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

$$ f(x)=(x_1^p + x_2^p)^{1/p} \ \ for\ \ 0<p <1 $$

I tried to solve it and found that $$TC(y) = \left\{ \begin{array}{ll} w_1y & \quad w_1 < w_2 \\ wy & \quad w_1=w_2 \\ w_2y & \quad w_2 <w_1 \end{array} \right. $$

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

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

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

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  • $\begingroup$ 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 $\endgroup$
    – victor
    Dec 27 '20 at 16:52
  • $\begingroup$ but your answer seem to be very great and Thank you very much $\endgroup$
    – victor
    Dec 27 '20 at 16:52
  • $\begingroup$ How then do you solve constrained optimization? $\endgroup$ Dec 27 '20 at 18:02
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    $\begingroup$ Tangency condition, no doubt. Start with (1) in Jesper's answer! $\endgroup$
    – Brennan
    Dec 28 '20 at 3:47
  • $\begingroup$ 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 ! $\endgroup$
    – victor
    Jan 5 at 10:06

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