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I'm considering a typical profit-maximization problem:

\begin{equation} \label{optimization1} \begin{aligned} & \underset{K,L}{\text{max}} & & P Y - r K - w L \\ \end{aligned} \end{equation}

where $r$ is the interest rate and $w$ is the wage rate.

The production function can be Cobb-Douglas,

$Y=AK^{\alpha}L^{\beta}$

where $0\leq \alpha \leq 1$ and $0\leq \beta \leq 1$. Or it can be a CES production function.

I would like to find the solution for $K$ and $L$ in either case of production function.

My code in the case of Cobb-Douglas, for example, is:

Y = A k^a L^b;
PROF = P Y - r k - w L;
z1 = D[PROF, k];
z2 = D[PROF, L];
Simplify[Solve[{z1 == 0, z2 == 0}, {k, L}], {0 <= a <= 1 && 0 <= b <= 1 && 
A > 0 && k > 0 && L > 0 && P > 0 && r > 0 && w > 0 && PROF > 0}]

And I get a very weird solution as follows. enter image description here

Can anyone help in finding what went wrong? Thanks!

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  • $\begingroup$ Why are you so convinced that anything went wrong? However, you have inequality constraints on k and L, so ypu need to solve the full set of KKT conditions. And those can usually only be solved when numerical values for all parameters are given. $\endgroup$ Commented Aug 10, 2019 at 4:05
  • $\begingroup$ @Henrik: You can sense that the solution is weird once you solve it manually and compare. (In fact, it is easy to find on the internet a note that describes the process of solving this kind of maximization problem.) $\endgroup$
    – ppp
    Commented Aug 12, 2019 at 9:25
  • $\begingroup$ The point is this: You have to make this problem intersting to other users, not me. It is unlikely that users perform long computations before they decide to help you or not. $\endgroup$ Commented Aug 12, 2019 at 13:11

1 Answer 1

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Try

Y = A k^a L^b;
PROF = P Y - r k - w L;
z1 = D[PROF, k];
z2 = D[PROF, L];
sol=Together[Simplify[Solve[{z1 == 0, z2 == 0}, {k, L}], {0 <= a <= 1 && 0 <= b <= 1 && a + b + c == 1 && A > 0 && k > 0 && L > 0 && P > 0 && r > 0 &&  w > 0 && PROF > 0}][[1]]]
{k, L} /. sol // FullSimplify
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    $\begingroup$ It's a bit nicer, imo, if we introduce another parameter c and add the assumption a + b + c == 1 to Simplify. (+1) $\endgroup$
    – Michael E2
    Commented Aug 12, 2019 at 15:28
  • $\begingroup$ @MichaelE2 Edited according to your hint. Thanks. $\endgroup$
    – Cesareo
    Commented Aug 12, 2019 at 15:57

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