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The specific problem I have is of the following type.

$$\begin{array}{ll} \text{maximize} & \frac{1}{1-a}x^{1-a} + \frac{1}{1-b}y^{1-b} +\frac{1}{1-c}z^{1-c}\\ \text{subject to} & x+y+z=K\end{array}$$

where $x,y,z$ are the optimization variables and $a, b, c > 1$ are constants.

I have tried using a Lagrange multiplier but it didn't work.

Is there a way to solve problems of this type? If not, is there a numerical method for approximating?

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  • $\begingroup$ $K$ is a positive constant. $\endgroup$ – Firzul Firaz Nov 24 '16 at 17:07
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    $\begingroup$ What did you get with Lagrange multipliers? $\endgroup$ – user856 Nov 24 '16 at 17:40
  • $\begingroup$ $x^{-a}=\lambda$; $y^{-b}=\lambda$; $z^{-c}=\lambda$; $x+y+z=K$ Wasn't able to solve for any. $\endgroup$ – Firzul Firaz Nov 24 '16 at 18:14
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    $\begingroup$ In that case $x=\lambda^a$ and similarly for $y$ and $z$, so you get $\lambda^a+\lambda^b+\lambda^c=K$, which given specific values of $a,b,c,K$ you can solve numerically for $\lambda$. $\endgroup$ – user856 Nov 24 '16 at 18:46
  • $\begingroup$ No algebraic solutions possible? $\endgroup$ – Firzul Firaz Nov 24 '16 at 18:53
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you will get $$x^{-a}+\lambda=0$$, $$y^{-b}+\lambda=0$$, $$z^{-c}+\lambda=0$$ and $$x+y+z=K$$ solve this system!

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  • $\begingroup$ I reached there too. It appears to be solvable when I count the unknowns and equations. But for some reason, it cannot be solved. $\endgroup$ – Firzul Firaz Nov 24 '16 at 18:23
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  1. Represent x, y, z with $\lambda$ and substitute them into the fourth equation.

  2. Then use bisection method to numerically solve for $\lambda$.

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