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In the following post an example is given of the corner solution for a concave utility function. I tried solving it but got stuck. I have no idea how these types of problems are solved so if you could please point me in the right direction.

Here's my work so far:

$U(x_1, x_2)=x_1+\ln(x_2)$

s.t.

$x_1p_1+x_2p_2\leq w$

$x_1\geq0;\; x_2\geq0$

\begin{alignat*}{3} % #1 L(x_1, x_2,x_3,&\lambda,\mu_1,\mu_2)=x_1+\ln(x_2) +\\ +&\lambda[w-(x_1p_1+x_2p_2)]+\mu_1x_1+\mu_2x_2 \end{alignat*}

$\frac{\partial L}{\partial x_1}=1-\lambda p_1+\mu_1 \leq 0$

$\frac{\partial L}{\partial x_2}=\frac{1}{x_2}-\lambda p_2+\mu_2 \leq 0$

$\frac{\partial L}{\partial \lambda}=w-(x_1p_1+x_2p_2) \leq 0$

$\frac{\partial L}{\partial \mu_1}=x_1 \leq 0$

$\frac{\partial L}{\partial \mu_2}=x_2 \leq 0$

Assuming the top constraints are binding. We can say:

$\lambda = \frac{1+\mu_1}{p_1}$

$\lambda = \frac{\frac{1}{x_2}+\mu_2}{p_2}$

$\frac{p_2(1+\mu_1)}{p_1}=\frac{1}{x_2}+\mu_2$

$\frac{p_2+p_2\mu_1-\mu_2p_1}{p_1}=\frac{1}{x_2}$

$x_2=\frac{p_1}{p_2+p_2\mu_1-\mu_2p_1}$

putting this into the budget constraint I get:

$x_1p_1+\frac{p_1p_2}{p_2+p_2\mu_1-\mu_2p_1}=w$

$x_1=\frac{w}{p_1}-\frac{p_2}{p_2+p_2\mu_1-\mu_2p_1}$

$x_2=\frac{p_1}{p_2+p_2\mu_1-\mu_2p_1}$

when $p_2\mu_1-\mu_2p_1$ is equal to 0 I get the solution for $w>p_1$, but I have no idea how they got the second half. So this is where I got stuck. Many thanks in advance to the math experts.

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  • $\begingroup$ An alternative way to solve this problem is posted here: qr.ae/pvYE8Q $\endgroup$
    – Amit
    May 27 at 2:37

1 Answer 1

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This is the problem we want to solve:

\begin{eqnarray*} \max_{x_1, x_2} & x_1 +\ln x_2 \\ \text{s.t.} & \ p_1 x_1 + p_2x_2 \leq w \\ \text{and} & \ x_1\geq 0, x_2>0 \end{eqnarray*}

Here $w>0$, $p_1>0$ and $p_2>0$.

First set up the Lagrangian function:

$\mathcal{L}(x_1, x_2)= x_1 +\ln x_2 - \lambda(p_1 x_1 + p_2x_2 - w) + \mu_1x_1$

First-order necessary conditions are:

$\dfrac{\partial\mathcal{L}}{\partial x_1} = 1 - \lambda p_1 + \mu_1 = 0$

$\dfrac{\partial\mathcal{L}}{\partial x_2} = \dfrac{1}{x_2} - \lambda p_2 = 0$

$p_1 x_1 + p_2x_2 \leq w, \ \lambda \geq 0, \ \lambda(p_1 x_1 + p_2x_2 - w) = 0$

$x_1 \geq 0, \ \mu_1 \geq 0, \ \mu_1x_1 = 0$

$x_2>0$

Since $\mathcal{L}$ is concave, if $(x_1^d, x_2^d)$ satisfies the first-order conditions, it is also a solution of the utility maximization problem.

Solving the above system we get the optimal values of $x_1^d$, $x_2^d$as:

\begin{eqnarray*} (x_1^d, x_2^d)(p_1, p_2, w)=\begin{cases} \left(\frac{w-p_1}{p_1},\frac{p_1}{p_2}\right) & \text{if } p_1 \leq w \\ \left(0,\frac{w}{p_2}\right) & \text{if } p_1 > w \end{cases} \end{eqnarray*}


Added Later

This is how we can solve the following system:

$\dfrac{\partial\mathcal{L}}{\partial x_1} = 1 - \lambda p_1 + \mu_1 = 0$

$\dfrac{\partial\mathcal{L}}{\partial x_2} = \dfrac{1}{x_2} - \lambda p_2 = 0$

$p_1 x_1 + p_2x_2 \leq w, \ \lambda \geq 0, \ \lambda(p_1 x_1 + p_2x_2 - w) = 0$

$x_1 \geq 0, \ \mu_1 \geq 0, \ \mu_1x_1 = 0$

$x_2>0$

Based on the above system, we can divide it into the following possibilities for the solution:

  • $p_1 x_1 + p_2x_2 < w$ This would imply that $\lambda = 0$, but that would mean that $\mu_1=-1$ which is not consistent with the system. So, there is no solution to the above system where $p_1 x_1 + p_2x_2 < w$.
  • $p_1 x_1 + p_2x_2 = w$ In this case there are two possibilities: $x_1>0$ and the other is $x_1=0$. Both of them are possible depending on the values of $p_1$, $p_2$ and $w$:

For the case of $x_1>0$, we get $\mu_1=0$, and that would imply that $\lambda=\frac{1}{p_1}$. So, $x_2=\frac{p_1}{p_2}$ and the corresponding value of $x_1= \frac{w-p_1}{p_1}$. Clearly, this is the only solution to the above system when $p_1< w$.

For the case of $x_1=0$, $x_2= \frac{w}{p_2}$ and the corresponding value of $\lambda=\frac{1}{w}$ and therefore, $\mu_1 = \frac{p_1-w}{p_1}$. Clearly, this is the only solution to the above system when $p_1 \geq w$.


To see how to solve a similar problem with $u(x, y) = 2\sqrt{x} + y$, you may refer to: https://youtu.be/l8vHgCv70h0

Related posts (Alternative way of finding demand for $u(x, y) = 2\sqrt{x} + y$): https://qr.ae/pGJuvH

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  • $\begingroup$ Thank you very much for the detailed answer. The examples were also very helpful. Essentially, solving with KKT conditions involves trying out different cases and finding out the ones in which the system holds true. I should probably open a new post about this, but would you happen to know if this problem can be solved with non-linear/linear programming? Additionally, what if I end up with multiple solutions within the same domain, like in the case of u(x1,x2)=x1^2+x2^2 with a corner solution? Should I evaluate all three and take the max for a utility, or is there a more rigorous method? $\endgroup$ May 27 at 7:03
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    $\begingroup$ Most welcome. The objective function is non-linear, so it is not a linear programming problem. KT Method discussed in this answer is a way to solve non-linear programming problems (including the other ways you saw). For solving a utility maximisation problem with $u = x_1^2+x_2^2$, the quick way is using the graph, but if you want to see the K-T method then you can refer to this: youtu.be/eFe25HVJSWg Since K-T conditions are no longer sufficient for the optimum when the objective function is $u = x_1^2+x_2^2$, you may get more solutions to the K-T conditions including the optimal one. $\endgroup$
    – Amit
    May 27 at 9:11

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