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I have a system of equations that looks like this:

$$\begin{array}{rl} a_1 b_1 c_1+a_2 b_2 c_2+a_3 b_3 c_3&=1000\\ a_1+a_2+a_3&=1\\ a_2&=0.6 \,a_1\\ b_1+b_2+b_3&=500 \end{array}$$

and $$a_1 a_2 a_3 b_1 b_2 b_3 c_1 c_2 c_3 > 0$$

$c_1,c_2,c_3$ are free.

I have no experience with linear programming, some in linear algebra. How do I go about finding the optimal solution so that $a_2b_2c_2$ is maximized?

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The first equation is not linear. –  copper.hat Jan 25 '13 at 6:17
    
My mistake, I've updated the title. –  mirai Jan 25 '13 at 6:17
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Optimal in what sense? –  Max Jan 25 '13 at 6:20
    
@Max optimal in the sense that it maximizes $a_2$ –  mirai Jan 25 '13 at 14:41
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I formatted the equations, removed the redundant one, check them. I'm not sure if the positive condition is for the product or for each variable. –  leonbloy Jan 25 '13 at 15:03
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1 Answer

up vote 1 down vote accepted

If the coefficients $a_i,b_i,c_i$ are not constrained to be non-negative, $a_2$ can be made as large as possible.

Consider $b_1 = b_2 = b_3 = \frac{500}{3}$. Let $a_2 =M$ be any positive number. $a_1 = \frac{5}{3}M$. Also, let $c_3 = -1$. Then, $c_1 = c_2 = \frac{18+5M}{8M}$ will satisfy all the equations. If $M$ is sufficiently large$(>\frac{3}{8}), a_3<0$. So, the product of all terms will be positive.

So, $a_2$ can be made as large as wanted. In other words, it is unbounded.

If you assume that all the $a_i,b_i,c_i$ are non-negative, the problem is straight-forward. Define new variables $x_1 = a_1b_1c_1,x_2 = a_2b_2c_2,x_3 = a_3b_3c_3$ in the first relation. Then, solve the simple LP that arises.

Try any online solver or matlab or mathematica for a sloution. You need not know how to solve a LP to get a solution.

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I realized that before reading your answer. Please see updated question (apologies for being unclear). –  mirai Jan 25 '13 at 22:03
    
What changes have you made. Can you please specify the ranges of all the variables - positive, non-negative or reals? –  dexter04 Jan 27 '13 at 7:53
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