<p>The production function</p>
<p><span class="math-container">$$F(x) = x_1^{\alpha_1}x_2^{\alpha_2}x_3^{\alpha_3},$$</span></p>
<p>has the derivative with respect to <span class="math-container">$x_j$</span> where for sake of example I choose <span class="math-container">$j=1$</span></p>
<p><span class="math-container">$$\frac{\partial F(x)}{\partial x_1} = \alpha_1x_1^{\alpha_1-1}x_2^{\alpha_2}x_3^{\alpha_3},$$</span></p>
<p>however, it is better (in my opinion) to write it as</p>
<p><span class="math-container">$$\frac{\partial F(x)}{\partial x_1} = x_1^{\alpha_1}x_2^{\alpha_2}x_3^{\alpha_3} \frac{\alpha_1}{x_1},$$</span>
because then the production function reappears in the derivative as the factor <span class="math-container">$x_1^{\alpha_1}x_2^{\alpha_2}x_3^{\alpha_3}$</span>.</p>
<p>By setting up the Lagrangian you get first-order conditions</p>
<p><span class="math-container">$$p_j = \lambda \frac{\partial F(x)}{\partial x_j} = \lambda x_1^{\alpha_1}x_2^{\alpha_2}x_3^{\alpha_3} \frac{\alpha_j}{x_j}$$</span></p>
<p>for <span class="math-container">$j=1,2,3$</span> and where <span class="math-container">$p_j$</span> is either <span class="math-container">$w$</span>,<span class="math-container">$r$</span> or <span class="math-container">$q$</span>. By using indexation you can write three equations as one (you save paper and thereby the rainforest - which is one way mathematicians are helping us be more ecofriendly).</p>
<p>Now it is easy to see that <span class="math-container">$\lambda$</span> must be positive for any non-zero production level (prices are positive, <span class="math-container">$\alpha_j$</span>'s are positive and the production level is positive <span class="math-container">$x_1^{\alpha_1}x_2^{\alpha_2}x_3^{\alpha_3}$</span> is positive. Hence, the constraint must be binding <span class="math-container">$F(x) = y$</span>.</p>
<p>Use this and rewrite FOC to get</p>
<p><span class="math-container">$$(1) \ \ p_jx_j  = \lambda y \alpha_j,$$</span></p>
<p>take the sum over <span class="math-container">$j=1,2,3$</span> and get</p>
<p><span class="math-container">$$C= \lambda y \bar \alpha,$$</span></p>
<p>where <span class="math-container">$C=\sum_j p_j x_j$</span> the total costs and <span class="math-container">$\bar \alpha = \sum_j \alpha_j$</span> which is 1 in the case of constant returns to scale. You now know that</p>
<p><span class="math-container">$$ \lambda = \frac{C}{y \bar \alpha},$$</span></p>
<p>which you can back substitute into (1) to get</p>
<p><span class="math-container">$$(2) \ \ x_j  = \frac{\alpha_j C}{\bar \alpha p_j},$$</span></p>
<p>you then substitute this into the constraint <span class="math-container">$F(x) = y$</span> to get</p>
<p><span class="math-container">$$ y = \left(\frac{\alpha_1 C}{\bar \alpha p_1}\right)^{\alpha_1}\left(\frac{\alpha_2 C}{\bar \alpha p_2}\right)^{\alpha_2}\left(\frac{\alpha_3 C}{\bar \alpha p_3}\right)^{\alpha_3} =\frac{\alpha_1^{\alpha_1}\alpha_2^{\alpha_2}\alpha_3^{\alpha_3} C^{\bar \alpha}}{p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3}\bar \alpha^{\bar \alpha}},$$</span></p>
<p>implying that</p>
<p><span class="math-container">$$C = \left( \frac{p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3} }{\alpha_1^{\alpha_1}\alpha_2^{\alpha_2}\alpha_3^{\alpha_3} } \right)^\frac{1}{\bar \alpha}\bar \alpha y^\frac{1}{\bar \alpha},$$</span></p>
<p>which is the cost function <span class="math-container">$C(p,y)$</span>. You can then back insert this into (2) to get
the conditional factor demand
<span class="math-container">$$x_j  = \frac{\alpha_j /\bar \alpha}{p_j} \left[\left( \frac{p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3} }{\alpha_1^{\alpha_1}\alpha_2^{\alpha_2}\alpha_3^{\alpha_3} } \right)^\frac{1}{\bar \alpha} \bar \alpha y^\frac{1}{\bar \alpha}\right],$$</span></p>
<p>where you could remove <span class="math-container">$\bar \alpha$</span> but the point of leaving it there is that the term in the square parenthesis is simply the cost function so it can be seen that the firm use a fixed share <span class="math-container">$\alpha_j/\bar \alpha$</span> of its total cost to finance us of factor <span class="math-container">$j$</span>.</p>

CC BY-SA 4.0
2022-02-17T17:00:27.890Z
2022-02-17T17:00:27.890Z
Jesper Hybel
<p>How can I determine the cost-minimizing input bundle with a standard Cobb-Douglas production function with three inputs. despite its simple process, the algebra becomes very hard as you go through the calculations. <strong>specifically, I want to find the closed form solutions for k, m and l in the below problem:</strong></p>
<p><span class="math-container">$$ min: \ wl +rk+qm   $$</span>
<span class="math-container">$$ st: y&lt; Ak^{\alpha}l^{\beta}m^{\gamma}     $$</span></p>
<p>Thanks.</p>

CC BY-SA 4.0
2022-02-17T15:23:58.920Z
please show us what you tried so far, on this site we require some attempt by the user before providing a solution. Even if you dont know how to arrive at a correct solution just show us your attempt.
1muflon1
<optimization><cost><cobb-douglas>
Calculating cost-minimizing inputs with 3 inputs in production function
Seyed Mahdi
# Calculating cost-minimizing inputs with 3 inputs in production function

(Asked by: username_0 on 2022-02-17)

How can I determine the cost-minimizing input bundle with a standard Cobb-Douglas production function with three inputs. despite its simple process, the algebra becomes very hard as you go through the calculations. **specifically, I want to find the closed form solutions for k, m and l in the below problem:**

$$ min: \ wl +rk+qm $$
$$ st: y< Ak^{\alpha}l^{\beta}m^{\gamma} $$

Thanks.

**Tags:** optimization, cost, cobb-douglas
> **Comments:**
> **username_1**: please show us what you tried so far, on this site we require some attempt by the user before providing a solution. Even if you dont know how to arrive at a correct solution just show us your attempt.  

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## 1 answers

### Answer by username_2 on 2022-02-17. Score: 5 | Accepted answer

The production function

$$F(x) = x\_1^{\alpha\_1}x\_2^{\alpha\_2}x\_3^{\alpha\_3},$$

has the derivative with respect to $x\_j$ where for sake of example I choose $j=1$

$$\frac{\partial F(x)}{\partial x\_1} = \alpha\_1x\_1^{\alpha\_1-1}x\_2^{\alpha\_2}x\_3^{\alpha\_3},$$

however, it is better (in my opinion) to write it as

$$\frac{\partial F(x)}{\partial x\_1} = x\_1^{\alpha\_1}x\_2^{\alpha\_2}x\_3^{\alpha\_3} \frac{\alpha\_1}{x\_1},$$
because then the production function reappears in the derivative as the factor $x\_1^{\alpha\_1}x\_2^{\alpha\_2}x\_3^{\alpha\_3}$.

By setting up the Lagrangian you get first-order conditions

$$p\_j = \lambda \frac{\partial F(x)}{\partial x\_j} = \lambda x\_1^{\alpha\_1}x\_2^{\alpha\_2}x\_3^{\alpha\_3} \frac{\alpha\_j}{x\_j}$$

for $j=1,2,3$ and where $p\_j$ is either $w$,$r$ or $q$. By using indexation you can write three equations as one (you save paper and thereby the rainforest - which is one way mathematicians are helping us be more ecofriendly).

Now it is easy to see that $\lambda$ must be positive for any non-zero production level (prices are positive, $\alpha\_j$'s are positive and the production level is positive $x\_1^{\alpha\_1}x\_2^{\alpha\_2}x\_3^{\alpha\_3}$ is positive. Hence, the constraint must be binding $F(x) = y$.

Use this and rewrite FOC to get

$$(1) \ \ p\_jx\_j = \lambda y \alpha\_j,$$

take the sum over $j=1,2,3$ and get

$$C= \lambda y \bar \alpha,$$

where $C=\sum\_j p\_j x\_j$ the total costs and $\bar \alpha = \sum\_j \alpha\_j$ which is 1 in the case of constant returns to scale. You now know that

$$ \lambda = \frac{C}{y \bar \alpha},$$

which you can back substitute into (1) to get

$$(2) \ \ x\_j = \frac{\alpha\_j C}{\bar \alpha p\_j},$$

you then substitute this into the constraint $F(x) = y$ to get

$$ y = \left(\frac{\alpha\_1 C}{\bar \alpha p\_1}\right)^{\alpha\_1}\left(\frac{\alpha\_2 C}{\bar \alpha p\_2}\right)^{\alpha\_2}\left(\frac{\alpha\_3 C}{\bar \alpha p\_3}\right)^{\alpha\_3} =\frac{\alpha\_1^{\alpha\_1}\alpha\_2^{\alpha\_2}\alpha\_3^{\alpha\_3} C^{\bar \alpha}}{p\_1^{\alpha\_1}p\_2^{\alpha\_2}p\_3^{\alpha\_3}\bar \alpha^{\bar \alpha}},$$

implying that

$$C = \left( \frac{p\_1^{\alpha\_1}p\_2^{\alpha\_2}p\_3^{\alpha\_3} }{\alpha\_1^{\alpha\_1}\alpha\_2^{\alpha\_2}\alpha\_3^{\alpha\_3} } \right)^\frac{1}{\bar \alpha}\bar \alpha y^\frac{1}{\bar \alpha},$$

which is the cost function $C(p,y)$. You can then back insert this into (2) to get
the conditional factor demand
$$x\_j = \frac{\alpha\_j /\bar \alpha}{p\_j} \left[\left( \frac{p\_1^{\alpha\_1}p\_2^{\alpha\_2}p\_3^{\alpha\_3} }{\alpha\_1^{\alpha\_1}\alpha\_2^{\alpha\_2}\alpha\_3^{\alpha\_3} } \right)^\frac{1}{\bar \alpha} \bar \alpha y^\frac{1}{\bar \alpha}\right],$$

where you could remove $\bar \alpha$ but the point of leaving it there is that the term in the square parenthesis is simply the cost function so it can be seen that the firm use a fixed share $\alpha\_j/\bar \alpha$ of its total cost to finance us of factor $j$.


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