Did I understand this question well: You have to find largest number which can be represented as $xy$ s.t. $x+y=1976$? Because if that is a question then it is trivial: $x+y\geq 2\sqrt{xy}$ so $(x+y)^2\geq 4xy$ so that number is $\frac{1976^2}4=976144$ m@re

You can get a number which is MUCH bigger. For example, you could get $2^{988}$

The biggest number you can find is $2.3^{658}$. Proof: Assume there was any factor 5 or more. we can split 5=2+3 which restults in a factor 6>5. Same for numbers larger than 5. So there are none. Thus we have the optimal number is of the form $2^m.3^n$. And since $3^2>2^3$, we need as much as possible into 3's, thus we end up with $2.3^{658}$.

Well, that's why I asked did I understand this question

Maybe we can find more: If $x_1,x_2,...,x_n$ are reals with $\sum x_i = 1976$, what is the maximum value $\prod x_i$ can obtain? Or maybe easier to start with, what is the value of $n$?

Simple computations gives that when $n$ is given, it's optimal to take $x_1=x_2=...=x_n$. So we want $(\frac{a}{n})^n$ ($a=1976$ here) to be as big as possible. Deriving $f(x)=(\frac{a}{x})^x=e^{x \ln \frac{a}{x}}$ gives $f'(x)=e^{x \ln \frac{a}{x}} (ln(\frac{a}{x})-1)$, so it attains it's only maximum at $x=\frac{a}{e}$. Now $n$ has to be the next greater or smaller integer (it's monotonous on both sides).

I just base cashed using AM-GM; $ \left(\frac{1976}{n}\right)^n \ge a_1 a_2 ... a_n$ Then, as the product was an integer n must be a factor of 1976 and the prime factors of 1976 are; 2,2,2,13,19 then consider all combinations of these. However, I don't know if I am missing something - it just seems a bit too simple

Exactly the same except that i explained a bit more on why we chose (2 and 3 and 3)not (4 and 4) or (5 and 3) or(1 and 4 and 3)

how do you know you must do an additional 2? how do you know that we should not separate the 2's and distribute them onto 3's?

what we know that is that $3^4>4^3$, but this does not mean $3\times3\times2>4\times4$ (without calculation)

So what is the answer here?

Peter wrote: Proof: Assume there was any factor 5 or more. we can split 5=2+3 which restults in a factor 6>5. Same for numbers larger than 5. So there are none. For this we first need to prove this claim. Anyone here please prove this.

The critical part is it says about positive integers. If positive reals, then the answer is $e^{\frac{1976}{e}}$. As it says for positive integers then have to work a little. We get from AM-GM that $\left(\frac{1976}{n}\right)^n\geq t$ where $t$ is the product of $n$ integers. To maximize the value of $f(x)=\left(\frac{1976}{x}\right)^x$ we have to find the first derivative of this then have to make it zero. Applying chain rule we have $f'(x)=e^{xln\frac{1976}{x}}\cdot f'(xln\frac{1976}{x})=e^{xln\frac{1976}{x}}\cdot (ln\frac{1976}{x}-1)$. Hence we get $ln\frac{1976}{x}-1=0\Longrightarrow x=\frac{1976}{e}$ We have to do some case work. We have find the integer solutions of the plot of $f(x)$ and the highest $y$ value is finally the answer. The value $e^{\frac{1976}{e}}$ is useful for the reason that the nearest values of it will be actually the answer (I think). The rest is little boring therefore I didn't find.

If @above's solution is correct, then we have a nice way of giving explicit computations for $\lfloor e^{\frac{n}{e}}\rfloor$ for all positive integers.

starchan wrote: If @above's solution is correct, then we have a nice way of giving explicit computations for $\lfloor e^{\frac{n}{e}}\rfloor$ for all positive integers. I have found error in my solution. Whatever. you can show the way of computing the value of $\lfloor e^{\frac{n}{e}}\rfloor$ what you have found.

Abdullahil_Kafi wrote: The critical part is it says about positive integers. If positive reals, then the answer is $e^{\frac{1976}{e}}$. As it says for positive integers then have to work a little. We get from AM-GM that $\left(\frac{1976}{n}\right)^n\geq t$ where $t$ is the product of $n$ integers. To maximize the value of $f(x)=\left(\frac{1976}{x}\right)^x$ we have to find the first derivative of this then have to make it zero. Applying chain rule we have $f'(x)=e^{xln\frac{1976}{x}}\cdot f'(xln\frac{1976}{x})=e^{xln\frac{1976}{x}}\cdot (ln\frac{1976}{x}-1)$. Hence we get $ln\frac{1976}{x}-1=0\Longrightarrow x=\frac{1976}{e}$ We have to do some case work. We have find the integer solutions of the plot of $f(x)$ and the highest $y$ value is finally the answer. The value $e^{\frac{1976}{e}}$ is useful for the reason that the nearest values of it will be actually the answer (I think). The rest is little boring therefore I didn't find. Umm you cannot have a nonintegral number of reals. $x$ must remain integer no matter what. You cannot have $x = \frac{1976}{e}$ because you cannot have those many positive reals. (this number should be an integer no matter what)

Suppose some positive integer $k\ge 5$ is there. Then replace $k$ with $2$ and $k-2$. We have $2(k-2)=2k-4>k$, so it is more optimal. If some $k=4$ is there, then replace it with two $2$'s. Thus, optimally, all these positive integers are either $2$ or $3$. Now we can replace three twos with two threes, because $3^2>2^3$. So the greatest is $\boxed{2\cdot 3^{658}}$.

Let $x_1, x_2, \dots, x_n$ be positives summing to $1976$ and $M$ be the max product of $x_1x_2 \dots x_n$. By observation we find that $M$ is of the form $2^r3^s$ implying that $2r+3s=1976$. It's obvious that to get max we must have as many 3's more than 2's possible (since $3^2> 2^3$). So, $2r \equiv 1976 ( \text{mod } 3) \implies (k,l)= (1,658) \implies M=2 \cdot 3^{658}.$ $\square$