<p>In an edgeworth box model the total amount of goods must be constant. In your problem this translates to</p>

<p><span class="math-container">\begin{eqnarray}
x_{A_1} + x_{B_1} &amp;=&amp; \omega_{A_1} + \omega_{B_1} = 10 + 4 = 14 \tag{1}\\
x_{A_2} + x_{B_2} &amp;=&amp; \omega_{A_2} + \omega_{B_2} = 6 + 2 = 8 \tag{2}\\
\end{eqnarray}</span></p>

<p>So let's consider the first intercept</p>

<p><span class="math-container">$$
(x_{A_1}, x_{A_2}) = (0, 38/5) \tag{3}
$$</span></p>

<p>which you properly derived. If you replace these values in Eq. (1) you get</p>

<p><span class="math-container">$$
0 + x_{B_1} = 14 ~~~\Rightarrow~~~ x_{B_1} = 14 \tag{4}
$$</span></p>

<p>and if you replace in Eq. (2) you get</p>

<p><span class="math-container">$$
\frac{38}{5} + x_{B_2} = 8 ~~~\Rightarrow~~~ x_{B_2} = \frac{2}{5} \tag{5}
$$</span></p>

<p>So there you have it</p>

<p><span class="math-container">$$
(x_{A_1}, x_{A_2}) = (0,38/5), ~~~(x_{B_1}, x_{B_2}) = (14,2/5)
$$</span></p>

<p>You can do the same for the other intercept, I will leave that for you to complete</p>

CC BY-SA 4.0
2019-01-11T00:24:38.267Z
2019-01-11T00:24:38.267Z
caverac
<p>We are given the following two-consumer, two-goods pure exchange economy.</p>
<p><span class="math-container">$u_1(x_1,y_1)=2x_1+5y_1$</span>,
<span class="math-container">$u_2(x_2,y_2)=3x_2+7y_2$</span> </p>
<p><span class="math-container">$\omega_1=(\omega_1^X,\omega_1^Y)=(4,6)$</span> and <span class="math-container">$\omega_2=(\omega_2^X,\omega_2^Y)=(10,2)$</span></p>
<p>Set of feasible allocations is given by:</p>
<p><span class="math-container">$\mathcal{F}=\{((x_1,y_1),(x_2,y_2))\in\mathbb{R}^2_+\times\mathbb{R}^2_+|x_1+x_2=14 \ \wedge \ y_1+y_2=8\}$</span></p>
<p>Set of Pareto efficient allocations is given by:</p>
<p><span class="math-container">$\mathcal{P}=\{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|x_1=0 \ \vee \ y_1 = 8\}$</span></p>
<p>Core is the set of efficient allocations that satisfy the property that both the individuals get at least as much satisfaction as their respective endowments. Observing the graph, we get the core as:</p>
<p><span class="math-container">\begin{eqnarray*} \mathcal{C} &amp; = &amp; \left\{((x_1,y_1),(x_2,y_2))\in\mathcal{P}|2x_1+5y_1\geq 38 \ \wedge \ 3x_2+7y_2\geq 44\right\} \\ &amp; = &amp; \left\{((x_1,y_1),(x_2,y_2))\in\mathcal{P}|x_1=0 \ \wedge \ \frac{38}{5}\leq y_1 \leq \frac{54}{7}\right\} \end{eqnarray*}</span></p>
<p>Here is the picture:
<a href="https://i.sstatic.net/4UAWa.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/4UAWa.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-09-28T10:07:56.650Z
2023-09-28T10:34:59.200Z
2023-09-28T10:34:59.200Z
Amit
<p>Consider an exchange economy with two goods (Good 1 and 2) and two individuals (<span class="math-container">$A$</span> and <span class="math-container">$B$</span>):</p>

<ul>
<li><p><span class="math-container">$A$</span>'s utility function is <span class="math-container">$u_A(x_{A_1}, x_{A_2}) = 2x_{A_1} + 5x_{A_2}$</span> </p></li>
<li><p><span class="math-container">$B$</span>'s utility function is <span class="math-container">$u_B(x_{B_1}, x_{B_2}) = 3x_{B_1} + 7x_{B_2}$</span> </p></li>
<li><p><span class="math-container">$A$</span>'s initial endowment is <span class="math-container">$(\omega_{A_1}, \omega_{A_2}) = (4, 6)$</span></p></li>
<li><p><span class="math-container">$B$</span>'s initial endowment is <span class="math-container">$(\omega_{B_1}, \omega_{B_2}) = (10, 2)$</span></p></li>
</ul>

<p>The core condition requires:</p>

<p><span class="math-container">$$
2x_{A_1} + 5x_{A_2} \geq 2\omega_{A_1} + 5\omega_{A_2} = 2 \times 4 + 5 \times 6 = 38
$$</span></p>

<p>and</p>

<p><span class="math-container">$$
3x_{B_1} + 7x_{B_2} \geq 3\omega_{B_1} + 7\omega_{B_2} = 3 \times 10 + 7 \times 2 = 44
$$</span></p>

<p>The line with  equation <span class="math-container">$2x_{A_1} + 5x_{A_2} = 38$</span> hits the left boundary at:</p>

<p><span class="math-container">$$
(x_{A_1}, x_{A_2}) = (0,38/5), ~~~~(x_{B_1}, x_{B_2}) = (14, 2/5)
$$</span></p>

<p>The line with equation <span class="math-container">$3x_{B_1} + 7x_{B_2} = 44$</span> hits the left boundary at:</p>

<p><span class="math-container">$$ 
(x_{A_1}, x_{A_2}) = (0,54/7) ~~~ (x_{B_1}, x_{B_2}) = (14, 2/7)
$$</span></p>

<p>I don't understand how these intercepts have been calculated. </p>

<p>I've used the MRS and endowment point to calculate the <span class="math-container">$(x_{A_1}, x_{A_2})$</span>
intercepts correctly:</p>

<p><span class="math-container">$$x_{A_2} = -\frac{2}{5}x_{A_1} + \frac{38}{5}$$</span></p>

<p>But get an incorrect <span class="math-container">$(x_{B_1}, x_{B_2})$</span> intercept:</p>

<p><span class="math-container">$$
x_{A_1} = 14, ~~ x_{A_2} = 2 
$$</span></p>

<p>Please can someone enlighten me? No worries if you can't.</p>

<p>Alex</p>

<microeconomics><free-trade><exchange>
How to calculate the intercept of an indifference curve in an Edgeworth box?
Alex Bridges
# How to calculate the intercept of an indifference curve in an Edgeworth box?

(Asked by: username_0 on 2019-01-09)

Consider an exchange economy with two goods (Good 1 and 2) and two individuals ($A$ and $B$):

* $A$'s utility function is $u\_A(x\_{A\_1}, x\_{A\_2}) = 2x\_{A\_1} + 5x\_{A\_2}$
* $B$'s utility function is $u\_B(x\_{B\_1}, x\_{B\_2}) = 3x\_{B\_1} + 7x\_{B\_2}$
* $A$'s initial endowment is $(\omega\_{A\_1}, \omega\_{A\_2}) = (4, 6)$
* $B$'s initial endowment is $(\omega\_{B\_1}, \omega\_{B\_2}) = (10, 2)$

The core condition requires:

$$
2x\_{A\_1} + 5x\_{A\_2} \geq 2\omega\_{A\_1} + 5\omega\_{A\_2} = 2 \times 4 + 5 \times 6 = 38
$$

and

$$
3x\_{B\_1} + 7x\_{B\_2} \geq 3\omega\_{B\_1} + 7\omega\_{B\_2} = 3 \times 10 + 7 \times 2 = 44
$$

The line with equation $2x\_{A\_1} + 5x\_{A\_2} = 38$ hits the left boundary at:

$$
(x\_{A\_1}, x\_{A\_2}) = (0,38/5), ~~~~(x\_{B\_1}, x\_{B\_2}) = (14, 2/5)
$$

The line with equation $3x\_{B\_1} + 7x\_{B\_2} = 44$ hits the left boundary at:

$$
(x\_{A\_1}, x\_{A\_2}) = (0,54/7) ~~~ (x\_{B\_1}, x\_{B\_2}) = (14, 2/7)
$$

I don't understand how these intercepts have been calculated.

I've used the MRS and endowment point to calculate the $(x\_{A\_1}, x\_{A\_2})$
intercepts correctly:

$$x\_{A\_2} = -\frac{2}{5}x\_{A\_1} + \frac{38}{5}$$

But get an incorrect $(x\_{B\_1}, x\_{B\_2})$ intercept:

$$
x\_{A\_1} = 14, ~~ x\_{A\_2} = 2
$$

Please can someone enlighten me? No worries if you can't.

Alex

**Tags:** microeconomics, free-trade, exchange


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## 2 answers

### Answer by username_1 on 2019-01-11. Score: 1

In an edgeworth box model the total amount of goods must be constant. In your problem this translates to

\begin{eqnarray}
x\_{A\_1} + x\_{B\_1} &=& \omega\_{A\_1} + \omega\_{B\_1} = 10 + 4 = 14 \tag{1}\\
x\_{A\_2} + x\_{B\_2} &=& \omega\_{A\_2} + \omega\_{B\_2} = 6 + 2 = 8 \tag{2}\\
\end{eqnarray}

So let's consider the first intercept

$$
(x\_{A\_1}, x\_{A\_2}) = (0, 38/5) \tag{3}
$$

which you properly derived. If you replace these values in Eq. (1) you get

$$
0 + x\_{B\_1} = 14 ~~~\Rightarrow~~~ x\_{B\_1} = 14 \tag{4}
$$

and if you replace in Eq. (2) you get

$$
\frac{38}{5} + x\_{B\_2} = 8 ~~~\Rightarrow~~~ x\_{B\_2} = \frac{2}{5} \tag{5}
$$

So there you have it

$$
(x\_{A\_1}, x\_{A\_2}) = (0,38/5), ~~~(x\_{B\_1}, x\_{B\_2}) = (14,2/5)
$$

You can do the same for the other intercept, I will leave that for you to complete


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### Answer by username_2 on 2023-09-28. Score: 1

We are given the following two-consumer, two-goods pure exchange economy.

$u\_1(x\_1,y\_1)=2x\_1+5y\_1$,
$u\_2(x\_2,y\_2)=3x\_2+7y\_2$

$\omega\_1=(\omega\_1^X,\omega\_1^Y)=(4,6)$ and $\omega\_2=(\omega\_2^X,\omega\_2^Y)=(10,2)$

Set of feasible allocations is given by:

$\mathcal{F}=\{((x\_1,y\_1),(x\_2,y\_2))\in\mathbb{R}^2\_+\times\mathbb{R}^2\_+|x\_1+x\_2=14 \ \wedge \ y\_1+y\_2=8\}$

Set of Pareto efficient allocations is given by:

$\mathcal{P}=\{((x\_1,y\_1),(x\_2,y\_2))\in\mathcal{F}|x\_1=0 \ \vee \ y\_1 = 8\}$

Core is the set of efficient allocations that satisfy the property that both the individuals get at least as much satisfaction as their respective endowments. Observing the graph, we get the core as:

\begin{eqnarray\*} \mathcal{C} & = & \left\{((x\_1,y\_1),(x\_2,y\_2))\in\mathcal{P}|2x\_1+5y\_1\geq 38 \ \wedge \ 3x\_2+7y\_2\geq 44\right\} \\ & = & \left\{((x\_1,y\_1),(x\_2,y\_2))\in\mathcal{P}|x\_1=0 \ \wedge \ \frac{38}{5}\leq y\_1 \leq \frac{54}{7}\right\} \end{eqnarray\*}

Here is the picture:
[![enter image description here](https://i.sstatic.net/4UAWa.png)](https://i.sstatic.net/4UAWa.png)


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