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I am given the production function

$y=x_1^\alpha x_2^{1-\alpha}$, where $0< \alpha <1$ I found the demand functions for minimum production cost to be

$ x_1^{*}(w_1,w_2,y)=\left ( \frac{w_2}{w_1}\frac{\alpha}{\beta} \right )^{\frac{\beta}{\alpha +\beta}}y^\frac{1}{{\alpha +\beta}} \; \wedge \; x_2^{*}(w_1,w_2,y)=\left ( \frac{w_1}{w_2}\frac{\beta}{\alpha} \right )^{\frac{\alpha}{\alpha +\beta}}y^\frac{1}{{\alpha +\beta}}$

Problem

Now, I have to find the elasticity of $(x_2^*/x_1^*)$ wrt. $w_2/w_1$. I found that $(x_2^*/x_1^*)=\frac{\beta w_1}{\alpha w_2}$ and thus

$\epsilon = \frac{\partial \frac{\beta w_1}{\alpha w_2}}{\partial (w_2/w_1)} \cdot \frac{w_2/w_1}{\frac{\beta w_1}{\alpha w_2}}$

  1. Here I end up stuck and do not know how to reduce/evaluate this expression
  2. What does the elasticity mean for the total production cost on input 1?
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For elasticity calculation why don't you try this:

\begin{align} \frac{x_2^*}{x_1^*}&=\frac{\beta w_1}{\alpha w_2} \\ \ln\bigg(\frac{x_2^*}{x_1^*}\bigg) &= c - \ln\bigg(\frac{w_2}{w_1}\bigg) \tag{$c=\ln(\beta/\alpha)$} \end{align}

It's easy to see from above that elasticity is $-1$

For second part, cost of production:

\begin{align} C&=w_1x_1^* + w_2x_2^* \\ &=w_1x_1^*(1+\beta/\alpha) \\ &=\frac{w_1x_1^*}{\alpha} \tag{$\beta=1-\alpha$} \end{align}

So the production cost per unit of input 1 is just the function of $w_1$ (given $\alpha$).

EDIT: As requested for clarification in comments: Note that elasticity of $y$ w.r.t $x$ is defined as:

\begin{align} \epsilon &= \frac{\partial y/y}{\partial x/x} \\ &= \frac{\partial \ln y}{\partial \ln x} \end{align}

So elasticity is simply the slope on the log-scale for $y$ and $x$.

Now substitute $y=x_2^*/x_1^*$ and $x=w_2/w_1$ in your case.

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  • $\begingroup$ Not my question but what effect does this have on production cost for input 1? $\endgroup$
    – user31331
    Dec 11 '20 at 3:39
  • $\begingroup$ @bymathformath: Added some part in answer for second part of the question. $\endgroup$
    – Dayne
    Dec 11 '20 at 4:26
  • $\begingroup$ Greetings. i am not a 100% confideny on how to use the log of the expression to find the derivative term in the elasticity $\endgroup$ Dec 11 '20 at 13:28
  • $\begingroup$ Elasticity is of $y$ w.r.t. $x$ is $d(\ln y)/d(\ln x)$. So it's just the slope on the log scale. $\endgroup$
    – Dayne
    Dec 11 '20 at 14:16
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    $\begingroup$ Yes I see. I am - normally - pretty good with derivatives but not that faimiliar with log transformation. I will try to simplify my newly found equation to $-1$. Thanks for the help! $\endgroup$ Dec 11 '20 at 15:31

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