<p>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 $r$.</p>

<p>\begin{eqnarray*} \min_{L, K} &amp;&amp; wL + rK \\ \text{s.t.} &amp;&amp; \left(\min(K, L)\right)^\beta \geq y \\  &amp;&amp; K \geq 0, \ L \geq 0 \end{eqnarray*}</p>

<p>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*}</p>

CC BY-SA 3.0
2017-04-25T13:18:55.563Z
2017-04-26T00:28:51.337Z
2017-04-26T00:28:51.337Z
Amit
<p>This is not a weird case, but a Leontief production function which is <em>not</em> 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.</p>

<p>Consider 
$$Q_b=[a K^{-\rho} +(1-a) L^{-\rho}  ]^{-\frac{b}{\rho}},\;\; b&gt;0$$</p>

<p>$$\Rightarrow Q_b = \frac 1{[a (1/K^{\rho}) +(1-a) (1/L^{\rho})  ]^{\frac{b}{\rho}}}$$</p>

<p>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.</p>

<p>Without loss of generality, assume $K\geq L \Rightarrow (1/K^{\rho})\leq (1/L^{\rho})$. We also have $K, L &gt;0$. Then we verify that the following inequality holds:</p>

<p>$$(1-a)^{b/\rho}(1/L^{b})\leq  Q_b^{-1} \leq (1/L^{b}) $$</p>

<p>$$\rightarrow (1-a)^{b/\rho}(1/L^{b})\leq  [a (1/K^{\rho}) +(1-a) (1/L^{\rho})  ]^{\frac{b}{\rho}} \leq (1/L^{b}) \tag{1}$$</p>

<p>by raising throughout to the $\rho/b$ power to get</p>

<p>$$(1-a)(1/L^{\rho}) \leq a (1/K^{\rho}) +(1-a) (1/L^{\rho})  \leq (1/L^{\rho}) \tag {2}$$
which indeed holds, obviously, given the assumptions. Then go back to the first row of $(1)$ and</p>

<p>$$\lim_{\rho\rightarrow \infty} (1-a)^{b/\rho}(1/L^{b}) =(1/L^{b})$$</p>

<p>which sandwiches the middle term in $(1)$ to $(1/L^{b})$ , so </p>

<p>$$\lim_{\rho\rightarrow \infty}Q_b = \frac {1}{1/L^b} = L^b = \big[\min\{K,L\}\big]^{b} \tag{3}$$</p>

<p>For more, see <a href="https://economics.stackexchange.com/a/399/61">this answer</a>.</p>

CC BY-SA 3.0
2016-06-07T19:04:55.087Z
2016-06-07T19:04:55.087Z
2017-04-13T12:51:38.253Z
Alecos Papadopoulos
<p>No problem. $$Q =\left(\min\{K, L\} \right)^b$$</p>

<p>It just means that first you compare $K$ and $L$ and your quantity $Q$ will be equal the lower one to the power $b$. </p>

<p>Example: $b=2$, $K=3$ and $L=7 \implies Q = 3^2 = 9$.</p>

CC BY-SA 3.0
2016-07-07T20:14:35.857Z
2016-07-07T20:14:35.857Z
snoram
<p>Define $Q=q^b$, where $q=min\{K,L\}$.</p>

<p>For $b&gt;0$, $Q$ 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 $Q$.</p>

CC BY-SA 3.0
2017-04-25T19:58:50.533Z
2017-04-25T19:58:50.533Z
luchonacho
<p>I am solving some micro related exercises and I came across this weird Leontief production function: $$Q =\left(\min\{K, L\} \right)^b$$  </p>

<p>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? </p>

<macroeconomics><leontief>
Weird Leontief production function
Ronaldo777
# Weird Leontief production function

(Asked by: username_0 on 2016-06-07)

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?

**Tags:** macroeconomics, leontief


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## 4 answers

### Answer by username_2 on 2016-06-07. Score: 5

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](https://economics.stackexchange.com/a/399/61).


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### Answer by username_1 on 2017-04-25. Score: 4

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\*}


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### Answer by username_3 on 2016-07-07. Score: 3

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 $L=7 \implies Q = 3^2 = 9$.


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### Answer by username_4 on 2017-04-25. Score: 1

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 $Q$.


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