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Consider a consumer whose preferences can be represented by the following utility function: $$u(x_1,x_2)=\dfrac{x_2}{(1+x_1)^2}.$$

  1. 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.
  2. 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

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

enter image description here

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.

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Consider a consumer whose preferences can be represented by the following utility function: $$u(x_1,x_2)=\dfrac{x_2}{(1+x_1)^2}.$$

  1. 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.
  2. 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:

enter image description here

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 $x_1 = 1$.

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  • $\begingroup$ Really perfect answer. Thanks a lot. I have a short question. We can see that he cannot use all income in optimum when we look at the graph. And I can see the optimum bundle is (1,1) in the part four. However, is there a way to show this result (1,1) with mathematical solution? For example when I use lagrangian, I obtain (2.5, 2.5). But I think is not possible to show this result (1,1) in that way. Right? The only way is to use plot. ? $\endgroup$ – user315 May 3 '18 at 9:14
  • $\begingroup$ And the another question from the part 5 is that the optimum $x_2$ = 2.5 in this case. And also the reason for $x_1=0$ is the good 1 is bad good. Right? $\endgroup$ – user315 May 3 '18 at 9:28
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    $\begingroup$ @B11b I've updated the answer with Lagrangian method to solve the same problem. $\endgroup$ – Amit May 3 '18 at 11:13
  • $\begingroup$ @Amit Thank you so much. I have also very close question to this. Please look at my answer. You are really good instructor. I did the question. But I am not sure. Thank you. economics.stackexchange.com/questions/21820/… $\endgroup$ – user315 May 4 '18 at 17:32
  • $\begingroup$ The link provided by you isn't working. $\endgroup$ – Amit May 5 '18 at 0:32
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Hi this seems and interesting question ! What I would do is this

$ Max $ $ u(x_1,x_2)=\frac{x_2}{(1+x_1)^2} $

subject to : $p_1*x_1+p_2*x_2 = y$

$ x_1 = x_2 $

The intuition I feel is, as you see the utility, the good 1 is bad, he doesn't want to buy and consume it as he gets more of it, his utility will decrease, but in this case, he has to buy it in order to consume the good 2. so for every unit he bought from the good 1 he wants immediately to buy good 2 since good 2 increase his utility, that's why I put (x1=x2). For other instance, when you put $p_1*x_1+p_2*x_2+p_1*x_2 = y$, I feel you are saying you have to pay twice the price of good 1 if you want to get the good 2, and that's not true.

For the second answer, he will not consume the good 1, just the good 2, but the optimum will be the one you find in the first problem maximization but in this case $x_1 = 0 $.

Hope this convince you ! :) , maybe its wrong what i said, but intuitively I feel is ok .

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  • $\begingroup$ Firstly thank you so much. I’m going to study on what you said and then we will discuss. Please keep in touch. Thanks a lot again dear Gutierrez $\endgroup$ – user315 May 1 '18 at 20:13
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    $\begingroup$ Amit is a kick ass !! My respect, but as he consumes everything he buys is $ x_1 + x_2 = 5 $ in the point 1, so you have to consider the constraint as a equality. that is similar to mine :P. The second point he develops it as a master. I answer with equalities cause in class we take because there's a Walras theorem that says the consumer will spent all the money he had. $\endgroup$ – Miguel Gutierrez May 4 '18 at 4:04

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