Concave utility functions solution example

Question

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.

Answer 1

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*}

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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 $p_1 \geq w$.

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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