IIT JEE , AIEEE Forum - Mathematics , Algebra - Write 271 as the sum of positive real numbers so as to maximize their product

little_genius

Write 271 as the sum of positive real numbers so as to maximize their product

rudra.panda

271=136+135
136*135=18360. Couldn't find any other.

nitigya

rudra ur ans i swrng as it says real no n not integers !!!!!!!!!

no the answer should be 135.5

as 135.5*135.5 = 18360.25 and that is the max product

rate if satisfied!

vasanth

hey

u c itz positive real nos......

so d first thing u shud look at is

x + y/2 >= rt(xy)

so itz obviously (135.5)^2........

in maximisin probs the value u luk fer mostly lies in middle

vasanth

oops........stupid me!

it s not mentioned 2 + real nos...is it........

let me assume the number of numbers to b 'n'

so

x₁ + x₂ + ....+ x_n >= n (x₁x₂.....x_n)^1/n

.......pdt....<= (271/n)ⁿ

so pdt is a function of number of terms

so p(n) = (271/n)ⁿ.........

logp = n log(271) - nlogn

1/p * dp/dn = log271 - logn - 1

dp/dn = p[log (271 / n) - 1]

dp/dn = 0 ------->> log(271/n) = 1 or p=0

n = 271/e or p = 0

but p cant be zero so

n = 271/e

n x₁ = nx₂ = .......= nx_n = 271

so...

x₁ = x₂ = x₃=.......=x_n = e

so the positive real numbers required are:

271/e................n times..........

x₁ = x₂ = x₃=.......=x_n = e.........

uhhhhhh........now anothr prob.....,m gettin not gettin n as an integer.......ne1 help.....

sachinguptaiit

Let the numbers be x and 271-x

P(x) = Product of two numbers

P(x) = x ( 271-x )

P'(x) = 271 - 2x

Since P(x) is max hence P'(x) = 0

x = 135.5

vasanth

@sachin gupta.....

dats d mistake we all made d last tym.........

its not mentioned 2 nos.--got it!

accordin to ma soln.....the max pdt is e^271/e ~~~approx....(2.7)¹⁰⁰.....which is way higher than ur 135.5^2

xperts......reply

rudra.panda

@vasanth i think you are correct. It is no where mentioned that it has to be 2 numbers. It can be more that that.

hsbhatt

Call the product for n integers as f(n)

Then $f(n) = \left(\frac{271}{n}\right)^n$

You will observe that the function initially increases. We must find out, up to what n this happens.

$f(n) < f(n+1) \Rightarrow \left(\frac{271}{n} \right )^{n} < \left(\frac{271}{n+1} \right )^{n+1} \\ \\<br/>\text{or} \ \left(1+\frac{1}{n} \right)^n (n+1) < 271$

Now, we can use tha fact that the function $\left(1+\frac{1}{n}\right)^n$ approaches e rapidly. So, we can approximate it by 2.7

This gives n = 100 as a good estimate for the largest such n. Thereafter it is a decreasing function.

I checked with Excel (that's the best I have!) and 100 is right

mukulaish

let the numbers b x and 271-x

their product is equal to 271x-x^2
now f(x)=271x-x^2

for the function to be maximum or minimum d/dx{f(x)=0

i.e 271-2x=0
=>x=271/2=135.5

now finding the double derivative of f(x)

it is equal to -2<0
this proves that x=135.5 is a point of maximum....

thus the two numbers are 135.5 and 135.5

further for a competitive point of view all the questions of this kind the answer is always half the number given.....

rate if useful....

feynmann

ans all are 2.71

Proof : Using the old A.M>= G.M

G.M. <= ( 271/n) ^n ( it attains the maximum value when all xi's are equal )

Now consider the characteristics of ( 271/x)^x ( let x be a real no )

Now we have that it attains its maximum value at x= 271/e= 99.6959

we need to consider only n = 99 or, n =100

Trough actual calculation it comes out that for x= 100 it attains the maximum value

Hence all the lenghths should be 2.71

hsbhatt

@mukulaish: The question has not placed a restriction on the number of real numbers that can be there.

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