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Let $y=x_1^\alpha x_2^\beta$ where $\beta=1-\alpha$ be a Cobb-Douglas production function.

Find the elasticity of the optimal demand functions (for minimizing production cost) for both goods wrt. $w_2/w_1$ ($w_1,w_2$ are the respective prices of the inputs). What does this say for our spending on $x_1$ compared to the total cost?

My try

I found the elasticity of $x_2^*/x_1^*$ wrt $w_2/w_1$ to be 1. Now, I have to find a value for the ammount spend on good $x_1$. How can I do that with my elasticity? I think it has to be constant, but still no clue on how to find such a value even though I have found the elasticity to be 1.

edit: I know it has to be constant.

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    $\begingroup$ What have you done already? You have to derive conditional demand as it is called - not optimal demand - could you please show your derivations? $\endgroup$
    – bomadsen
    Dec 22 '20 at 12:30
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The first order conditions equate marginal revenue per factor to the price of that factor:

\begin{align} p\cdot\alpha\frac{y}{x_1} &= w_1\\ p\cdot\beta\frac{y}{x_2} &= w_2, \end{align}

Where I used the property of power function $(x^n)'_n = n \frac{x^n}{x}$.

Divide the second FOC by the first to get the relation between the relative prices and the relative factor demands:

$$\frac{\beta}{\alpha}\frac{x_1}{x_2} = \frac{w_2}{w_1}. \tag{A}$$

From this relation we can draw two conclusions:

  1. Rewrite (A) in log form:

$$- \ln \frac{x_2}{x_1} + \ln \frac{\beta}{\alpha} =\ln \frac{w_2}{w_1},$$

And using the log definition of elasticity $\epsilon_y^x = \frac{\mathrm{d}\ln y}{\mathrm{d}\ln x}$ we come to the conclusion, that relative factor demand is decreasing in relative factor prices with unit elasticity:

$$\frac{\mathrm{d}\ln x_2/ x_1}{\mathrm{d}\ln w_2/w_1} = -1.$$

  1. Multiply both sides of (A) by $\frac{x_2}{x_1}$ :

$$\frac{\beta}{\alpha} = \frac{w_2x_2}{w_1x_1}. \tag{B}$$

Rearrangement (B) says that the expenditures on different factors are proportional to their respective input elasticities, i.e. if our total spending on factor 1 is $\\\$\alpha$ then we must spend $\\\$\beta$ on factor 2.

The total cost $C$ is allocated in the same proportion, i.e. for a general Cobb-Douglas production function, spending on factor 1 is

$$w_1 x_1 = \frac{\alpha}{\alpha+\beta}C,$$

or simply $w_1 x_1 =\alpha C$ if $\alpha+\beta=1$, i.e. if the production function is homogeneous of degree 1.

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  • $\begingroup$ I found the $MP$ and multiplied it by $x_1$ to aquire the total cost which gave me $\alpha \cdot y$ and thus the total factor used is just $\alpha$. Is this not correct and much faster? $\endgroup$
    – user31331
    Dec 23 '20 at 17:10
  • $\begingroup$ $MP$ of $x_1$ that is $\endgroup$
    – user31331
    Dec 23 '20 at 17:12
  • $\begingroup$ MP, if you mean marginal product, is measured in real terms, units of output per units of factor. If you multiply this amount by $x_1$ you get something measured in units of output. Cost and spending on the other hand is always measured in monetary units, so you lack a price (that of the final product). It might be that in a certain problem the price is equal to 1, so you make no mistake in math, but it is better to pay due to the accounting logic. $\endgroup$
    – Konstantin
    Dec 23 '20 at 17:36
  • $\begingroup$ Found this for better explanation of what I mean; Kevin Clinton, September 2004, "A useful production function: Cobb-Douglas". His second paragraph explains what I mean and thus that $MP_1(x_1)$ would give the usage of that input relative to all the usage. This is what I think is intended here. (the pdf is just a copy paste on google and should pop right up). $\endgroup$
    – user31331
    Dec 23 '20 at 17:39
  • $\begingroup$ well, if no one is asking for a proof, you totally should use the properties of Cobb-Douglas function. If you have a Cobb-Douglas function $x_1^\alpha x_2^\beta x_3^\gamma ...$ with $\alpha+\beta+\gamma + .. = 1$ then the total production cost is allocated between factors in proportion $\alpha : \beta : \gamma : ...$. Elasticity from the first part does not play any role here. $\endgroup$
    – Konstantin
    Dec 23 '20 at 17:55

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