Utility maximization question setting up. Consider a consumer whose preferences can be represented by the following utility function: $$u(x_1,x_2)=\dfrac{x_2}{(1+x_1)^2}.$$
  
  
    Assume the agent's income is $y=5$. The price of one unit of good $1$ is $p_1=1$. For each unit of good $1$ the agent buys, he qualifies to buy up to one unit of good $2$ at an additional price of $p_2=1$. In other words, to buy one unit of good $2$ the agent has to first buy one unit of good $1$. The agent must consume everything he buys. Using this information, sketch the feasible set. Is it convex? Derive the utility maximizing bundle.
    How does your answers to question 4. change if the agent does not have to consume everything he buys ("free disposal")?
     


For question (4)

The budget constraint is 

$$x_1p_1+x_2(p_1+p_2)=y$$



My question is 

1) my budget constraint is true? 

2) what is the difference when budget constraint allow free disposal and not allow?

I don’t understand the part (5). 

My solution also is posted. 

Thanks a lot.
Consider a consumer whose preferences can be represented by the following utility function: $$u(x_1,x_2)=\dfrac{x_2}{(1+x_1)^2}.$$
  
  
    Assume the agent's income is $y=5$. The price of one unit of good $1$ is $p_1=1$. For each unit of good $1$ the agent buys, he qualifies to buy up to one unit of good $2$ at an additional price of $p_2=1$. In other words, to buy one unit of good $2$ the agent has to first buy one unit of good $1$. The agent must consume everything he buys. Using this information, sketch the feasible set. Is it convex? Derive the utility maximizing bundle.
    How does your answers to question 4. change if the agent does not have to consume everything he buys ("free disposal")?
     


For Q 4 :

Utility maximization problem of the consumer is :

\begin{eqnarray*} \max_{x_1, x_2} &amp; \ \ \frac{x_2}{(1+x_1)^2} \\ \text{s.t.} &amp; \ \ x_1+x_2 \leq 5 \\ \text{and} &amp;  \ \ 0 \leq x_2 \leq x_1  \end{eqnarray*}

Here is the constraint set of the consumer, along with a few indifference curves:



Observe that the constraint set is convex and the consumer does not spend all his income in optimum. His optimal consumption bundle is $(x_1, x_2) = (1,1)$.

For Q 5 :

Utility maximization problem (with free disposal) of the consumer is :

\begin{eqnarray*} \max_{x_1, x_2, b_1, b_2} &amp; \ \ \frac{x_2}{(1+x_1)^2} \\ \text{s.t.} &amp; \ \ b_1+b_2 \leq 5 \\  &amp;  \ \ 0 \leq b_2 \leq b_1  \\ \text{and} &amp;  \ \ 0 \leq x_1 \leq b_1, 0 \leq x_2 \leq b_2\end{eqnarray*}

Here $b_1$, $b_2$ denotes the amount of the two commodities bought by the consumer, and $x_1$, $x_2$ denotes the amount consumed. In this case, the consumer will try and maximize his consumption of commodity 2 $(x_2)$ by buying as much amount of commodity 2 $(b_2)$ as he can. Clearly, the solution to this utility maximization problem is
$b_1 = b_2 = x_2 = 2.5, x_1 = 0$.



For Q 4, here is a way to solve the optimization problem using Lagrangian method :

Given the utility maximization problem of the consumer :

\begin{eqnarray*} \max_{x_1, x_2} &amp; \ \ \frac{x_2}{(1+x_1)^2} \\ \text{s.t.} &amp; \ \ x_1+x_2 \leq 5 \\ \text{and} &amp;  \ \ 0 \leq x_2 \leq x_1  \end{eqnarray*}

We set up the Lagrangian as follows:

$\mathcal{L}(x_1, x_2) = \dfrac{x_2}{(1+x_1)^2} - \lambda(x_1+x_2-5)  +\mu_1(x_1-x_2)+ \mu_2x_2 $

Necessary conditions for optimality are as follows :

$\dfrac{\partial \mathcal{L}}{\partial x_1} = \dfrac{-2x_2}{(1+x_1)^3} - \lambda + \mu_1 = 0$

$\dfrac{\partial \mathcal{L}}{\partial x_2} = \dfrac{1}{(1+x_1)^2} -\lambda - \mu_1 + \mu_2 = 0$

$x_1+x_2 \leq 5$, $\lambda \geq 0$ and $\lambda(x_1+x_2-5) = 0$

$x_1 \geq x_2$, $\mu_1 \geq 0$ and $\mu_1(x_1-x_2) = 0$

$x_2 \geq 0$, $\mu_2 \geq 0$ and $\mu_2x_2 = 0$

Solving the above system, we get

$x_1 = 1$, $x_2 = 1$, $\mu_1 = \frac{1}{4}$, $\mu_2=0$, $\lambda = 0$



Alternatively for Q 4, the problem can also be converted to a single variable optimization problem. This is because the individual will always consume equal amounts of $x_1$ and $x_2$ in equilibrium. Substituting $x_2=x_1$ yields :

\begin{eqnarray*} \max_{x_2} &amp; \ \ \frac{x_2}{(1+x_2)^2} \\ \text{s.t.}  &amp;  \ \ 0 \leq x_2 \leq 2.5  \end{eqnarray*}

Differentiating $\dfrac{x_2}{(1+x_2)^2}$ with respect to $x_2$ gives the following necessary condition

$\dfrac{(1+x_2)^2 - 2(1+x_2)x_2}{(1+x_2)^4} = \dfrac{1-x_2^2}{(1+x_2)^4} = 0$

which yields $x_2 = 1$, and the corresponding value of $x_1 = 1$.