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I have the following optimization problem I need to solve, with the goal being to express $p_1$ in terms of solely $x_1$ and $x_2$ and $p_2$ in terms of solely $x_1$ and $x_2$ $$\max_{p_1,p_2} \ c_{1}p_1 + 2c_{2}(p_1p_2)^{1/2} + c_{3}p_2 \\ \text{ subject to } \\ p_1x_1 + p_2x_2 = 1 \\ p_1, p_2 > 0$$

Using Lagrangian techniques I know:

$$ \mathcal{L} = c_{1}p_1 + 2c_{2}(p_1p_2)^{1/2} + c_{3}p_2 - \lambda(p_1x_1 + p_2x_2 = 1)$$

and the first order conditions are:

\begin{align} c_1 + c_2p_1^{-1/2}p_2^{1/2} - \lambda x_1 = 0 \\ c_3 + c_2p_1^{1/2}p_2^{-1/2} - \lambda x_2 = 0 \\ p_1x_1 + p_2x_2 = 1 \end{align}

where again I want to solve for $p_1$ in terms of $x_1$ and $x_2$. and I want to solve for $p_2$ in terms of $x_1$ and $x_2$. My problem is I am unable to do so. Clearly this should be doable as I have 3 equations in 3 unknowns, but I dont know how to proceed. Any help would be greatly appreciated.

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Aren't there some sign constraints on the fixed constants ($x_1,x_2,c_1,c_2,c_3$) in your problem ? If there aren't, the maximal value of your objective function could be infinite.

Here is an example. I shall assume your $p_1,p_2\in \mathbf{R}_+ $ since you take the square root of them in your objective function. Take $c_1=c_2=c_3=p_1=1,p_2=-1$. Then the set of feasible points lies on the line $p_1 = 1+p_2$, which is graphed here: http://www.wolframalpha.com/input/?i=plot+y-x%3D1 . In particular as $p_1 \rightarrow \infty$, so too does $p_2 \rightarrow \infty$. On the other hand, your objective function is $(\sqrt{p_1}+\sqrt{p_2})^2$. Obviously as $p_1$ and $p_2$ go to infinity, the objective function increases without bound. So the maximum value of the objective is $\infty$ in this case.

After you have sorted out the sign constraints, it might be easier to solve your system of linear equations by doing the replacement $p_1 \leftarrow y_1^2$, $p_2 \leftarrow y_2^2$, where $y_1,y_2 \in \mathbf{R}$, which is valid because of my assumption above. After that, get two different expressions for $\frac{y_1}{y_2}$ by manipulating the first two of your equations. Solve for $\lambda$ by equating these two different expressions. Lastly substitute into the last equation.

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    $\begingroup$ awesome, thanks! yes it has to be the case that the $p_i$ are positive, I forgot to add the context of the equation so forgot this condition. I will try the change of variable you suggested $\endgroup$
    – celrin
    May 16, 2018 at 21:24

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