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I am studying and going through some old exams for a microeconomic analysis class. I am just looking for some clarification regarding one of the answers given. The question is as follows

Suppose that a firm's production function is given by $y=lnx_{1}+2lnx_{2}$ where $x_{1}$ and $x_{2}$ are respectively the amounts of input 1 and input 2, and $y$ is the output. Let $w_{1}$ and $w_{2}$ be the prices for inputs 1 and 2. Calculate the cost function for this firm.

I know that the cost function is given by

$c(\mathbf{w}, y)=\min\; \; w_{1}x_{1}+w_{2}x_{2}$

such that $y=f(\mathbf{x})$

In this case it will be

$c(w_{1},w_{2}, y)=\min\; \; w_{1}x_{1}+w_{2}x_{2}$

such that $y=f(x_1,x_2)=\ln x_{1}+2\ln x_{2}$

I set up the Lagrangian:

$L=w_{1}x_{1}+w_{2}x_{2}+\lambda (y-\ln x_{1}+2\ln x_{2}) $

Solving for the first order conditions:

$\frac{\partial L}{\partial x_1}=w_1 - \frac {\lambda }{x_1}=0$ $\frac{\partial L}{\partial x_2}=w_2 - \frac {2\lambda }{x_2}=0$ $w_1 = \frac {\lambda }{x_1}$ $w_2 = \frac {2\lambda }{x_2}$

From here I divide $w_1$ by $w_2$ to find

$x_2=\frac {2x_1w_1}{w_2}$ $x_1=\frac {w_2x_2}{2w_1}$

I am okay up until this point. In the given answer the professor substitutes $x_2$ into the production function, so that

$y=\ln x_1+2\ln{(\frac{2x_1w_1}{w_2})}$

From here he finds (after substituting in $x_1$ as well)

$x_1=\left ( \frac{e^yw_{2}^{2}}{4w_{1}^{2}} \right )^{1/3}$ and $x_2=\left ( \frac{e^y2w_{1}}{w_{2}} \right )^{1/3}$

I am having trouble arriving at the same thing. If anyone could show me his steps I would very much appreciate it. I wont show the different things I tried, but I of course started with $e^y=...$

1 Answers 1

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Are any of these steps difficult?

$y=\ln x_1+2\ln{\left(\frac{2x_1w_1}{w_2}\right)}$

$y=\ln x_1+2\ln x_1 +2 \ln{\left(\frac{2w_1}{w_2}\right)}$

$3\ln x_1 = y- 2 \ln{\left(\frac{2w_1}{w_2}\right)}$

$\ln x_1 = \frac{1}{3} \left(y- {2} \ln{\left(\frac{2w_1}{w_2}\right)}\right)$

$ x_1 = \left(\exp(y){\left(\frac{2w_1}{w_2}\right)^{-2}}\right)^\frac{1}{3}$

$ x_1 = {\left(\frac{e^y w_2^2}{4w_1^2}\right)}^\frac{1}{3}$