<p>For elasticity calculation why don't you try this:</p>
<p><span class="math-container">\begin{align}
\frac{x_2^*}{x_1^*}&amp;=\frac{\beta w_1}{\alpha w_2}  \\
\ln\bigg(\frac{x_2^*}{x_1^*}\bigg) &amp;= c - \ln\bigg(\frac{w_2}{w_1}\bigg) \tag{$c=\ln(\beta/\alpha)$}
\end{align}</span></p>
<p>It's easy to see from above that elasticity is <span class="math-container">$-1$</span></p>
<p>For second part, cost of production:</p>
<p><span class="math-container">\begin{align}
C&amp;=w_1x_1^* + w_2x_2^* \\
&amp;=w_1x_1^*(1+\beta/\alpha) \\
&amp;=\frac{w_1x_1^*}{\alpha} \tag{$\beta=1-\alpha$}
\end{align}</span></p>
<p>So the production cost per unit of input 1 is just the function of <span class="math-container">$w_1$</span> (given <span class="math-container">$\alpha$</span>).</p>
<p>EDIT: As requested for clarification in comments:
Note that elasticity of <span class="math-container">$y$</span> w.r.t <span class="math-container">$x$</span> is defined as:</p>
<p><span class="math-container">\begin{align}
\epsilon &amp;= \frac{\partial y/y}{\partial x/x} \\
&amp;= \frac{\partial \ln y}{\partial \ln x}
\end{align}</span></p>
<p>So elasticity is simply the slope on the log-scale for <span class="math-container">$y$</span> and <span class="math-container">$x$</span>.</p>
<p>Now substitute <span class="math-container">$y=x_2^*/x_1^*$</span> and <span class="math-container">$x=w_2/w_1$</span> in your case.</p>

CC BY-SA 4.0
2020-12-11T04:26:03.987Z
@bymathformath: Added some part in answer for second part of the question.
Dayne
CC BY-SA 4.0
2020-12-11T13:28:06.973Z
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
CC BY-SA 4.0
2020-12-11T14:16:55.253Z
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
CC BY-SA 4.0
2020-12-11T14:23:15.907Z
$y$ wrt to $x$? are you saying that $\frac{\partial \frac{\beta w_1}{\alpha w_2}}{\partial (w_2/w_1)} =d(ln(y))/d (ln (x))$? Could you please show how you compute this because I am fairly confused now.
mathstudent23
CC BY-SA 4.0
2020-12-11T14:29:22.130Z
I added some details in the answer for clarification
Dayne
CC BY-SA 4.0
2020-12-11T14:44:37.113Z
But in the clarification I end up with the same source for my confusion (which I should have clarified). I dont know how to compute such a derivative i.e $\frac{\partial ln(x_2^*/x_1^*)}{\partial ln(w_2/w_1)}$. (the fact that the  derivative is of respect to a variable that is not included confuses me (since $w_2/w_1$ and not $w_1/w_2$ as in the  $(x_2^*/x_1^*)$)
mathstudent23
CC BY-SA 4.0
2020-12-11T14:53:50.710Z
$$\frac{x_2^*}{x_1^*}=\frac{\beta w_1}{\alpha w_2} = \frac{\beta/\alpha}{w_2/w_1}$$. Now is it clear?
Dayne
CC BY-SA 4.0
2020-12-11T15:21:17.860Z
yes I think! I would then end up with the elasticity $-\frac{\beta}{\alpha(w_2/w_1))^2}\cdot\frac{w_2/w_1}{(\beta w_1)/(\alpha w_2)}$. Is this correct (I have not found it to $-1$ yet)
mathstudent23
CC BY-SA 4.0
2020-12-11T15:26:23.897Z
The whole point of taking $\ln$ was to avoid this calculation. Just take $\ln$ on both sides of the equation that I have mentioned in my previous comment. In any case if you are not familiar with log transformation and derivatives that much then you can use your approach. On simplifying your expression further you will get $-1$.
Dayne
CC BY-SA 4.0
2020-12-11T15:31:35.863Z
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
CC BY-SA 4.0
2020-12-11T18:12:42.670Z
By the chain rule $\frac{\partial \log f(x)}{\partial x} = \frac{\partial f(x)}{\partial log(x)} \frac{\partial \log x}{ \partial x}$ this implies that $\frac{1}{f(x)} \frac{\partial f(x)}{\partial x} = \frac{\partial \log f(x)}{\partial \log x} \frac{1}{x}$ therefore $\frac{\partial \log f(x)}{\partial \log x} = \frac{x}{f(x)} \frac{\partial f(x)}{\partial x}$ which is definition of elasticity of $f(x)$ with respect to $x$. Offcourse $x>0$ to take $\log$.
Jesper Hybel
CC BY-SA 4.0
2020-12-12T02:07:52.560Z
@Dayne the second question that is formulated as "What does the elasticity signify for the portion of total production cost that is used on input 1"  I see your "cost of prodution" but where is the elasticity used and/or explained since I am having a hard time relating the elasticity to the portion of total production cost that is used on input 1.
mathstudent23
CC BY-SA 4.0
2020-12-11T02:22:33.620Z
2020-12-11T14:29:11.730Z
2020-12-11T14:29:11.730Z
Dayne
<p>I am given the production function</p>
<p><span class="math-container">$y=x_1^\alpha x_2^{1-\alpha}$</span>, where <span class="math-container">$0&lt; \alpha &lt;1$</span> I found the demand functions for minimum production cost to be</p>
<p><span class="math-container">$ 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}}$</span></p>
<p><strong>Problem</strong></p>
<p>Now, I have to find the elasticity of <span class="math-container">$(x_2^*/x_1^*)$</span> wrt. <span class="math-container">$w_2/w_1$</span>. I found that <span class="math-container">$(x_2^*/x_1^*)=\frac{\beta w_1}{\alpha w_2}$</span> and thus</p>
<p><span class="math-container">$\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}}$</span></p>
<ol>
<li>Here I end up stuck and do not know how to reduce/evaluate this expression</li>
<li>What does the elasticity mean for the total production cost on input 1?</li>
</ol>

<elasticity><production-function><cobb-douglas>
Elasticity Cobb-Douglas production function
mathstudent23
# Elasticity Cobb-Douglas production function

(Asked by: username_0 on 2020-12-11)

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?

**Tags:** elasticity, production-function, cobb-douglas

---

## 1 answers

### Answer by username_1 on 2020-12-11. Score: 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.

> **Comments:**
> **username_1**: @bymathformath: Added some part in answer for second part of the question.  
> **username_0**: Greetings. i am not a 100% confideny on how to use the log of the expression to find the derivative term in the elasticity  
> **username_1**: 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.  
> **username_0**: $y$ wrt to $x$? are you saying that $\frac{\partial \frac{\beta w_1}{\alpha w_2}}{\partial (w_2/w_1)} =d(ln(y))/d (ln (x))$? Could you please show how you compute this because I am fairly confused now.  
> **username_1**: I added some details in the answer for clarification  
> **username_0**: But in the clarification I end up with the same source for my confusion (which I should have clarified). I dont know how to compute such a derivative i.e $\frac{\partial ln(x_2^*/x_1^*)}{\partial ln(w_2/w_1)}$. (the fact that the  derivative is of respect to a variable that is not included confuses me (since $w_2/w_1$ and not $w_1/w_2$ as in the  $(x_2^*/x_1^*)$)  
> **username_1**: $$\frac{x_2^*}{x_1^*}=\frac{\beta w_1}{\alpha w_2} = \frac{\beta/\alpha}{w_2/w_1}$$. Now is it clear?  
> **username_0**: yes I think! I would then end up with the elasticity $-\frac{\beta}{\alpha(w_2/w_1))^2}\cdot\frac{w_2/w_1}{(\beta w_1)/(\alpha w_2)}$. Is this correct (I have not found it to $-1$ yet)  
> **username_1**: The whole point of taking $\ln$ was to avoid this calculation. Just take $\ln$ on both sides of the equation that I have mentioned in my previous comment. In any case if you are not familiar with log transformation and derivatives that much then you can use your approach. On simplifying your expression further you will get $-1$.  
> **username_0**: 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!  
> **username_2**: By the chain rule $\frac{\partial \log f(x)}{\partial x} = \frac{\partial f(x)}{\partial log(x)} \frac{\partial \log x}{ \partial x}$ this implies that $\frac{1}{f(x)} \frac{\partial f(x)}{\partial x} = \frac{\partial \log f(x)}{\partial \log x} \frac{1}{x}$ therefore $\frac{\partial \log f(x)}{\partial \log x} = \frac{x}{f(x)} \frac{\partial f(x)}{\partial x}$ which is definition of elasticity of $f(x)$ with respect to $x$. Offcourse $x>0$ to take $\log$.  
> **username_0**: @Dayne the second question that is formulated as "What does the elasticity signify for the portion of total production cost that is used on input 1"  I see your "cost of prodution" but where is the elasticity used and/or explained since I am having a hard time relating the elasticity to the portion of total production cost that is used on input 1.  

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