# Applying maximum products to real life solutions?

• Feb 15th 2009, 06:49 AM
RedGeorgie
Applying maximum products to real life solutions?
I've been calculating all the different products from the splitting of pairs or triples or quadruples of numbers that sum to a specific number e.g. For number 12 - the pairs 11 and 1 = product of 11, the pairs 10 and 2 = product 20 with a maximum product in that range of pairs of 6 and 6 = 36. Then for the number 15 - the triples of 13 and 1 and 1 = product 13; then 12, 2, and 1 = 24 with a max product combination of 5,5, and 5 = 125.

Then I made a formula for this pattern of identifying maximum products which for pairs is (n/2) to power of 2; for triples is (n/3) to power of three and so on.

1. Does that make sense?

2. What could be real life applications of being able to determine maximum products in this way - pairs, triples and something more numerically exotic?

Thanks for advice and best wishes.

Red.
• Feb 15th 2009, 09:12 AM
ThePerfectHacker
Quote:

Originally Posted by RedGeorgie
I've been calculating all the different products from the splitting of pairs or triples or quadruples of numbers that sum to a specific number e.g. For number 12 - the pairs 11 and 1 = product of 11, the pairs 10 and 2 = product 20 with a maximum product in that range of pairs of 6 and 6 = 36. Then for the number 15 - the triples of 13 and 1 and 1 = product 13; then 12, 2, and 1 = 24 with a max product combination of 5,5, and 5 = 125.

Then I made a formula for this pattern of identifying maximum products which for pairs is (n/2) to power of 2; for triples is (n/3) to power of three and so on.

1. Does that make sense?

2. What could be real life applications of being able to determine maximum products in this way - pairs, triples and something more numerically exotic?

Thanks for advice and best wishes.

Red.

If you want to maximize $x_1x_2 ... x_k$ (of positive numbers) subject with the condition $x_1 + x_2 + ... + x_k = n$ for a given integer $n$ then you need to choose $x_1 = x_2 = ... = x_k = \tfrac{n}{k}$. One way to prove this result is by using Lagrange's multipliers. Define $f(x_1,...,x_k) = x_1x_2...x_k$ and $g(x_1,...,x_k) = x_1 + ... + x_k$. Thus, you want to maximize $f$ subject with $g(x_1,...,x_k) = n$. This means you need to satisfy $\nabla f = k \nabla g \text{ and }g=n$. From here solve for $x_1,...,x_k$.