# Finding the greatest product

• Nov 8th 2011, 01:34 AM
gurgehclosp
Finding the greatest product
Hi all,

I came up with an interesting question for myself, but I can't find the answer. Here are the rules:

You take any number, let's say we take 9.

You're then allowed to use any group of positive numbers whose sum is 9, and try to get the biggest product out of them.

For example, you might try 4*5 or 3*3*3 or 4.5*4.5, etc. There is no limit to how many multiplications you use.

That's pretty much it. Basically I wanted to see if there was some sort of formula for always finding the greatest product for any number while abiding by these rules, or any way of showing what the greatest product was even for a particular number.

I came up with this: (N^0.5)^(N^0.5) which as far as I know abides by the rules above - although I don't know how to write it out with normal multiplications - but it turns out not to be the answer after all. Any help or comments much appreciated!
• Nov 8th 2011, 03:27 AM
earboth
Re: Finding the greatest product
Quote:

Originally Posted by gurgehclosp
Hi all,

I came up with an interesting question for myself, but I can't find the answer. Here are the rules:

You take any number, let's say we take 9.

You're then allowed to use any group of positive numbers whose sum is 9, and try to get the biggest product out of them.

For example, you might try 4*5 or 3*3*3 or 4.5*4.5, etc. There is no limit to how many multiplications you use.

That's pretty much it. Basically I wanted to see if there was some sort of formula for always finding the greatest product for any number while abiding by these rules, or any way of showing what the greatest product was even for a particular number.

I came up with this: (N^0.5)^(N^0.5) which as far as I know abides by the rules above - although I don't know how to write it out with normal multiplications - but it turns out not to be the answer after all. Any help or comments much appreciated!

1. Let n denotes the number you choose (in your example n = 9)

2. Then the 1st factor of the product is x and the 2nd consequently (n - x)

3. The product p can be written as

$p(x)=x(n-x)$

4. This is a quadratic function opening down whose maximum is at it's vertex which is

$V\left(\frac n2 , \frac{n^2}{4} \right)$

5. So the maximum value is $\frac{n^2}4$ (with your example $x = 4.5, p_{max} = 20.25$)
• Nov 8th 2011, 06:13 AM
Ackbeet
Re: Finding the greatest product
Quote:

Originally Posted by earboth
1. Let n denotes the number you choose (in your example n = 9)

2. Then the 1st factor of the product is x and the 2nd consequently (n - x)

3. The product p can be written as

$p(x)=x(n-x)$

4. This is a quadratic function opening down whose maximum is at it's vertex which is

$V\left(\frac n1 , \frac{n^2}{4} \right)$

5. So the maximum value is $\frac{n^2}4$ (with your example $x = 4.5, p_{max} = 20.25$)

Hmm. Not sure I agree with this solution. According to the rules in the OP, you're allowed to use as many multiplications as you want (could be more than two). $3 \times 3 \times 3 = 27 > 20.25.$ I think, by symmetry, you can say that, given $n$ multiplicands, the maximum product you can get would be given by $x = 9/n$, multiplied $n$ times. That is, if you allow two multiplicands, then the max occurs at $9/2 = 4.5,$ as earboth found. If you have three multiplicands, then I think the maximum would occur at $9/3 = 3,$ which gives the product $27.$ So, given this way of choosing the maximum product, given the number of multiplicands, which number of multiplicands gives the maximum product? Obviously, $1^9 = 1$ is less than $3^3 = 27$, so there is a maximum in there somewhere. The formula to use is $(9/n)^n$. I admit to using a bit of calculus to maximize this function. The max occurs at $9/e$. The closest integer (you have to have an integer number of multiplicands) is $3,$ so I think the maximum is $27$ in this case. As another example, if you're solving the problem with, say, $13$ instead of $9,$ then the max occurs at $13/e.$ So, since $2 I would try taking the number and dividing by $2,$ and then dividing by $3$ to see which one of them is greater when plugged into your function $(13/n)^n$. Probably dividing by $3$ is going to give you the maximum most of the time, if not all the time.
• Nov 8th 2011, 11:29 PM
gurgehclosp
Re: Finding the greatest product
Thanks for the replies chaps.

I had also reasoned that dividing by the number of multiplicands always yielded the highest product. It's because of the rule that everything always has to sum to the same amount (I think).

That's also why I thought of putting it to a power rather than using multiplication, so I could use the same technique for numbers that didn't divide neatly into integers. I.e. (N^0.5)^(N^0.5). Apologies for the awkward formatting.

(Actually, could someone just clarify whether this still falls within the rules or not? I'm working on the assumption that powers are just a different way of expressing normal multiplication)

In conclusion, it doesn't seem like there's a particular formula for getting at the highest product (within this rule set), unless you set the number of multiplicands. Is that right?
• Nov 9th 2011, 07:06 AM
veileen
Re: Finding the greatest product
$\frac{\sum_{i=1}^{n}a_i}{n} \geq \sqrt[n]{\prod_{i=1}^{n}a_i}$, with equality if and only if $a_1=a_2=...=a_n$ (inequality of arithmetic and geometric means)

Your sum is: $\frac{\sum_{i=1}^{n}a_i}{n}=\frac {9}{n}$. Obviously, the product maxim value is $\left ( \frac {9}{n} \right )^n$.