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} & \ \ \frac{x_2}{(1+x_1)^2} \\ \text{s.t.} & \ \ x_1+x_2 \leq 5 \\ \text{and} & \ \ 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} & \ \ \frac{x_2}{(1+x_1)^2} \\ \text{s.t.} & \ \ b_1+b_2 \leq 5 \\ & \ \ 0 \leq b_2 \leq b_1 \\ \text{and} & \ \ 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} & \ \ \frac{x_2}{(1+x_1)^2} \\ \text{s.t.} & \ \ x_1+x_2 \leq 5 \\ \text{and} & \ \ 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} & \ \ \frac{x_2}{(1+x_2)^2} \\ \text{s.t.} & \ \ 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 $x_1 = 1$.
