# Question about budget constraint and utility maximization [closed]

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.

Edit

• your description of the problem is insufficient. is the consumer able to go to just one store or to both? May 9, 2018 at 14:14
• @saguru the question say nothing about this. But I guess the consumer can go to both. May 9, 2018 at 19:33
• 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? May 10, 2018 at 8:22
• 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 May 10, 2018 at 8:23
• @saguru this is the only one part of my question. I post whole question. Thanks for helps:) May 10, 2018 at 8:52

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 : $x_{11}^* = 4, x_{12}^* = 0, x_{22}^* = 4, \lambda^* = 8, \mu_1^* = 0,\mu_2^* = 12$.