Maximize parts of 271

Call the number of summands $n$ and the summands themselves $a_1, a_2, a_3,\dots, a_n$ .

By the AM-GM inequality, we have $$\frac{a 1+a 2+a 3+\cdots+a n}{n} \geq \left( a 1 a 2 a 3\cdots a n \right)^{1/n},$$ with equality holding when $a_1=a_2=a_3=...=a_n$ . Noting that $a_1+a_2+a_3+\cdots+a_n=271$ , and raising both sides to the power $n$ , we get $$\left(\frac{271}{n} \right)^n \geq a 1 a 2 a 3\cdots a n.$$ For a given $n$ , therefore, $a_1 a_2 a_3\cdots a_n$ is maximized when $a_1=a_2=a_3=...=a_n$ .

Letting $a_1=a_2=a_3=...=a_n$ , we have the product $$p(n)=a 1 a 2 a 3\cdots a n=\left(\frac{271}{n} \right)^n,$$ which we wish to maximize. Note, that we cannot maximize this function using calculus because it is not a function of a real variable, but rather a function of a whole number variable. So instead we maximize the function $f: (0,\infty) \to \mathbb{R}, \ f(x)= \left(\frac{271}{x} \right)^x,$ which we will do by calculus. To find the maximum, we must first take the derivative of $f$ , which can be done using logarithmic differentiation: $$\ln f(x) = x \ln \left( \frac{271}{x} \right)$$ $$\ln f(x)= x \ln 271 - x \ln x.$$ Differentiating implicitly with respect to $x$ , we get $$\frac{f'(x)}{f(x)}= \ln 271 - \ln x - 1$$ $$\frac{f'(x)}{f(x)}= \ln \left( \frac{271}{e\cdot x} \right)$$ $$f'(x)= \left(\frac{271}{x} \right)^x \ln \left( \frac{271}{e\cdot x} \right).$$ To maximize $f$ , we must find where $f'$ switches from positive to negative. Setting $f'(x)=0$ and solving gives us $$ \left(\frac{271}{x} \right)^x \ln \left( \frac{271}{e\cdot x} \right) = 0$$ $$\ln \left( \frac{271}{e\cdot x} \right)=0$$ $$ \frac{271}{e\cdot x}=1 $$ $$x= \frac{271}{e}.$$ It is easy to verify that $f'(x)>0$ when $x<\frac{271}{e}$ , and $f'(x)<0$ when $x>\frac{271}{e}$ . Thus the maximum of $f$ occurs at $x= \frac{271}{e}$ . Because $e$ is a little bigger than $2.71$ , we have that $\frac{271}{e}$ is a little less than $100$ , i.e., $99 < \frac{271}{e} < 100$ .

Because $f(x)$ has its maximum here, $p(n)$ should have its maximum here as well. But because $n$ is a whole number, $p(n)$ will have its maximum either at $n=99$ or $n=100$ . Calculating $p(99)$ and $p(100)$ shows that $p(100)>p(99)$ , which means that $p(n)$ has its maximum at $n=\boxed{100}$ .

I want to find a general solution for all value of * " $a$ " *

let the summands are $b_{1}, b_{2}, b_{3},...b_{x-1}$ and $b_{x}$ . where $x$ is the number of required summands.

such that $b_{1}+ b_{2}+ b_{3}+...b_{x-1}+b_{x}=a$ .

By AM-GM inequality we come to the conclusion, that the product of the summands will be maximized if and only if every summands are equal.

so we get, $b_{1}=b_{2}= b_{3}=...b_{x-1}=b_{x}$ and $b_{m}=\frac{a}{x}$ for $m={1,2,3,4...,x}$

And the product of the summands are: $\prod_{1\leq j\leq x}^{ } b_{j} = \left ( \frac{a}{x} \right )^{x}$

So all we need to do is to maximize $\left ( \frac{a}{x} \right )^{x}$

let, $y=\left ( \frac{a}{x} \right )^{x}$ ,

$\Rightarrow ln (y) = x ln(a) - x ln (x)$

$\Rightarrow \frac{1}{y} \times \frac{dy}{dx} = ln (a) - (1+ ln (x))$

$\Rightarrow \frac{dy}{dx} = y[ ln \left (\frac{a}{x} \right ) - 1]$

Now, we want to find out when $\frac{dy}{dx}$ becomes 0, As $y= \left ( \frac{a}{x} \right )^{x} \neq 0$ , then $[ ln \left ( \frac{a}{x} \right ) - 1]$ must be 0.**

So we get,

$[ ln \left ( \frac{a}{x} \right ) - 1]=0$

$\Rightarrow ln \left (\frac{a}{x} \right )=1=ln e$

$\Rightarrow \frac{a}{x}=e$

$\Rightarrow x=\frac{a}{e}$ , where e defines the base of Natural logarithm.

Now, $\frac{dy}{dx}=y ln\left (\frac{a}{x} \right) - y$

$\Rightarrow \frac{d^{2}y}{dx^{2}}=\frac{dy}{dx}[ ln \left ( \frac{a}{x} \right ) - 1] - \frac{1}{ax}$

Now plugging $x=\frac{a}{e}$ we get,

$\frac{d^{2}y}{dx^{2}}=\frac{dy}{dx}[ ln(e) - 1] - \frac{1}{ax}=0-\frac{1}{ax}=-\frac{1}{ax}$

So, for $x=\frac{a}{e}$ the value of $\frac{d^{2}y}{dx^{2}}$ becomes Negative . That is to say, for $x=\frac{a}{e}$ , the value of $\left ( \frac{a}{x} \right )^{x}$ is maximized.

So 271 will require $x=\frac{a}{e}=\frac{271}{e}$ summands, or approximately $\boxed{100}$ summands, to be the product maximized.

More of a thinking process than a solution, but here goes...

Pretend you are breaking down the number 10 rather than 271. You could try 2 2 2 2 2=32, or 5 5 = 25, but those are both smaller than 3 3*4=36. You'll notice a couple of things while thinking about 10.

Let's think about breaking down 9. If you think about it, you'll realize 3 3 3 seems to be the best option.

1. Choosing 5 as a summand is a bad choice (2*3 will give you 6 and still only add 5 to the sum).
2. Choosing 4 as a summand is the same as choosing 2 (2*2 and 4 both add 4 to the sum).
3. Choosing 3 seems slightly better than choosing 2.
4. Choosing the same number as all your summands leads to a likely maximum product.

But wait, we're dealing with real numbers, not just integers here...so what now...

If we break down a number n into pieces of x, we want to maximize the following function. x ^ (n/x)

If you want, take the derivative and maximize that function (or be lazy and use wolframalpha...I don't judge).

The maximum occurs at x = e = 2.718...

271 / 2.718 is about 100. Your last summand is probably a little smaller than e, but it should still be a maximum.

Thus, there should be 100 summands to maximize the product.

Lets suppose we divide 271 into n parts. such that, $a_{1} + a_{2} + \ldots + a_{n} = 271$

Now, let us apply AM-GM Inequality to this relation of n real and positive numbers. Thus,

$\frac {a_{1} + a_{2} + \ldots + a_{n}}{n} \geq \sqrt[n] {a_{1} a_{2} \ldots a_{n}}$

Using the fact that the L.H.S is $\frac {271}{n}$ , and raising to the power of n on both sides, we get,

$\frac {271^n}{n^n} \geq a_{1} a_{2} \ldots a_{n}$

The R.H.S is the required product of the summands. Hence, to find the maximum value of this product, we clearly need to maximize $\frac {271^n}{n^n}$ Therefore, let us consider,

f(x) = $\frac {271^n}{n^n}$

Differentiating, rearranging and equating the expression to 0. We find,

n = $\frac {271}{e}$ which on assuming e $\approx$ 2.71, we get,

n = 100

We want to maximize $x^{\frac{271}{x}}$ to do so we will find when it's derivative equals 0. Using logarithmic differentiation we find that the derivative is $271x^{\frac{271}{x}}(1-ln(x))$ . Setting this equal to 0 we find that $x=0$ or $e$ . Since optimally we want an integer power of x we want either $x=\frac{271}{100}$ or $\frac{271}{99}$ . Using a calculator it is easy to verify that $x=\frac{271}{100}$ is larger, thus our answer is $\boxed{100}$ .

If the sum $S = a+b$ is given, the product $a\cdot b$ is maximized by making $a = b = \tfrac12S$ . This is easy to check, and generalizes immediately to $n$ factors, so that we maximize the product of $n$ summands by taking $a_1 = a_2 = \cdots = a_n = S/n$ .

The only question, then, is how much $n$ should be.

Now the product of the summands is $P = a_1 \cdot a_2 \cdot \cdots \cdot a_n = \left(\frac S n\right)^n.$ To maximize this, I will maximize the logarithm by settings its derivative equal to zero. $\frac{dP}{dn} = \frac{d}{dn}\left[n(\ln S - \ln n)\right] = \ln S - \ln n - 1 = 0.$ The solution is $\ln n = \ln S - 1 \ \ \ \therefore\ \ \ \ n = \frac S e = \frac{271}{2.71\dots},$ which lies between 99 and 100. Therefore the solution is either 99 or 100.

We check $99(\ln 721 - \ln 99) = 196.57; \\100(\ln 721 - \ln 100) = 197.55.$ The latter is slightly greater, so the optimal number of summands is $n = \boxed{100}$ .

A.M - G.M. inequality:

[ (x1 + x2 + ... + xn) /n ] ^ n >= (x1 * x2 * .... * xn)

Let x1 + x2 + ... +xn =271 and let P = x1 * x2 * ... * xn

We have now, P <= (271/n)^n. So we have to find the value of n for which (271/x)^x is maximum.

Let f(x) = (271/x)^x. Taking log and differentiating we get that

d(f(x))/dx = (271/x)^x * [ ln(271/x) - 1].

d(f(x))/dx = 0 at x = 271/e.....................[ ln(271/x) - 1=0 implies ln(271/ex) =1 implies ex= 271]

f(x) is increasing for x < 271/e and decreasing for x > 271/e and attains it's maximum at 271/e.

Now, 271/e is approximately equal to 99.6953 and the required number of summands is 100.