Question

If log [a] + log [b] + log [c] = 0, then maximum and minimum values of a+b+c

Answer 1

Given Log a + Log b + Log c = 0
Log (a b c) = 0
So a ×b× c = 1
c = 1 /ab

a, b and c are real positive numbers.
Let Z = a + b + c = a + b + 1/ab

Differentiate partially wrt a and b keeping the other variable as constant . Equate them to 0 to find minimized Z.

dZ/da = 1 - 1/ba^2 = 0
So b × a^2 = 1. .... (1)

dZ/ db = 1 - 1/ab^2 = 0
So a×b^2 = 1. .... (2)

Solving these equation s we get
a = b = 1 and so c = 1.

Second partial derivative of Z
d^2 Z/da^2 = 2/ba^3 > 0
d^2 Z/db^2 = 2/ab^3 > 0

Hence Z is minimum when a=b=c=1.
Minimum sum = 3.

Answer 2

Given :

log( a ) + log ( b ) + log ( c ) = 0

By using : log(m) + log(n) = log ( m × n )

log ( a × b × c ) = log 1 [ log ( 1 ) = 0 ]

Removing log :

a b c = 1

Now :

Maximum value

a = b = c = 1

Only then a + b + c = 1 + 1 + 1 = 3

          a + b + c can be maximum only when a = b = c = 1

Maximum value is 3

Minimum value

Imagine a to be  - 1

b = -1

c = 1

Here a + b + c = - 1 - 1 + 1

                       = - 1

Minimum value can be - 1

Hope it helps :)

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