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In The Keynesian economy,

I have the following model

enter image description here

Here, the production function $F(N,K) = Y/K = A(N/K)^{1.10}$ and by the labor demand function $F_N(K,N)= N/K= b_0(w/p)^{b_1}$.

right?

That is, what is the F(K,N)=? and $F_N(N,K)=?$

Here , I mean that $F_N(N,K)$ is marginal product of labor.

enter image description here

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To get back to the original production function just multiply both sides with capital. Here:

$$Y/K = A (N/K)^{1.1} = F(N/K,1) \implies Y = AN^{1.1}K^{-0.1} =F(N,K)$$

Also, the marginal product of labor here will be:

$$F_N'(N,K) = 1.1 A N^{0.1}K^{-0.1} = 1.1 A (N/K)^{0.1}$$

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  • $\begingroup$ Okay, I see right now. Thanks a lot. Now, I don’t want to have a direct derivative of the function $F$. As you know, MPL(marginal product of labor) is equal to the real wage (w/p). That’s, $F_N(N,K)=w/p$. So, in this case, $F_N(N/K,1)= b_0(w/p)^{b_1}$ then, the original one is $F_N(N,K)= b_0(w/p)^{b_1}K$. Am I right? $\endgroup$ – B11b Nov 17 '20 at 12:32
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    $\begingroup$ @B11b It is true that marginal product of labor, in competitive equilibrium, is equal to real wage but also marginal product of labor is by definition $\partial F/\partial N$. So in this case it would be appropriate to say that MPL = $1.1A(N/K)^{0.1} = w/p$ and this implies that $N/K = (\frac{1}{1.1A})^{1/0.1}(w/p)^{1/0.1}$ but you can simply state that $\frac{1}{1.1A})^{1/0.1} = \beta_0$ and that $1/0.1 = \beta_1$ and get $N/K = \beta_0 (w/p)^{\beta_1}$ $\endgroup$ – 1muflon1 Nov 17 '20 at 12:39
  • $\begingroup$ But, in your case, $\beta_1$ is not negative. Yet, question says it is negative. $\endgroup$ – B11b Nov 17 '20 at 13:58
  • $\begingroup$ @B11b that is actually good point, can you please give me reference for source where you got the model above? I think that there must be some explanation for this otherwise math checks out nicely. I cant see any obvious mistake in the marginal product derivation. $\endgroup$ – 1muflon1 Nov 17 '20 at 14:17
  • $\begingroup$ I posted the full question. It is from Keynesian model problem set. I agree with you. Your math side is correct I think so. $\endgroup$ – B11b Nov 17 '20 at 15:13

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