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Math Help - Applying maximum products to real life solutions?

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by RedGeorgie View Post
    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.
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