A two-person two commodity economy has social endowment of x = 1 unit of food and y = 1 unit of wine. Agents preferences are increasing in own consumption but decreasing in wine consumption of the other person. Preferences of agents A and B are as follows, $$u_A(x_A, y_A, y_B) = x_A[1 + max(y_A − y_B, 0)],$$ $$u_B(x_B, y_B, y_A) = x_B[1 + max(y_B − y_A, 0)],$$ where A consumes $x_A$ and $y_A$ units of x and y respectively, similarly B’s consumption is $x_B$ and $y_B$. Which of the following is a Pareto optimum allocation.

$A. (1/4,1/2) \ (3/4,1/2)$

$B. (1/2,1/4) \ (1/2,3/4)$

$C. (1/4,1) \ (3/4,0)$

$D. (1/4,0) \ (3/4,1)$

Answer is supposed to be the option D

My approach: if we take out the utilities associated with the given allocations then we'll get that Option B and C are providing the same utilities and for Agent B, option D is providing the best satisfaction. enter image description here

the answer must be a boundary solution according to me

i absolutely do not understand how to proceed after that. can someone please guide me? thanks a lot.

  • 1
    $\begingroup$ Allocation in option D does not exhaust all of $y$. Is it $\left(\left(\frac{1}{4}, 0\right), \left(\frac{3}{4}, 1\right)\right)$ ? $\endgroup$
    – Amit
    May 31, 2022 at 19:14

1 Answer 1


Allocation $((x_A, y_A), (x_B, y_B)) = \left(\left(\frac{1}{4}, 0\right), \left(\frac{3}{4}, 1\right)\right)$ is Pareto superior to $\left(\left(\frac{1}{4}, \frac{1}{2}\right), \left(\frac{3}{4}, \frac{1}{2}\right)\right)$, so option A is ruled out.

Allocation $\left(\left(\frac{1}{2}, 0\right), \left(\frac{1}{2}, 1\right)\right)$ is Pareto superior to $\left(\left(\frac{1}{2}, \frac{1}{4}\right), \left(\frac{1}{2}, \frac{3}{4}\right)\right)$, so option B is ruled out.

Allocation $\left(\left(\frac{1}{2}, 0\right), \left(\frac{1}{2}, 1\right)\right)$ is Pareto superior to $\left(\left(\frac{1}{4}, 1\right), \left(\frac{3}{4}, 0\right)\right)$, so option C is ruled out.

So, allocations in options A, B and C are not Pareto efficient.

Here is the set of all efficient allocations:

enter image description here

Set of feasible allocations is $\mathcal{F} = \left\{((x_A,y_A),(x_B,y_B))\in\mathbb{R}^2_+\times\mathbb{R}^2_+|x_A+x_B=1\wedge y_A+y_B=1\right\}$

Set of efficient allocations is $\mathcal{E} = \left\{((x_A,y_A),(x_B,y_B))\in\mathcal{F}|\left(y_A=0\wedge 0\leq x_A\leq \frac{2}{3}\right)\vee\left(y_A=1\wedge \frac{1}{3}\leq x_A\leq 1\right)\right\}$

  • $\begingroup$ i was trying to solve this through edgeworth box, could you please see to it, i have just added that. $\endgroup$ May 31, 2022 at 20:45
  • 1
    $\begingroup$ Shape of the IC looks correct to me. If you want to answer this MCQ using Edgeworth box, then draw ICs of the two individuals passing through the allocation in option D and check that it is efficient in the usual way. $\endgroup$
    – Amit
    May 31, 2022 at 23:35
  • 2
    $\begingroup$ Also note that not all boundaries of the Edgeworth box are efficient. First argue that if a feasible allocation is efficient then it must satisfy $y_A=0$ or $y_A=1$. Note that the converse is not true (see option C). $\endgroup$
    – Amit
    May 31, 2022 at 23:43

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