My function is $y=x_1^\alpha x_2^\beta$ with $\beta={1-\alpha}$.

I found: the minimization problem for demand to be

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

and the cost function to be

$C=w_1^{\frac{\alpha}{\alpha+\beta}}w_2^{\frac{\beta}{\alpha+\beta}}y^{\frac{1}{\alpha+\beta}}\left ( \left ( \frac{\alpha}{\beta} \right )^\frac{\beta}{\alpha+\beta}+ \left ( \frac{\alpha}{\beta} \right )^\frac{-\alpha}{\alpha+\beta} \right )$.

Now, how would I find Marginal cost (MC)? Why does it not depend on $y$ and only $w_1,w_2$?

  • $\begingroup$ Just substitute $\alpha + \beta=1$ and you'll see that $C=ay$, where $a$ is a term that depends only on $w_1, w_2, \alpha$. So the marginal cost $C^\prime(q)$ is just $a$. $\endgroup$
    – Dayne
    Commented Dec 13, 2020 at 2:37
  • $\begingroup$ ohh okay. So simplify, take the derivative with respect to y $\endgroup$
    – user31331
    Commented Dec 13, 2020 at 16:49
  • $\begingroup$ @Dayne why does MC not depend on $y$? Does it have to do with $AC$? $\endgroup$
    – user31331
    Commented Dec 13, 2020 at 16:59
  • $\begingroup$ No it has to do with the production function. For example if $\alpha +\beta <1$, MC will be increasing in $y$. In fact, since there is some input which cannot be changed, in the short term $\alpha +\beta <1$ is a fair assumption and thus generally we assume MC as increasing in short run $\endgroup$
    – Dayne
    Commented Dec 13, 2020 at 17:28
  • $\begingroup$ @Dayne but why does the marginal cost which I have derived only depend on the variables $w_1,w_2$ and not the production function? $\endgroup$
    – user31331
    Commented Dec 13, 2020 at 18:01

1 Answer 1


The Cobb Douglas production function with constants returns to scale

$$y = \prod_i x_i^{\alpha_i} = A \prod_i \left(\frac{x_i}{\alpha_i}\right)^{\alpha_i} ,$$

where $A:= \prod_i \alpha_i^{\alpha_i}$ annoying constant.

Cost minimization with perfect competition

$$\min_x \ \ p^\top x\ \lvert \ y = \prod \left(\frac{x_i}{\alpha_i}\right)^{\alpha_i},$$

implies FOC

$$p_j - \lambda \left[A\prod_i \left(\frac{x_i}{\alpha_i}\right)^{\alpha_i}\right] \left(\frac{x_j}{\alpha_j}\right)^{-1}=0 \Leftrightarrow \frac{\lambda y}{p_j} =\frac{x_j}{\alpha_j},$$

insert in production function to get

$$y = A\prod_i \left( \frac{\lambda y}{p_i}\right)^{\alpha_i} \Leftrightarrow \lambda = \frac{\bar p}{A},$$

where $\bar p := \prod_i p_i^{\alpha_i}$ which is a price index for production factors. Reinsert in FOC to get

$$p_j - \frac{\bar p}{A}\left[A\prod_i \left(\frac{x_i}{\alpha_i}\right)^{\alpha_i}\right] \left(\frac{x_j}{\alpha_j}\right)^{-1}=p_j - \frac{\bar p}{A}y \left(\frac{x_j}{\alpha_j}\right)^{-1}=0,$$

from which you find

$$(1) \ \ \ p_j x_j = \frac{\alpha_j \bar p}{A} \cdot y,$$

using this formula several implications follow. First take sum over $j$ and use $\sum_j \alpha_j = 1$ to get cost function

$$(2) \ \ C(p,y)= \frac{\bar p}{A} y.$$

Set $y=1$ to get unit cost, diffrentiate with $y$ to get marginal cost, divide with y to get average cost and it follows that

$$(3) \ \ C(p,1) = \frac{\partial C(p,y)}{\partial y} = \frac{C(p,y)}{y} = \frac{\bar p}{A}.$$ Finally from (1) divide with $x_j$ to get

$$(4) \ \ x_j^\star(p,y) = \frac{\alpha_j C(p,y)}{p_j} = \frac{\alpha_j}{p_j} \frac{\bar p}{A} y,$$ stating that the share $\alpha_j$ of the costs are used on production factor $x_j$ and under perfect competition there is zero profit so revenue equals costs, hence $\alpha_j$ of the revenue must be used (price of $y$ must be unit cost, marginal cost, average cost = $C(p,1)$).

With only two production factors the cost function can be written as

$$C(p,y) = C(p_1,p_2,y) = \frac{\bar p}{A} y = \frac{p_1^{\alpha} p_2^{1-\alpha}}{\alpha^\alpha (1-\alpha)^{(1-\alpha)}} y,$$ hence marginal cost are easily seen to be

$$\frac{\partial C(p,y)}{\partial y} = \frac{p_1^{\alpha} p_2^{1-\alpha}}{\alpha^\alpha (1-\alpha)^{(1-\alpha)}}.$$

  • $\begingroup$ Hello! This is a bit advanced for me but I got a few questions. Are my results so far correct? I am having a hard time following this. $\endgroup$
    – user31331
    Commented Dec 13, 2020 at 15:38
  • $\begingroup$ From my approach; how would I find the marginal cost? $\endgroup$
    – user31331
    Commented Dec 13, 2020 at 15:55

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