Algebra

abhishek gupta's Avatar
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11 Jul 2008 23:16:25 IST
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Write 271 as the sum of positive real numbers so as to maximize their product
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Algebra

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



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®µD®A's Avatar

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11 Jul 2008 23:21:56 IST
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271=136+135
136*135=18360. Couldn't find any other.
nitigya's Avatar

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11 Jul 2008 23:35:54 IST
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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_mech pilani's Avatar

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12 Jul 2008 00:19:53 IST
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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_mech pilani's Avatar

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13 Jul 2008 16:49:45 IST
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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 + ....+ xn   >=  n (x1x2.....xn)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 = .......= nxn = 271


so...


x1 = x2 = x3=.......=xn = e


so the positive real numbers required are:


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


x1 = x2 = x3=.......=xn = e.........


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


Scorching goIITian

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13 Jul 2008 17:10:16 IST
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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_mech pilani's Avatar

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13 Jul 2008 17:15:56 IST
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@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 e271/e ~~~approx....(2.7)100.....which is way higher than ur 135.5^2


 


xperts......reply

®µD®A's Avatar

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13 Jul 2008 17:29:44 IST
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@vasanth i think you are correct. It is no where mentioned that it has to be 2 numbers. It can be more that that.
Hari Shankar's Avatar

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13 Jul 2008 19:12:55 IST
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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

Aishwarya Vardhan Chaturvedi's Avatar

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13 Jul 2008 19:23:22 IST
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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....
abhishek sinha's Avatar

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13 Jul 2008 19:25:31 IST
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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

Hari Shankar's Avatar

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13 Jul 2008 19:26:22 IST
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@mukulaish: The question has not placed a restriction on the number of real numbers that can be there.




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