Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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
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
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.