# Elasticity Cobb-Douglas production function

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?

## 1 Answer

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.

• Not my question but what effect does this have on production cost for input 1? – user31331 Dec 11 '20 at 3:39
• @bymathformath: Added some part in answer for second part of the question. – Dayne Dec 11 '20 at 4:26
• Greetings. i am not a 100% confideny on how to use the log of the expression to find the derivative term in the elasticity – mathstudent23 Dec 11 '20 at 13:28
• 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. – Dayne Dec 11 '20 at 14:16
• 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! – mathstudent23 Dec 11 '20 at 15:31