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}}$


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?

1 Answer 1


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.

  • $\begingroup$ Not my question but what effect does this have on production cost for input 1? $\endgroup$
    – user31331
    Dec 11, 2020 at 3:39
  • $\begingroup$ @bymathformath: Added some part in answer for second part of the question. $\endgroup$
    – Dayne
    Dec 11, 2020 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, 2020 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, 2020 at 14:16
  • 1
    $\begingroup$ 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, 2020 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.