# Cost function from CES production function

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?

• 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 $c(w,y) = \sum_k w_k x_k^\star(w,y)$. Dec 26 '20 at 10:44
• can you accept this answer or state what is missing? Dec 27 '20 at 16:28

If you are interested in the case where $$\rho \geq 1$$ then look at the post CES $$\ \ \rho \geq 1$$. 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

$$C(w_1,w_2,y) = a^{\rho -1} y = \left(w_1^{\frac{\rho}{\rho -1}} +w_2^{\frac{\rho}{\rho -1}}\right)^{\frac{\rho - 1}{\rho}} y$$

• 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 Dec 27 '20 at 16:52
• but your answer seem to be very great and Thank you very much Dec 27 '20 at 16:52
• How then do you solve constrained optimization? Dec 27 '20 at 18:02