IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

Sushmi

In how many parts can 271 be divided so that their product is maximum?

a.) 100 b.) 90

c.) 10 d.) none of these

spideyunlimited

Answer is 100.

If 271 is divided into n parts, each part will be 271/n
and their product will be (271/n)ⁿ

Differentiate this (and then equate to zero) :

[ nⁿ(271ⁿ .log 271) - 271ⁿnⁿ(1 + log n) ] / n²ⁿ

= 0

[ nⁿ(271ⁿ .log 271) - 271ⁿnⁿ(1 + log n) ] = 0

271ⁿnⁿ( log 271 - (1 + log n) ) = 0

log 271 - (1 + log n) = 0 ....................Since 271ⁿnⁿcan never be 0.

log 271 - 1 = log n

log 271 - log e = log n

log (271/e) = log n

log (271 / 2.71) = log n

log 100 = log n

So n= 100

Sushmi

Yes good solution could be done in a little shorter way

x1 + x2 + x3 + ...+ xn = 271

x1.x2.x3.x4.....xn = Product = P

P will be max when x1 = x2 = x3 =...... xn

So P = (271/n) ^ n

ln P = n (ln 271 - ln n)

Differentiating,

1/P dP/dn = ln 271 - ln n - 1

or ln (271/en) = 0

271 = en , so n = 100 for prod to be max.

sandeepramesh

this question was asked by my friend a few days ago :)

for any k, the general answer is [k/e] where k = the number to be divided, e = Napiers number and [.] = GIF :)

Ask & Discuss Questions with Community & Experts