<p>The steps to optimally go from the Production function to the Cost function is to solve the cost-minimization problem for the firm, which determines the input mix given prices and some level of output, and in turn allows us to express one input with respect to the other (in this simplest two-inputs case).</p>

<p>When the production function is of the Leontief variety, optimal input mix is indeed determined by the equality stated by the OP, since by simple reasoning, any amount of one input above the amount satisfying the equality will incur the cost to  buy it but not any extra output.</p>

<p>Inserting the equality </p>

<p>$$\frac{L^*}{2a}=\frac{K^*}{4b} \implies L^* = \frac {a}{2b}K^*$$ </p>

<p>into the cost function we get Cost expressed as a function of one of the inputs only, $C^* = g(K^*)$ </p>

<p>In order to eliminate it too, we turn again to the production function which can also be expressed as a function of one input at the optimal solution, </p>

<p>$$Q = h(K^*) \implies K^* = h^{-1}(Q)$$</p>

<p>Inserting $h^{-1}(Q)$ into the Cost function we obtain</p>

<p>$$C = g\left[h^{-1}(Q)\right]$$</p>

<p>where also the various exogenous constants and prices will be present. In fact, this is easier done when the production function is Leontief, than when it is Cobb-Douglas for example, and so it should be easy for the OP to arrive at the correct solution.</p>

CC BY-SA 3.0
2017-12-14T13:53:40.960Z
So what is the computed $h^{-1}(Q)$ here?
Shreya Bhandari
CC BY-SA 3.0
2017-12-14T13:54:28.823Z
@ShreyaBhandari That's very easy to obtain, don't you think?
Alecos Papadopoulos
CC BY-SA 3.0
2017-12-14T13:56:14.227Z
I don't know, tell me!
Shreya Bhandari
CC BY-SA 3.0
2017-12-14T13:58:21.277Z
@ShreyaBhandari You already know. Given the equality that you have already wrote in your post, and that will hold at the optimal solution, how can you express $Q$ as a function of $K$?
Alecos Papadopoulos
CC BY-SA 3.0
2017-12-14T14:05:24.320Z
No idea, that was my question, my reason for confusion, cuz I'm getting the answer in terms of $K$ and the options are in terms of $Q$. There is something I am missing there.
Shreya Bhandari
CC BY-SA 3.0
2017-12-14T14:06:27.973Z
@ShreyaBhandari Once you express $Q$ as a function of $K$ what is stopping you from moving the terms around and express $K$ as a function of $Q$?
Alecos Papadopoulos
CC BY-SA 3.0
2017-12-14T14:08:14.870Z
Because the function is a min function. If you know, just type it down please.
Shreya Bhandari
CC BY-SA 3.0
2017-12-14T14:20:05.473Z
@ShreyaBhandari No I won't type it down. We are not here to solve your econ problems, we are here to help you understand how to solve them. Sometimes we succeed, sometimes we don't. One last try: the production function may be a $min$ function in general, but at the optimal the equality $\frac{L}{2a}=\frac{K}{4b}$ holds. If it holds, then what happens to the $min$ function?
Alecos Papadopoulos
CC BY-SA 3.0
2017-12-14T14:30:19.230Z
At optimality we find the intersection of the cost function and the IC. Min function simply implies that the optimality would be alond the equality line $\frac{L}{2a}=\frac{K}{4b}$. By solving the way i did, I got the intersection equality. Nd don't know how to proceed further. So if you tell me, I'd surely understand.
Shreya Bhandari
CC BY-SA 3.0
2017-12-14T14:32:41.553Z
@ShreyaBhandari What I asked you to think about is the following: If $\frac{L}{2a}=\frac{K}{4b}$, then $\min \left(\frac{L}{2a}, \frac{K}{4b}\right) = ?$
Alecos Papadopoulos
CC BY-SA 3.0
2017-12-14T14:42:30.403Z
Won't that depend on where $a$ is greater than or less than $b$. The min function would be equal to the the smaller value among the two right?
Shreya Bhandari
CC BY-SA 3.0
2017-12-14T14:44:00.070Z
@ShreyaBhandari But we start by assuming that the two ratios will be equal at the optimum, so...
Alecos Papadopoulos
CC BY-SA 3.0
2017-12-14T14:45:48.340Z
So it could be any of the two since they are supposedly equal
Shreya Bhandari
CC BY-SA 3.0
2017-12-14T14:49:35.043Z
@ShreyaBhandari Correct (and not "supposedly" -they are certainly equal under cost-minimizing behavior). So you are allowed to eliminate the $min$ from the picture and write $Q = L/2a$ or equivalently $Q=K/4b$. Isn't it now possible to express $K$ as a function of $Q$?
Alecos Papadopoulos
CC BY-SA 3.0
2017-12-14T14:54:25.253Z
So it would come out to be $2Q(wa+2br)$. So the correct option would be D? Or am I still missing something?
Shreya Bhandari
CC BY-SA 3.0
2017-12-14T14:57:09.053Z
@ShreyaBhandari This is indeed the correct cost function.
Alecos Papadopoulos
CC BY-SA 3.0
2017-12-14T15:01:46.037Z
Interesting! Got it!
Shreya Bhandari
CC BY-SA 3.0
2017-12-14T15:06:40.943Z
@ShreyaBhandari Good. Now please check my answer with the green mark, in order for your question to leave the "Unanswered" queue.
Alecos Papadopoulos
CC BY-SA 3.0
2017-12-14T15:10:43.070Z
Done, removing the down vote would also be a good idea!
Shreya Bhandari
CC BY-SA 3.0
2017-12-14T15:12:07.723Z
@ShreyaBhandari This is up to the person that has cast the downvote.
Alecos Papadopoulos
CC BY-SA 3.0
2017-12-14T15:13:01.890Z
And you are not that person?
Shreya Bhandari
CC BY-SA 3.0
2017-12-14T15:13:23.040Z
@ShreyaBhandari No.
Alecos Papadopoulos
CC BY-SA 3.0
2017-12-14T15:16:34.857Z
Haha okay, I wish that person would give me the reason! I am pretty new here!
Shreya Bhandari
CC BY-SA 3.0
2017-12-14T15:21:08.997Z
@ShreyaBhandari The person is me! Please read our [policy on homework questions](https://economics.stackexchange.com/help/on-topic).
Giskard
CC BY-SA 3.0
2017-12-14T15:31:02.460Z
@denesp Okay got it! Would keep that in mind! Will try to put the question in a more generalized manner.
Shreya Bhandari
CC BY-SA 3.0
2017-12-14T13:46:25.780Z
2017-12-14T13:46:25.780Z
Alecos Papadopoulos
<p>The cost function associated with this production function is
$A) 2awQ\\
B) 4brQ\\
C) (wa + 2br)Q\\
D) None\; of\; the\; above$</p>

<p>What I have tried is:
We have the cost function $wL+rK=C$.
Since, here, the goods are perfect compliments, therefore, $\frac{L}{2a}=\frac{K}{4b}$.
Putting the value of $L$ in cost function we would get $\left(\frac{K}{2b}(wa+2br)\right)=C$.
Which would be C option if $Q$ can taken as $\frac{K}{2b}$. 
But I'm not sure about that.
 Please confirm?</p>

CC BY-SA 3.0
2017-12-14T13:05:43.657Z
How did you arrive at $C$? Your work should be part of your post.
Alecos Papadopoulos
CC BY-SA 3.0
2017-12-14T13:24:12.663Z
Sure, check the edit!
Shreya Bhandari
<production-function><producer-theory>
Consider the following production function $Q=min \left(\frac{L}{2a}, \frac{K}{4b}\right)$. Let $w$ and $r$ be the wage and rental rate respectively
Shreya Bhandari
# Consider the following production function $Q=min \left(\frac{L}{2a}, \frac{K}{4b}\right)$. Let $w$ and $r$ be the wage and rental rate respectively

(Asked by: username_0 on 2017-12-13)

The cost function associated with this production function is
$A) 2awQ\\
B) 4brQ\\
C) (wa + 2br)Q\\
D) None\; of\; the\; above$

What I have tried is:
We have the cost function $wL+rK=C$.
Since, here, the goods are perfect compliments, therefore, $\frac{L}{2a}=\frac{K}{4b}$.
Putting the value of $L$ in cost function we would get $\left(\frac{K}{2b}(wa+2br)\right)=C$.
Which would be C option if $Q$ can taken as $\frac{K}{2b}$.
But I'm not sure about that.
Please confirm?

**Tags:** production-function, producer-theory

> **Comments:**
> **username_1**: How did you arrive at $C$? Your work should be part of your post.  
> **username_0**: Sure, check the edit!  

---

## 1 answers

### Answer by username_1 on 2017-12-14. Score: 0 | Accepted answer

The steps to optimally go from the Production function to the Cost function is to solve the cost-minimization problem for the firm, which determines the input mix given prices and some level of output, and in turn allows us to express one input with respect to the other (in this simplest two-inputs case).

When the production function is of the Leontief variety, optimal input mix is indeed determined by the equality stated by the OP, since by simple reasoning, any amount of one input above the amount satisfying the equality will incur the cost to buy it but not any extra output.

Inserting the equality

$$\frac{L^\*}{2a}=\frac{K^\*}{4b} \implies L^\* = \frac {a}{2b}K^\*$$

into the cost function we get Cost expressed as a function of one of the inputs only, $C^\* = g(K^\*)$

In order to eliminate it too, we turn again to the production function which can also be expressed as a function of one input at the optimal solution,

$$Q = h(K^\*) \implies K^\* = h^{-1}(Q)$$

Inserting $h^{-1}(Q)$ into the Cost function we obtain

$$C = g\left[h^{-1}(Q)\right]$$

where also the various exogenous constants and prices will be present. In fact, this is easier done when the production function is Leontief, than when it is Cobb-Douglas for example, and so it should be easy for the OP to arrive at the correct solution.

> **Comments:**
> **username_0**: So what is the computed $h^{-1}(Q)$ here?  
> **username_1**: @username_0 That's very easy to obtain, don't you think?  
> **username_0**: I don't know, tell me!  
> **username_1**: @username_0 You already know. Given the equality that you have already wrote in your post, and that will hold at the optimal solution, how can you express $Q$ as a function of $K$?  
> **username_0**: No idea, that was my question, my reason for confusion, cuz I'm getting the answer in terms of $K$ and the options are in terms of $Q$. There is something I am missing there.  
> **username_1**: @username_0 Once you express $Q$ as a function of $K$ what is stopping you from moving the terms around and express $K$ as a function of $Q$?  
> **username_0**: Because the function is a min function. If you know, just type it down please.  
> **username_1**: @username_0 No I won't type it down. We are not here to solve your econ problems, we are here to help you understand how to solve them. Sometimes we succeed, sometimes we don't. One last try: the production function may be a $min$ function in general, but at the optimal the equality $\frac{L}{2a}=\frac{K}{4b}$ holds. If it holds, then what happens to the $min$ function?  
> **username_0**: At optimality we find the intersection of the cost function and the IC. Min function simply implies that the optimality would be alond the equality line $\frac{L}{2a}=\frac{K}{4b}$. By solving the way i did, I got the intersection equality. Nd don't know how to proceed further. So if you tell me, I'd surely understand.  
> **username_1**: @username_0 What I asked you to think about is the following: If $\frac{L}{2a}=\frac{K}{4b}$, then $\min \left(\frac{L}{2a}, \frac{K}{4b}\right) = ?$  
> **username_0**: Won't that depend on where $a$ is greater than or less than $b$. The min function would be equal to the the smaller value among the two right?  
> **username_1**: @username_0 But we start by assuming that the two ratios will be equal at the optimum, so...  
> **username_0**: So it could be any of the two since they are supposedly equal  
> **username_1**: @username_0 Correct (and not "supposedly" -they are certainly equal under cost-minimizing behavior). So you are allowed to eliminate the $min$ from the picture and write $Q = L/2a$ or equivalently $Q=K/4b$. Isn't it now possible to express $K$ as a function of $Q$?  
> **username_0**: So it would come out to be $2Q(wa+2br)$. So the correct option would be D? Or am I still missing something?  
> **username_1**: @username_0 This is indeed the correct cost function.  
> **username_0**: Interesting! Got it!  
> **username_1**: @username_0 Good. Now please check my answer with the green mark, in order for your question to leave the "Unanswered" queue.  
> **username_0**: Done, removing the down vote would also be a good idea!  
> **username_1**: @username_0 This is up to the person that has cast the downvote.  
> **username_0**: And you are not that person?  
> **username_1**: @username_0 No.  
> **username_0**: Haha okay, I wish that person would give me the reason! I am pretty new here!  
> **username_2**: @username_0 The person is me! Please read our [policy on homework questions](https://economics.stackexchange.com/help/on-topic).  
> **username_0**: @denesp Okay got it! Would keep that in mind! Will try to put the question in a more generalized manner.  

---

