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Calculus Level 5

Let $s_1, \ldots, s_n$ , with $n$ being an integer $\geq 1$ , be a series of positive real numbers such that

$\displaystyle \sum_{i=1}^n s_i = 271.$

If the maximum value of

$\displaystyle \prod_{k=1}^n s_k = \dfrac {a^x}{b^y}$

where $a$ and $b$ are square-free coprime positive integers, and $x$ and $y$ being positive integers greater than 1, find $a+x+b+y$ .

Note: $n$ is variable.

1 solution

Rohith M.Athreya
Apr 8, 2016

we first use AM GM inequality and arrive at a maximum value for expression $\displaystyle \prod_{k=1}^n s_k = \dfrac {a^x}{b^y}$

this happens to be $(\frac{271}{n})^n$

on differentiating, we get $ln\frac{271}{n}=1$ to be satisfied

thus n is 100

max value is $((\frac{271^{100}}{100^{100}})) = e^{100} = \frac{271^{100}}{10^{200}}$

thus required answer is 581