Cost Minimization of $f(x) = min(x_1,x_2) + x_3$ The following production function is given,
$f(X) = min\{x_1,x_2\} + x_3$
There is a solution here https://math.stackexchange.com/questions/605925/constrained-maximization-of-leontif-utility-function-minx-1-x-2, which is similar to this function. However, it does not include an additive $x_3$.
My initial solution attempt
Redefine the production function, such that,
$f(X) = x_1 + x_3$ if $x_2 &lt; x_1$ and $f(x) = x_2 + x_3$ otherwise. This gives the following Lagrangian;
$L = p_1 x_1 + p_2 x_2 + p_3 x_3 - \lambda_1(x_1 +x_3 - Q) - \lambda_2(x_2 + x_3 - Q)$
With each $\lambda_i &gt; 0$ both constraints are binding, such that,
$Q=x_1+x_3$ and $Q=x_2 + x_3$; and then the constraints reduces to
$x_1 - x_2 = 0$ implying that $x_1 = x_2$. However, this contradicts the initial conditions that I made on the relationship between $x_1$ and $x_2$ and, thereby, the arising production functions.
I'm caught in a bit of confusion here. How should I proceed with this problem?
While not said explicitly in the question I am guessing from the Langrangian function you set up that the problem you intend to solve is
$$\min_{x_1,x_2,x_3} p_1x_1 + p_2x_2 + p_3x_3 \\[8pt]
h(x_1,x_2,x_3) = \min\{x_1,x_2\} + x_3\geq z$$
This assignment combines perfect complements in production $f(x_1,x_2) = \min \{x_1,x_2\}$ with perfect substitues $g(y_1,y_2) = y_1 + y_2$.
One approach is sometimes to solve problems in two stages. Intuitively, with $y_1:=\min\{x_1,x_2\}$ firm will produce $y_1$ at the unit cost $p_{y_1} = p_1+p_2$ for whatever level of $y_1$ is produced it can never be cost minimizing to use more of $x_1$ than $x_2$ or other way around. Hence $x_1=x_2 = \min\{x_1,x_2\} = y_1$ and the seperate cost for each unit of $y_1$ is therefore $p_1+p_2$.
Next, solve cost minimizing with production
$$g(y_1,y_2) = y_1 + y_2 = y_1 + x_3 = z$$
with input prices $p_{y_1}$ and $p_{3}$. Since factors $y_1$ and $x_3$ are equally productive (constant equal marginal productivity) production will happen with cheapest factor so cost must be $C(z,p) =\min\{p_{y_1},p_3\}z = \min\{p_1+p_2,p_3\}z$.
I avoid using Lagrangian multiplier method because the production function is non-differentiable.