Maximize this!

Algebra Level 5

Let ${ A }_{ 1 },{ A }_{ 2 }, { A }_{ 3 },...,{ A }_{ k }$ be a sequence of positive real numbers. Given that $\displaystyle \sum _{ i=1 }^{ k }{ { A }_{ i } } = 1000$

and that $\prod _{ i=1 }^{ k }{ { A }_{ i } }$ is maximized, determine $k$ .

The answer is 368.

1 solution

Madhavan V
Jul 10, 2015

For a given sum of k numbers, if we have to maximize the product of those numbers, then arithmetic mean of k numbers should be equal to geometric mean of those k numbers. Therefore from this inequality, we get product of k numbers=(1000/k)^1/k.Let a new function y be equal to :- y=(1000/k)^1/k.differentiating this function with respect to k, we get d(y)/dk=y*(ln(1000)-lnk-1),as y is never equal to 0 ,(at k=infinity it becomes equal to 1) therefore ln(1000)-ln(k)-1=0. Solving this we get value of k=1000/e (where e=2.7182). Hence k is equal to 367.87 ≃368

Maximize this!

The answer is 368.

1 solution

0 pending reports