I am solving some micro related exercises and I came across this weird Leontief production function: $$Q =\left(\min\{K, L\} \right)^b$$
I am not sure how to solve it. I have to find the inputs' demand, cost function etc.... Should I solve it as $K^b=L^B=Q$ and then proceed as usual with the standard Leontief?
This is not a weird case, but a Leontief production function which is not homogeneous of degree one, but homogeneous of degree $b$. You can see this if you use the connection between a C.E.S. production function and the Leontief one.
Consider $$Q_b=[a K^{-\rho} +(1-a) L^{-\rho} ]^{-\frac{b}{\rho}},\;\; b>0$$
$$\Rightarrow Q_b = \frac 1{[a (1/K^{\rho}) +(1-a) (1/L^{\rho}) ]^{\frac{b}{\rho}}}$$
Take the limit when $\rho \rightarrow \infty$. Since we are interested in the limit when $\rho\rightarrow \infty$ we can ignore the interval for which $\rho \leq0$, and treat $\rho$ as strictly positive.
Without loss of generality, assume $K\geq L \Rightarrow (1/K^{\rho})\leq (1/L^{\rho})$. We also have $K, L >0$. Then we verify that the following inequality holds:
$$(1-a)^{b/\rho}(1/L^{b})\leq Q_b^{-1} \leq (1/L^{b}) $$
$$\rightarrow (1-a)^{b/\rho}(1/L^{b})\leq [a (1/K^{\rho}) +(1-a) (1/L^{\rho}) ]^{\frac{b}{\rho}} \leq (1/L^{b}) \tag{1}$$
by raising throughout to the $\rho/b$ power to get
$$(1-a)(1/L^{\rho}) \leq a (1/K^{\rho}) +(1-a) (1/L^{\rho}) \leq (1/L^{\rho}) \tag {2}$$ which indeed holds, obviously, given the assumptions. Then go back to the first row of $(1)$ and
$$\lim_{\rho\rightarrow \infty} (1-a)^{b/\rho}(1/L^{b}) =(1/L^{b})$$
which sandwiches the middle term in $(1)$ to $(1/L^{b})$ , so
$$\lim_{\rho\rightarrow \infty}Q_b = \frac {1}{1/L^b} = L^b = \big[\min\{K,L\}\big]^{b} \tag{3}$$
For more, see this answer.
We are given the production function $Q = \left(\min(K, L)\right)^\beta$. Cost minimization problem of the producer is defined as finding the labor-capital combination that minimizes the cost of producing at least $y$ units of output given that the price of labor is $w$, and price of capital is $r$.
\begin{eqnarray*} \min_{L, K} && wL + rK \\ \text{s.t.} && \left(\min(K, L)\right)^\beta \geq y \\ && K \geq 0, \ L \geq 0 \end{eqnarray*}
Solution to this problem (also known as conditional input demand functions) satisfy: \begin{eqnarray*} L = K = y^{1/\beta} \end{eqnarray*} and the associated (optimal) cost function is therefore, \begin{eqnarray*} C(w,r,y) = (w+r)y^{1/\beta} \end{eqnarray*}
No problem. $$Q =\left(\min\{K, L\} \right)^b$$
It just means that first you compare $K$ and $L$ and your quantity $Q$ will be equal the lower one to the power $b$.
Example: $b=2$, $K=3$ and $L=7 \implies Q = 3^2 = 9$.
Define $Q=q^b$, where $q=min\{K,L\}$.
For $b>0$, $Q$ is a monotonic transformation of $q$. As such, the solution for $q$ is equivalent to the solution for $Q$. Simply solve your problem for $q$, and then rework it in terms of $Q$.
