$e = 2.71$

Number Theory Level pending

If $n$ positive real numbers add up to 27, find the maximum product of $n$ numbers so chosen (to the nearest integer).

12701 20589 16834 729 19683

1 solution

Shreyas Panasa
Feb 22, 2017

product increases when the no.s tend to e so the maximum product is(e)^( $\frac{27}{e}$ ) which is 20589. . $\frac{27}{e}$ has to be a number because no.of terms must be a natural number so we need to select the closest no. to $\frac{n}{e}$ in this case 10 is the nearest to $\frac{27r}{e}$ so we take it to be (e)^10

Yes, we are looking to maximize the function $f(x) = \left(\dfrac{27}{x}\right)^{x}$ , and then finding the closest integer value to $x$ to use as $n$ . This function is maximized when $x = 27/e \approx 9.93$ , so we take $n = 10$ and find that $2.7^{10} \approx 20589$ to the nearest integer.

Brian Charlesworth - 4 years, 3 months ago

@Shreyas Panasa Can you update your solution? Note that $e ^ { 27 / e } \approx 20593.79$ . The max that we can obtain is slightly lower, due to the constraint of having 10 terms instead of $27/e$ terms.

Calvin Lin Staff - 4 years, 3 months ago

ok i will update

Shreyas Panasa - 4 years, 3 months ago

e = 2.71 e = 2.71 e = 2 . 7 1

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$e = 2.71$