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Nisha Kumari

· started a discussion

· 1 Months ago

Question id-107962.0 (Question: 107962.0 )

I have done this question in a different way. My answer is different.

(1+a)(1+b)(1+c)=1+abc+(a+bc)+(b+ac)+(c+ab)=1+8+(a+8/a)+(b+8/b)+(c+8/c)

after differentiating the term a+8/a, we find that it has a minima at a=2^1/2
and the minimum value is 4*root(2)

Similarily other two expressions (b+8/b) &(c+8/c)
So the overall minima will be
9+12root(2)
what is wrong with this method?

Question:

If a + b + c = 1, then Find max value of  (1+a) (1+b) (1+c)?

Options:
A) 1.47
B) 2.87
C) 55 / 27
D) 2.37
Solution:
Ans: (d) Given,  a + b + c = 1

            

            a = \(\frac{1}{3}\), b = \(\frac{1}{3}\), c = \(\frac{1}{3}\)(for max value)

           a + b + c =1

           \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) = 1

           \(\frac{3}{3}\) = 1

            1 = 1

            put the value of a, b, c 

            (1 + a) (1 + b) (1 + c)

            (1 + \(\frac{1}{3}\) ) (1 + \(\frac{1}{3}\) ) (1 + \(\frac{1}{3}\) ) 

             \(\frac{4}{3}\) × \(\frac{4}{3}\) × \(\frac{4}{3}\)

             \(\frac{64}{27}\) = 2.37 

Knowledge Expert

· commented

· 1 Months ago

Dear Student The only thing missing in this solution is the explanation for why the equation.. (a+b)_>2(ab)^(1/2).............................................(2) The reason for this explanation is that we have to compute the value of (1+a)(1+b)(1+c) and thus we can keep one of the variables as 1 And thus for that sole convenience we have taken up this equation (or per se the operating equation as equation (2)) Keep learning with us! Thanks and regards Team TopRankers

Subhash Gangwar

· commented

· 1 Months ago

Yet to be approved!

Subhu Kumar

· commented

· 1 Months ago

First of all, it's a 3 variable function and u r doing partial differentiation. I don't know how did u get this value 4.root2. if there are 2 variable then we do partial differentiation and get p, q r, s, t. And then determine whether a minima or Maxima exist. Jyada hoshiyari mat dikhao, exam aa gya, aur tum experiment kar rha hai. SSC CGL me bas jyada practice karo, aur paas karo, dher engineering wala dimag mat lagao Chhoti.
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