<p>If you are interested in the case where <span class="math-container">$\rho \geq 1$</span> then look at the post <a href="https://economics.stackexchange.com/questions/8873/ces-production-function-with-rho1?rq=1">CES
<span class="math-container">$\ \ \rho \geq 1$</span></a>. For the standard case where <span class="math-container">$0 &lt; \rho &lt; 1$</span> you should get a result like this</p>
<p><span class="math-container">$$C(w_1,w_2,y)  = \left(w_1^{\frac{\rho}{\rho -1}} +w_2^{\frac{\rho}{\rho -1}}\right)^{\frac{\rho - 1}{\rho}} y.$$</span></p>
<p>To see this you should start by setting up the cost minimization problem</p>
<p><span class="math-container">$$\min_{x_1,x_2} \ \ w_1x_1 + w_2x_2 \\[8pt] 
s.t. \ \ (x_1^\rho + x_2^\rho)^{1/\rho} \geq y$$</span></p>
<p>for this problem the Lagrangian function is</p>
<p><span class="math-container">$$\mathcal L(x_1,x_2,\lambda) = w_1x_1 + w_2x_2 - \lambda((x_1^\rho + x_2^\rho)^{1/\rho} -y).$$</span></p>
<p>From the first order conditions of the Lagrangian you can show the constraint is binding in optimum <span class="math-container">$(x_1^\rho + x_2^\rho)^{1/\rho} = y$</span> and get MRS equal to relative prices</p>
<p><span class="math-container">$$(1) \ \ \frac{w_1}{w_2} = \frac{x_1^{\rho - 1}}{x_2^{\rho - 1}},$$</span>
given this information you should be able to solve for <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span> as a function of the parameters of the problem which in this case is <span class="math-container">$\rho,y,w_1,w_2$</span>.</p>
<p>Try to get <span class="math-container">$(x_1^\rho + x_2^\rho)^{1/\rho}$</span> to appear in the MRS equal to relative prices. So manipulate (1) to get</p>
<p><span class="math-container">$$w_1^{\frac{\rho}{\rho -1}}x_2^\rho =  w_2^{\frac{\rho}{\rho -1}}x_1^\rho,$$</span>
then add <span class="math-container">$w_2^{\frac{\rho}{\rho-1}}x_2^\rho$</span> to both sides of equation</p>
<p><span class="math-container">$$w_1^{\frac{\rho}{\rho -1}}x_2^\rho +w_2^{\frac{\rho}{\rho -1}}x_2^\rho =  w_2^{\frac{\rho}{\rho -1}}x_1^\rho + w_2^{\frac{\rho}{\rho -1}}x_2^\rho,$$</span></p>
<p>isolate factors on both sides and exponentiate with exponent <span class="math-container">$1/\rho$</span> to get</p>
<p><span class="math-container">$$\left(w_1^{\frac{\rho}{\rho -1}} +w_2^{\frac{\rho}{\rho -1}}\right)^{1/\rho}x_2 =   w_2^{\frac{1}{\rho -1}}(x_2^\rho + x_1^\rho)^{1/\rho} = w_2^{\frac{1}{\rho -1}} y ,$$</span></p>
<p>from here you can solve for conditional demand <span class="math-container">$x_2^\star(w_1,w_2,y)$</span>. However, it is easier to oberserve that the factor <span class="math-container">$\left(w_1^{\frac{\rho}{\rho -1}} +w_2^{\frac{\rho}{\rho -1}}\right)^{1/\rho}$</span> do not change when interchanging indexes - it is symmetric. So define <span class="math-container">$a := \left(w_1^{\frac{\rho}{\rho -1}} +w_2^{\frac{\rho}{\rho -1}}\right)^{1/\rho}$</span> and conclude that</p>
<p><span class="math-container">$$ax_1 = w_1^{\frac{1}{\rho -1}} y \\[8pt]
ax_2 = w_2^{\frac{1}{\rho -1}} y,$$</span>
multiply first equation with <span class="math-container">$w_1$</span> and second with <span class="math-container">$w_2$</span> and add them to get</p>
<p><span class="math-container">$$a(w_1x_1 + w_2x_2) = (w_1^{\frac{\rho}{\rho -1}}+w_2^{\frac{\rho}{\rho -1}})y = a^\rho y$$</span></p>
<p>solve for <span class="math-container">$(w_1x_1 + w_2x_2)$</span> which are the costs to get the result that</p>
<p><span class="math-container">$$C(w_1,w_2,y) = a^{\rho -1} y = \left(w_1^{\frac{\rho}{\rho -1}} +w_2^{\frac{\rho}{\rho -1}}\right)^{\frac{\rho - 1}{\rho}} y$$</span></p>

CC BY-SA 4.0
2020-12-27T16:52:07.420Z
It seems to be a great answer but tbh It is so difficult, I literraly never see lagrangien function and I just start economy this year.  It is why I'm a bit confuse
victor
CC BY-SA 4.0
2020-12-27T16:52:32.200Z
but your answer seem to be very great and Thank you  very much
victor
CC BY-SA 4.0
2020-12-27T18:02:32.990Z
How then do you solve constrained optimization?
Jesper Hybel
CC BY-SA 4.0
2020-12-28T03:47:52.400Z
Tangency condition, no doubt. Start with (1) in Jesper's answer!
Brennan
CC BY-SA 4.0
2021-01-05T10:06:14.530Z
Thank you, I just unterstand now, what you did, my teacher didn't explain us yet what you did  but now it's ok for me !
victor
CC BY-SA 4.0
2021-01-05T16:18:33.843Z
To be honest Lagrange is to me a magical process the math of why it works I do not understand, nevertheless it is fortunately usually enough to know when and how it works.
Jesper Hybel
CC BY-SA 4.0
2020-12-26T11:15:02.153Z
2021-01-05T22:09:03.327Z
2021-01-05T22:09:03.327Z
Jesper Hybel
<p>How can I find the cost function <span class="math-container">$c(w,p)$</span> given that the production is</p>
<p><span class="math-container">$$ f(x)=(x_1^p + x_2^p)^{1/p} \ \ for\ \  0&lt;p &lt;1 $$</span></p>
<p>I tried to solve it and found that
<span class="math-container">$$TC(y) = \left\{
        \begin{array}{ll}
            w_1y &amp; \quad w_1 &lt; w_2 \\
            wy &amp; \quad w_1=w_2 \\
            w_2y &amp; \quad w_2 &lt;w_1
        \end{array}
    \right.
$$</span></p>
<p>Can you say me if I'm on the right way?</p>

CC BY-SA 4.0
2020-12-26T10:44:08.193Z
It does not really look like it. You have to set up cost minimization problem and solve for conditional demands $x_k^\star(w,y)$ with these found you have the cost function as $c(w,y) = \sum_k w_k x_k^\star(w,y)$.
Jesper Hybel
CC BY-SA 4.0
2020-12-27T16:28:09.923Z
can you accept this answer or state what is missing?
Jesper Hybel
<microeconomics><cost-functions><ces>
Cost function from CES production function
victor
# Cost function from CES production function

(Asked by: username_0 on 2020-12-25)

How can I find the cost function $c(w,p)$ given that the production is

$$ f(x)=(x\_1^p + x\_2^p)^{1/p} \ \ for\ \ 0

I tried to solve it and found that
$$TC(y) = \left\{
\begin{array}{ll}
w\_1y & \quad w\_1 < w\_2 \\
wy & \quad w\_1=w\_2 \\
w\_2y & \quad w\_2

Can you say me if I'm on the right way?

**Tags:** microeconomics, cost-functions, ces
> **Comments:**
> **username_1**: It does not really look like it. You have to set up cost minimization problem and solve for conditional demands $x_k^\star(w,y)$ with these found you have the cost function as $c(w,y) = \sum_k w_k x_k^\star(w,y)$.  
> **username_1**: can you accept this answer or state what is missing?  

---

## 1 answers

### Answer by username_1 on 2020-12-26. Score: 6

If you are interested in the case where $\rho \geq 1$ then look at the post [CES
$\ \ \rho \geq 1$](https://economics.stackexchange.com/questions/8873/ces-production-function-with-rho1?rq=1). For the standard case where $0 < \rho < 1$ you should get a result like this

$$C(w\_1,w\_2,y) = \left(w\_1^{\frac{\rho}{\rho -1}} +w\_2^{\frac{\rho}{\rho -1}}\right)^{\frac{\rho - 1}{\rho}} y.$$

To see this you should start by setting up the cost minimization problem

$$\min\_{x\_1,x\_2} \ \ w\_1x\_1 + w\_2x\_2 \\[8pt]
s.t. \ \ (x\_1^\rho + x\_2^\rho)^{1/\rho} \geq y$$

for this problem the Lagrangian function is

$$\mathcal L(x\_1,x\_2,\lambda) = w\_1x\_1 + w\_2x\_2 - \lambda((x\_1^\rho + x\_2^\rho)^{1/\rho} -y).$$

From the first order conditions of the Lagrangian you can show the constraint is binding in optimum $(x\_1^\rho + x\_2^\rho)^{1/\rho} = y$ and get MRS equal to relative prices

$$(1) \ \ \frac{w\_1}{w\_2} = \frac{x\_1^{\rho - 1}}{x\_2^{\rho - 1}},$$
given this information you should be able to solve for $x\_1$ and $x\_2$ as a function of the parameters of the problem which in this case is $\rho,y,w\_1,w\_2$.

Try to get $(x\_1^\rho + x\_2^\rho)^{1/\rho}$ to appear in the MRS equal to relative prices. So manipulate (1) to get

$$w\_1^{\frac{\rho}{\rho -1}}x\_2^\rho = w\_2^{\frac{\rho}{\rho -1}}x\_1^\rho,$$
then add $w\_2^{\frac{\rho}{\rho-1}}x\_2^\rho$ to both sides of equation

$$w\_1^{\frac{\rho}{\rho -1}}x\_2^\rho +w\_2^{\frac{\rho}{\rho -1}}x\_2^\rho = w\_2^{\frac{\rho}{\rho -1}}x\_1^\rho + w\_2^{\frac{\rho}{\rho -1}}x\_2^\rho,$$

isolate factors on both sides and exponentiate with exponent $1/\rho$ to get

$$\left(w\_1^{\frac{\rho}{\rho -1}} +w\_2^{\frac{\rho}{\rho -1}}\right)^{1/\rho}x\_2 = w\_2^{\frac{1}{\rho -1}}(x\_2^\rho + x\_1^\rho)^{1/\rho} = w\_2^{\frac{1}{\rho -1}} y ,$$

from here you can solve for conditional demand $x\_2^\star(w\_1,w\_2,y)$. However, it is easier to oberserve that the factor $\left(w\_1^{\frac{\rho}{\rho -1}} +w\_2^{\frac{\rho}{\rho -1}}\right)^{1/\rho}$ do not change when interchanging indexes - it is symmetric. So define $a := \left(w\_1^{\frac{\rho}{\rho -1}} +w\_2^{\frac{\rho}{\rho -1}}\right)^{1/\rho}$ and conclude that

$$ax\_1 = w\_1^{\frac{1}{\rho -1}} y \\[8pt]
ax\_2 = w\_2^{\frac{1}{\rho -1}} y,$$
multiply first equation with $w\_1$ and second with $w\_2$ and add them to get

$$a(w\_1x\_1 + w\_2x\_2) = (w\_1^{\frac{\rho}{\rho -1}}+w\_2^{\frac{\rho}{\rho -1}})y = a^\rho y$$

solve for $(w\_1x\_1 + w\_2x\_2)$ which are the costs to get the result that

$$C(w\_1,w\_2,y) = a^{\rho -1} y = \left(w\_1^{\frac{\rho}{\rho -1}} +w\_2^{\frac{\rho}{\rho -1}}\right)^{\frac{\rho - 1}{\rho}} y$$

> **Comments:**
> **username_0**: It seems to be a great answer but tbh It is so difficult, I literraly never see lagrangien function and I just start economy this year.  It is why I'm a bit confuse  
> **username_0**: but your answer seem to be very great and Thank you  very much  
> **username_1**: How then do you solve constrained optimization?  
> **username_2**: Tangency condition, no doubt. Start with (1) in Jesper's answer!  
> **username_0**: Thank you, I just unterstand now, what you did, my teacher didn't explain us yet what you did  but now it's ok for me !  
> **username_1**: To be honest Lagrange is to me a magical process the math of why it works I do not understand, nevertheless it is fortunately usually enough to know when and how it works.  

---

