Maximizing by parts

Algebra Level 5

Given that

$27=\displaystyle \sum_{i=1}^{n} a_i$

Where each $a_i$ is a positive real number and $n$ is a positive integer.

Find the maximum value of

$P=\displaystyle \prod_{i=1}^n a_i$

Input your answer as the first 3 digits of $P$ .

The answer is 205.

1 solution

Karan Shekhawat
Apr 17, 2015

Since Sum is constant , So product is maximum when numbers are equal (By AM-GM) .... Hence Clearly... $\displaystyle{P(n)={ \left( \cfrac { 27 }{ n } \right) }^{ n }\\ P^{ ' }\left( n \right) =0\\ \Rightarrow n=\cfrac { 27 }{ e } \approx 10\\ { \left( P(n) \right) }_{ max }=P(10)={ \left( e \right) }^{ \cfrac { 27 }{ e } }\approx 20589.1\\ \boxed { Ans=205 } \\ }$

Moderator note:

This solution has been marked incomplete. Why must the statement "Since Sum is constant , So product is maximum when numbers are equal" be true? Moreover, you have only shown that the turning point of the function occurs at $\frac {27}{e}$ . You should show by the second derivative test that the value you had found is maximum.

By AM-GM, $P=\prod_{i=1}^na_i \le \left( \begin{array}{c}\sum\limits_{i=1}^na_i\\ \hline n \end{array}\right)^n=\left(\dfrac{27}{n}\right)^n$

Daniel Liu - 6 years, 1 month ago

its more of a calculus question than algebra.

A Former Brilliant Member - 3 years, 7 months ago

Maximizing by parts

The answer is 205.

1 solution

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