Pure exchange economy with free and non-free disposal Question is as follows:



My answers are

(I) 



Let k is constant utility level

Then 

$$k=\sqrt{x_A^1x_A^2}$$

$$k^2/x_A^1=x_A^2$$

The first derivative is negative. So the indifference curve is decreasing.
The second derivative is positive.

Since utility function is quasiconcave, then preferences are convex because his prefers averages to extremes.

For B



Similarly 

For constant utility level k 

$$k=x_B^1/(1+x_B^2)$$

Then $$x_B^2=x_B^1/k-1$$

This is straight line. 

Due to the linear indifference curve he is indifferent between averages and extremes. Thus not strictly convex preference. 

(II)

In order to analyze whether preferences are nonstaiated and strictly monotonic, we need to look at its MU’s.

For A,

since good 1 and 2 may decrease the marginal utility, preferences are not monotonic.  But preferences are non-satiated. 

For B,

Since marginal utility is increasing, the preferences are monotonic. But preferences not strictly monotonic because getting more x2 decreases utility. However preferences are locally nonsatiated because by giving the consumer a bit more $x_B^1$ holding $x_B^2$ constant, he strictly better off. 

(III)



I did first three parts in this way. However, I couldn’t proceed the last three parts that I marked with red pen. Please help me to do this. Thank you.
Consider a pure exchange economy with two players - A and B.

A's utility function is

\begin{eqnarray*} u_A = x_A^{\frac{1}{2}}y_A^{\frac{1}{2}}\end{eqnarray*} 

B's utility function is

\begin{eqnarray*} u_B = \frac{x_B}{1 + y_B}\end{eqnarray*} 

Endowments of A and B are respectively

\begin{eqnarray*} \omega_A = (2,5), \ \omega_B =(5,2)\end{eqnarray*} 


  Find the set of Pareto efficient allocations in this economy.


Set of Pareto efficient allocations consists of those feasible allocations in which B consumes $0$ unit of commodity $Y$.

$$\text{Pareto Set} = \left\{\left(\left(x_A, y_A\right), \left(x_B, y_B\right)\right) \in \mathbb{R}_+^2 \times \mathbb{R}_+^2 | x_A + x_B = y_A + y_B = 7, y_B = 0\right\}$$


  Assume there is no free disposal. Player A can make a take-it-or-leave-it offer to B. Player B can either accept or reject. If player B rejects, then each player stays with his or her current endowment. If player B accepts, A's proposal is implemented. Find the equilibrium allocation outcome $\left(\left(x_A^*, y_A^*\right), \left(x_B^*, y_B^*\right)\right)$.


In a subgame perfect equilibrium, B's strategy will be to accept an allocation proposal $\left(\left(x_A, y_A\right), \left(x_B, y_B\right)\right)$ if $u_B(x_B, y_B) \geq u_B(5, 2) = \frac{5}{3}$, and reject otherwise. Given B's strategy, A will choose a proposal by solving the following maximization problem :
\begin{eqnarray*} \max_{\left(\left(x_A, y_A\right), \left(x_B, y_B\right)\right)\in \mathbb{R}_+^2 \times \mathbb{R}_+^2} &amp; \ x_A^\frac{1}{2}y_A^\frac{1}{2} \\ \text{s.t.} &amp; \ x_A+ x_B = 7 \\ &amp; \ y_A + y_B = 7 \\  &amp; \ \frac{x_B}{1+y_B} \geq \frac{5}{3}\end{eqnarray*} 

Solving this problem we get the following proposal :
$\left(\left(x_A^*, y_A^*\right), \left(x_B^*, y_B^*\right)\right) =  \left(\left(\frac{16}{3}, 7\right), \left(\frac{5}{3}, 0\right)\right)$


  Assume there is now free disposal. Redo the last part allowing for free disposal.


Allowing for free disposal, in a subgame perfect equilibrium, B's strategy will be to accept an allocation proposal $\left(\left(x_A, y_A\right), \left(x_B, y_B\right)\right)$ if $u_B(x_B, y_B) \geq u_B(5, 0) = 5$, and reject otherwise. Given B's strategy, A will choose a proposal by solving the following maximization problem :
\begin{eqnarray*} \max_{\left(\left(x_A, y_A\right), \left(x_B, y_B\right)\right)\in \mathbb{R}_+^2 \times \mathbb{R}_+^2} &amp; \ x_A^\frac{1}{2}y_A^\frac{1}{2} \\ \text{s.t.} &amp; \ x_A+ x_B = 7 \\ &amp; \ y_A + y_B = 7 \\  &amp; \ x_B \geq 5\end{eqnarray*} 

Solving this problem we get the following proposal :
$\left(\left(x_A^*, y_A^*\right), \left(x_B^*, y_B^*\right)\right) =  \left(\left(2, 7\right), \left(5, 0\right)\right)$


  Find competitive equilibrium (if it exists) in both the cases : (a) no free disposal, (b) free disposal


Competitive equilibrium consists of an allocation $\left(\left(x_A, y_A\right), \left(x_B, y_B\right)\right)$ and prices $(p_x, p_y = 1)$ such that


$x_A = \dfrac{2p_x + 5}{2p_x}$, $y_A = \dfrac{2p_x + 5}{2}$
$x_B = \dfrac{5p_x + 2}{p_x}$, $y_B = 0$
$x_A+x_B = 7$, $y_A+y_B=7$


Solving it we get, $p_x = \dfrac{9}{2}$.
Therefore, competitive equilibrium allocation is $\left(\left(x_A, y_A\right), \left(x_B, y_B\right)\right)=\left(\left(\dfrac{14}{9}, 7\right), \left(\dfrac{49}{9}, 0\right)\right)$ and supporting prices are $(p_x, p_y) = \left(\dfrac{9}{2}, 1\right)$. Equilibrium is the same in both cases - with or without free disposal.