
2021-11-21T08:43:01.080Z
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.
Shaikh Ammar
<international-trade><production-function><general-equilibrium><edgeworth-box>
In a box diagram, why does efficiency locus lie on one side of the diagonal, if both sectors haves constant returns to scale function?
Shaikh Ammar
# In a box diagram, why does efficiency locus lie on one side of the diagonal, if both sectors haves constant returns to scale function?

(Asked by: username_0 on 2021-11-20)

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.

**Tags:** international-trade, production-function, general-equilibrium, edgeworth-box
> **Comments:**
> **username_1**: 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?  
> **username_0**: 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)  
> **username_1**: 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?  
> **username_0**: 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.  

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## 2 answers

### Answer by username_2 on 2023-08-14. Score: 3

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](https://i.sstatic.net/Y7P4i.png)](https://i.sstatic.net/Y7P4i.png)


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### Answer by username_0 on 2021-11-22. Score: 1

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.


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