# Math Help - Applying maximum products to real life solutions?

1. ## 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.

2. 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 $x_1,...,x_k$.