Prove that the largest product P of N numbers whose sum is S is $$\rm{P}=\left(\frac{\rm{S}}{\rm{N}}\right)^\rm{N}$$
For clarity:
1. An example of a specific case would be (so you understand what I meant): Find the largest product P of 2 numbers whose sum is 42, which then solved using diff eq to obtain $$x_1=x_2=21$$ and $$P=x_1\cdot x_2=21\cdot 21=441$$ and we can generalize that for N=2 $$P=\left(\frac{\rm{S}}{\rm{2}}\right)^\rm{2}$$, what I wanted to know ts if this would be true for every nonzero positive integer N
2. All numbers is a nonzero positive real number. $$x,\rm{N} \in \mathbb{R}^+$$
3. Function to be maximized: $$\rm{P}=\prod^N_{i=1} x_i$$ constraint: $$\rm{S}=\sum^N_{i=1} x_i$$

Question

Prove that the largest product P of N numbers whose sum is S is $$\rm{P}=\left(\frac{\rm{S}}{\rm{N}}\right)^\rm{N}$$ 
For clarity: 
1. An example of a specific case would be (so you understand what I meant): Find the largest product P of 2 numbers whose sum is 42, which then solved using diff eq to obtain $$x_1=x_2=21$$ and $$P=x_1\cdot x_2=21\cdot 21=441$$ and we can generalize that for N=2 $$P=\left(\frac{\rm{S}}{\rm{2}}\right)^\rm{2}$$, what I wanted to know ts if this would be true for every nonzero positive integer N
2. All numbers is a nonzero positive real number. $$x,\rm{N} \in \mathbb{R}^+$$ 
3. Function to be maximized: $$\rm{P}=\prod^N_{i=1} x_i$$  constraint: $$\rm{S}=\sum^N_{i=1} x_i$$

anonymous · Answer

$$P=\frac{S^n}{n^n} \  S=\sum^n_{i=1}x_i \ P= \prod^n_{i=1}x_i \ \text{for } x|x \in \bf{R}$$

anonymous · Answer

but I have no idea how to prove it's general form

anonymous · Answer

I would recommend clarifying your terminology. What does 'maximum product possible' mean?

I think what you are trying to prove is not right. I cannot judge fully because I do not understand the term 'maximum product possible'.

anonymous · Answer

ahh sorry, it's a variation of this question 'if the sum of 2(n) number is 42(S), what is the maximum product(P) possible of both numbers? using diff eq to find the maximum then the number is both 21 and the product would be 21 squared, for n>2 I used lagrange multiplier

anonymous · Answer

and after I solved for n=3,4,5 in it is clear that all numbers ($$x_1,x_2,...,x_n$$) always have the same value, which implies P=S^n/n^n

anonymous · Answer

I think you are extrapolating too much from a special case. Let me see if I follow what you are trying to figure out:

If the sum of $n$ numbers is $S$, what is the maximum product $P$ of all $n$ numbers?

I would immediately think the maximum product is just $\prod_{1 \leq i \leq n}x_i$. This thought, however, assumes all numbers are positive.

However! If you are working under the premise that $x_i=x_j$ for all $i$ and $j$, your conclusion is provable via algebra:

$$\begin{align}
P&=\frac{S^n}{n^n}\
&=\left(\sum_{1 \leq i \leq n}x_i\right)^n\frac{1}{n^n}\
&=\frac{(x_1+x_2+\dots+x_n)^n}{n^n}\
&=\frac{(nx_1)^n}{n^n}\
&=\frac{n^nx_1^n}{n^n}\
&=x_1^n
\end{align}$$

Once again, I should emphasize we are working with positive numbers because the answer is much more intuitive. With negative numbers, we would have to be more precise. With all of $\mathbb{R}$, we would need a much more critical analysis.

anonymous · Answer

yes I am only interested in positive S and x, but what I am actually curious is the proof that $$x_i=x_j$$ which eludes me. in a sense, maybe I'm extrapolating too much from a special case

anonymous · Answer

You can't really prove that. You could have, for example, $\{x_1,x_2,x_3\}=\{1,2,3\}$.

anonymous · Answer

One particular thing has me quite curious, however. How did Lagrange multipliers become relevant to this?

anonymous · Answer

it seems so though when I tried to solve n=5 
$$\text{constraint: }g(x,y,z,w,v)=x+y+z+w+v-\rm{S}=0 \ \text{maximize: }\rm{P}=xyzwv \f(x,y,z,w,v)=xyzwv+\lambda (x+y+z+w+v-\rm{S}) \\frac{\partial f}{\partial x}=yzwv+\lambda=0 \\frac{\partial f}{\partial y}=xzwv+\lambda=0 \\frac{\partial f}{\partial z}=xywv+\lambda=0 \\frac{\partial f}{\partial w}=xyzv+\lambda=0 \\frac{\partial f}{\partial v}=xyzw+\lambda=0 \\frac{\partial f}{\partial \lambda}=x+y+z+w+v-\rm{S}=0 \\text{which \implies }x=y=z=w=v  \text{ \therefore } \rm{P}=\frac{S^5}{5^5}$$

anonymous · Answer

I am not particularly educated on Lagrange multipliers. Pardon me if I sound stupid, but how can you use $0=0$ as a constraint? $g(x,y,z,w,v)$ evaluates to zero based on your definition of it.

I will try to read more critically about Lagrange multipliers if you can explain what your reasoning is here.

anonymous · Answer

because lagrange function is in the form f(x,y)=h(x,y)+$$f=h+\lambda g \ h \text{ is the function \to be maximized where }g \text{ is the constraint \in the form } g(x,y)=0 \text{ and } \lambda \text{ is the multiplier used \to solve the eqn} $$ as a side note it seems when I try to insert 'in' 'to' 'implies' 'therefore' in \text{...} they would have a backslash in front of them...

anonymous · Answer

so if the constraint S=x+y+z then g(x,y)=x+y+z-S=0

anonymous · Answer

as for why is that so, simple explanation could be found in http://www.karlscalculus.org/pdf/lagrange.pdf page 4-6

anonymous · Answer

I understand this now. However, how are you concluding (from the partial derivative equations) that $x=y=z=w=v$? Is this conclusion based on the fact that the partial derivative equations have no solution for $\lambda$ if this is not the case? ($\lambda$ has to have just one value, after all.)

Am I understanding your reasoning or not?

anonymous · Answer

no it's just I don't want to write long algebra, by solving the eqns simultanously.. yzwv=xzwv yzwv=xywv yzwv=xyzv yzwv=xyzw then you would get x=y x=z x=w x=v ,

anonymous · Answer

That's pretty much what I just said.

But I am rather confused right now. It is clearly impossible that all of the numbers are simultaneously equal in all cases. I think your method has shown you that your product is the highest if all of the numbers are equal. It is essentially saying that exponentiation grows faster than multiplication.

Do you see how this differs from saying that $P$ is always equal to the form you derived? I believe you can also show, by the way, that this is true of all such products as you have maximized them. You could use rather lengthy algebra, essentially.

anonymous · Answer

Thank you for posting this question, by the way. I greatly enjoy questions where I am motivated to learn new things.

anonymous · Answer

I believe you gave me some insight too (especially in communicating my ideas..) I'm sorry though if my writing is kinda confusing, thanks for your help

anonymous · Answer

though I still do not see in why would all numbers being equal as impossible such as your example {x1,x2,x3}={1,2,3} , the sum would be 6 and according to the equation xi=xj x1=x2=x3=2 therefore the largest product is 8

anonymous · Answer

In my example, they are unequal by definition. I was using set notation to say $x_1=1$, $x_2=2$, and $x_3=3$. My point was that you could find any such set in all of $\mathbb{R}$. I believe, however, that your conclusion was correct. You concluded that the *maximum* product occurs when all the numbers -are- equal.

I had originally thought that you were presuming the product always took the form which you had posted--which is not always true. (The form is true if and only if all the numbers are equal.)

anonymous · Answer

Now I think you understand how bad I am with written arguments T_T , as a side note, where do you learn mathematics online?

anonymous · Answer

Everywhere. I teach myself everything and occasionally ask for help from college students.

anonymous · Answer

I am learning Algebra right now from B.L. van der Waerden's Algebra. Currently I am on chapter two.