<p>Set up your Lagrangian:</p>
<p><span class="math-container">\begin{gather}
L(x_1, x_2, \lambda) = a ln x_1 + blnx_2 - \lambda(p_1x_1 + p_2x_2 - W)
\end{gather}</span></p>
<p>Derive your first order conditions:
<span class="math-container">\begin{gather}
\frac{\partial L}{\partial x_1} = \frac{a}{x_1} - \lambda p_1 = 0 \\
\frac{\partial L}{\partial x_2} = \frac{b}{x_2} - \lambda p_2 = 0 \\
\frac{\partial L}{\partial \lambda} = p_1x_1 + p_2x_2 - W = 0
\end{gather}</span></p>
<p>Now you have your system of equations, which you can solve by yourself. Use the <span class="math-container">$\lambda$</span> from the first two conditions to get a relationship between <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span>, which you can then use on your third condition.</p>

CC BY-SA 4.0
2020-11-06T05:07:40.307Z
2020-11-06T05:07:40.307Z
Alex Ruiz
<p>Is this solvable using the Lagrange method? If so, Does anyone know how to solve this using the Lagrange method? Utility is <span class="math-container">$a\ln x_1 + \ln x_2$</span>. Budget constraint is <span class="math-container">$P_1x_1+P_2x_2= W$</span></p>

CC BY-SA 4.0
2020-11-06T05:36:15.517Z
Is this homework? If so, please add a self-study tag.
Dayne
<microeconomics><mathematical-economics>
How to maximize utility using the Lagrange method
Ryan Hammond
# How to maximize utility using the Lagrange method

(Asked by: username_0 on 2020-11-06)

Is this solvable using the Lagrange method? If so, Does anyone know how to solve this using the Lagrange method? Utility is $a\ln x\_1 + \ln x\_2$. Budget constraint is $P\_1x\_1+P\_2x\_2= W$

**Tags:** microeconomics, mathematical-economics

> **Comments:**
> **username_1**: Is this homework? If so, please add a self-study tag.  

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## 1 answers

### Answer by username_2 on 2020-11-06. Score: 1

Set up your Lagrangian:

\begin{gather}
L(x\_1, x\_2, \lambda) = a ln x\_1 + blnx\_2 - \lambda(p\_1x\_1 + p\_2x\_2 - W)
\end{gather}

Derive your first order conditions:
\begin{gather}
\frac{\partial L}{\partial x\_1} = \frac{a}{x\_1} - \lambda p\_1 = 0 \\
\frac{\partial L}{\partial x\_2} = \frac{b}{x\_2} - \lambda p\_2 = 0 \\
\frac{\partial L}{\partial \lambda} = p\_1x\_1 + p\_2x\_2 - W = 0
\end{gather}

Now you have your system of equations, which you can solve by yourself. Use the $\lambda$ from the first two conditions to get a relationship between $x\_1$ and $x\_2$, which you can then use on your third condition.


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