# Calculating cost-minimizing inputs with 3 inputs in production function [closed]

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.

• 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
Feb 17, 2022 at 15:23

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 $$j$$.

• Thanks a lot. It was Really helpful. Feb 19, 2022 at 6:58
• You are welcome. Feb 19, 2022 at 14:14