<p>The bottom right origin is actually not in the Pareto set.  At that point, <span class="math-container">$(x_A,y_A)=(8,0)$</span>, so <span class="math-container">$U_A(x_A,y_A)=0$</span>.  Similarly, <span class="math-container">$(x_B,y_B)=(0,4)$</span>, so <span class="math-container">$U_B(x_B,y_B)=0$</span>.</p>

<p>As an example, <span class="math-container">$B$</span> could give one unit of <span class="math-container">$y$</span> to <span class="math-container">$A$</span> and thereby raise <span class="math-container">$A$</span>'s utility to 8 without hurting <span class="math-container">$B$</span>'s own utility (which would remain zero).  As a matter of fact, any allocation other than the top left origin is a Pareto improvement over the bottom right origin.</p>

CC BY-SA 4.0
2019-01-24T04:34:12.173Z
Really sorry, I meant to ask about bottom left corner. Have edited the original post.
PGupta
CC BY-SA 4.0
2019-01-23T02:34:47.347Z
2019-01-23T02:34:47.347Z
ExpelledFromParadise
<p>I believe you are correct. The points for which Ya = 0 and <span class="math-container">\begin{equation}
0\le Xa \le 4
\end{equation}</span>
will be all Pareto efficient points.</p>

<p>Proof: Consider an allocation like (2,0). The indifference curve through this point for Individual A is the positive x-axis. (Look at the graph for the IC of Individual B)</p>

<p><a href="https://i.sstatic.net/y3HDg.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/y3HDg.png" alt="enter image description here"></a></p>

<ul>
<li>Increasing the satisfaction level of A would require us to move to a point where Ya>0 but this would reduce the satisfaction level of B.</li>
<li>Increasing the satisfaction level of B would require us to move to a point where Ya&lt;0 but this is not possible as we cannot have a negative allocation of product y for individual A.</li>
</ul>

<p>Hence, all such points are also Pareto efficient in addition to the line of kinks for individual 2. </p>

CC BY-SA 4.0
2019-08-18T11:26:39.100Z
2019-08-28T17:06:00.840Z
2019-08-28T17:06:00.840Z
Rayhan Ali
<p>If <span class="math-container">$U_A(x_A,y_A)=x_Ay_A$</span> and <span class="math-container">$U_B(x_B,y_B)=min(x_B,y_B)$</span> and the total endowments are (8,4), is the Pareto set given by the line joining the kinks of B (black line shown in the diagram)? Shouldn't the bottom left origin and the flat portion on the horizontal axis also be Pareto efficient?
Suppose we are at the bottom left origin, there is no way to improve one person's utility without hurting the other. By this definition, the origin should be included in the Pareto set. Please point out any flaw in the logic.</p>

<p><a href="https://i.sstatic.net/nm8wq.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/nm8wq.png" alt="enter image description here"></a></p>

CC BY-SA 4.0
2019-02-01T18:03:53.680Z
With Cobb Douglas preferences, utility on any point on the horizontal segment is zero, so it shouldn't be possible to improve anyone's lot without hurting the other. Any thoughts?
PGupta
<general-equilibrium><pareto-efficiency><leontief>
Pareto set with Cobb-Douglas and Leontief preferences
PGupta
# Pareto set with Cobb-Douglas and Leontief preferences

(Asked by: username_0 on 2019-01-22)

If $U\_A(x\_A,y\_A)=x\_Ay\_A$ and $U\_B(x\_B,y\_B)=min(x\_B,y\_B)$ and the total endowments are (8,4), is the Pareto set given by the line joining the kinks of B (black line shown in the diagram)? Shouldn't the bottom left origin and the flat portion on the horizontal axis also be Pareto efficient?
Suppose we are at the bottom left origin, there is no way to improve one person's utility without hurting the other. By this definition, the origin should be included in the Pareto set. Please point out any flaw in the logic.

[![enter image description here](https://i.sstatic.net/nm8wq.png)](https://i.sstatic.net/nm8wq.png)

**Tags:** general-equilibrium, pareto-efficiency, leontief

> **Comments:**
> **username_0**: With Cobb Douglas preferences, utility on any point on the horizontal segment is zero, so it shouldn't be possible to improve anyone's lot without hurting the other. Any thoughts?  

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## 2 answers

### Answer by username_2 on 2019-08-18. Score: 3 | Accepted answer

I believe you are correct. The points for which Ya = 0 and \begin{equation}
0\le Xa \le 4
\end{equation}
will be all Pareto efficient points.

Proof: Consider an allocation like (2,0). The indifference curve through this point for Individual A is the positive x-axis. (Look at the graph for the IC of Individual B)

[![enter image description here](https://i.sstatic.net/y3HDg.png)](https://i.sstatic.net/y3HDg.png)

* Increasing the satisfaction level of A would require us to move to a point where Ya>0 but this would reduce the satisfaction level of B.
* Increasing the satisfaction level of B would require us to move to a point where Ya<0 but this is not possible as we cannot have a negative allocation of product y for individual A.

Hence, all such points are also Pareto efficient in addition to the line of kinks for individual 2.


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### Answer by username_1 on 2019-01-23. Score: 3

The bottom right origin is actually not in the Pareto set. At that point, $(x\_A,y\_A)=(8,0)$, so $U\_A(x\_A,y\_A)=0$. Similarly, $(x\_B,y\_B)=(0,4)$, so $U\_B(x\_B,y\_B)=0$.

As an example, $B$ could give one unit of $y$ to $A$ and thereby raise $A$'s utility to 8 without hurting $B$'s own utility (which would remain zero). As a matter of fact, any allocation other than the top left origin is a Pareto improvement over the bottom right origin.

> **Comments:**
> **username_0**: Really sorry, I meant to ask about bottom left corner. Have edited the original post.  

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