I have also following budget set

$$B=\{x=(x_1,x_2)\in R^2_+ \mid 2\sqrt{x_1}+x_2\le y\}$$

where y is income.

Assume that there are two stories. The agent can shop in both of them. The first store has a quantity discount on good1 as described above. And a fixed unit price of good 2. $p_2=1$. The other store charges fixed unit prices, $p_1=2$ and $p_2=1/2$ respectively. Assume that income y is 6. How can I write the feasible budget constraint for utility maximization $U(x_1, x_2)=x_1x_2$. And since there is two stores, how can I write Lagrangian function for calculation of the optimization problem.

I know I have to write my trials. But I cannot write properly any budget constraint and Lagrangian therefore, I cannot post my solution.

Any help is appreciated. Thank you.


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  • $\begingroup$ your description of the problem is insufficient. is the consumer able to go to just one store or to both? $\endgroup$
    – saguru
    May 9, 2018 at 14:14
  • $\begingroup$ @saguru the question say nothing about this. But I guess the consumer can go to both. $\endgroup$
    – studentp
    May 9, 2018 at 19:33
  • $\begingroup$ so the consumer is able to shop good 1 at one store and then go to the other for good 2? Then for good 2 it is obvious. Could you please add the quantity discount information too? $\endgroup$
    – saguru
    May 10, 2018 at 8:22
  • $\begingroup$ and you should probably reconsult your micro books to understand the connection between the budget set and the buget constraint. as you are given the budget set, the budget constraint is given too $\endgroup$
    – saguru
    May 10, 2018 at 8:23
  • $\begingroup$ @saguru this is the only one part of my question. I post whole question. Thanks for helps:) $\endgroup$
    – studentp
    May 10, 2018 at 8:52

1 Answer 1


Let $x_{ij}$ denote the amount of good $i$ bought by the consumer from store $j$. So, consumer's utility maximization problem is described as follows : \begin{eqnarray*} \max_{x_{11}, x_{12}, x_{21}, x_{22}} & (x_{11}+x_{12})(x_{21}+x_{22}) \\ \text{s.t.} & \ \ 2\sqrt{x_{11}} + x_{21} + 2x_{12} + \frac{1}{2} x_{22} \leq 6 \\ & x_{11} \geq 0, x_{12} \geq 0, x_{21} \geq 0, x_{22} \geq 0 \end{eqnarray*} Given that good 2 is cheaper in store 2, we know that the solution to this problem will satisfy $x_{21} = 0$. Consequently, it is clear from the objective that consumer will always choose $x_{22} > 0$. Also consumer's utility is an increasing function, so consumer will spend all his income in optimum. Taking them into account, we can rewrite the utility maximization problem of the consumer as : \begin{eqnarray*} \max_{x_{11}, x_{12}, x_{22}} & (x_{11}+x_{12})x_{22} \\ \text{s.t.} & \ \ 2\sqrt{x_{11}} + 2x_{12} + \frac{1}{2} x_{22} = 6 \\ & x_{11} \geq 0, x_{12} \geq 0, x_{22} > 0\end{eqnarray*} Now we set up the Lagrangian: \begin{eqnarray*} \mathcal{L}(x_{11}, x_{12}, x_{22}) = (x_{11}+x_{12})x_{22} - \lambda (2\sqrt{x_{11}} + 2x_{12} + \frac{1}{2} x_{22} - 6) + \mu_1 x_{11} + \mu_2 x_{12} \end{eqnarray*} Corresponding Kuhn-Tucker conditions are : \begin{eqnarray*} \frac{\partial \mathcal{L}}{\partial x_{11}} & = & x_{22} - \frac{\lambda}{\sqrt{x_{11}}} + \mu_1 = 0 \\ \frac{\partial \mathcal{L}}{\partial x_{12}} & = & x_{22} - 2\lambda + \mu_2 = 0 \\ \frac{\partial \mathcal{L}}{\partial x_{22}} & = & x_{11} + x_{12} - \frac{\lambda}{2} = 0 \\ \frac{\partial \mathcal{L}}{\partial \lambda} & = & 6 - 2\sqrt{x_{11}} - 2x_{12} - \frac{1}{2} x_{22} = 0, \ \lambda \geq 0 \\ \frac{\partial \mathcal{L}}{\partial \mu_1} & = & x_{11} \geq 0, \ \mu_1 \geq 0, \ \mu_1x_{11} = 0 \\ \frac{\partial \mathcal{L}}{\partial \mu_2} & = & x_{12} \geq 0, \ \mu_2 \geq 0, \ \mu_2x_{12} = 0\end{eqnarray*}

One of the solutions to the above conditions solve the optimization problem. Here is that solution : $x_{11}^* = 4, x_{12}^* = 0, x_{22}^* = 4, \lambda^* = 8, \mu_1^* = 0,\mu_2^* = 12$.

Therefore, consumer will buy 4 units of commodity 1 and 0 unit of commodity 2 from the first store; and 0 unit of commodity 1 and 4 units of commodity 2 from the other store.

  • $\begingroup$ Amit, I have learnt perfectly from your solution for this type of questions that I previously asked. I can deal with other parts, but I couldn’t find tha part 2 only. Please can you summarize your solution. I doNot need long answer. I can proceed it. economics.stackexchange.com/questions/22442/… $\endgroup$
    – b11bb
    Jun 13, 2018 at 4:59
  • $\begingroup$ Dear Amit I know I disturbed you too much but I have posted two questions that I could not solve. I have an exam again. Please help me. These two questions are last questions that I ask for you. But I need their solutions too much. Because this exam is my last chance as well. I will be happy if you help me. Please. Many many thanks. I don’t know how to give something like a gift toward your great helps. But this is last help! I really can give you Promise! :) $\endgroup$
    – studentp
    Jul 26, 2018 at 11:29
  • $\begingroup$ First question : economics.stackexchange.com/questions/23890/… $\endgroup$
    – studentp
    Jul 26, 2018 at 11:30
  • $\begingroup$ Second question : economics.stackexchange.com/questions/23891/… $\endgroup$
    – studentp
    Jul 26, 2018 at 11:30
  • $\begingroup$ Amit, if it is difficult and take much time to write them by latex in this website, I can send them by email and you can send them by your hand-writing, if you want. This is really last help.🙏🏻 $\endgroup$
    – studentp
    Jul 26, 2018 at 11:36

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