Question

If `a+b+c+d+e+f=12` then the greatest value of `ab+bc+cd+de+ef+fa` will be :- (A) 24,(B)36,(C) 30,(D) none of these,provided `a,b,c,d,e and f` are non negative real number.?

Answer 1

(B) 36

Explanation:

The simplest method seems to be to use Lagrange multipliers (sorry!)...

Let:

`{ (a = A^2), (b = B^2), (c = C^2), (d = D^2), (e = E^2), (f = F^2) :}`

where `A, B, C, D, E, F` are real numbers. This takes care of the condition that `a, b, c, d, e, f` be non-negative.

Then we want to maximise:

`f(A, B, C, D, E, F) = A^2B^2+B^2C^2+C^2D^2+D^2E^2+E^2F^2+F^2A^2`

subject to:

`g(A, B, C, D, E, F) = A^2+B^2+C^2+D^2+E^2+F^2 - 12 = 0`

Define the Lagrangian:

`L(A, B, C, D, E, F, lambda)`

`= f(A, B, C, D, E, F) - lambda g(A, B, C, D, E, F)`

`= A^2B^2+B^2C^2+C^2D^2+D^2E^2+E^2F^2+F^2A^2-lambda(A^2+B^2+C^2+D^2+E^2+F^2 - 12)`

Next we want to solve:

`grad L(A, B, C, D, E, F, lambda) = bb(0)`

This is, all of the partial derivatives with respect to `A, B, C, D, E, F, lambda` must be zero.

`0 = (del L)/(del A) = 2A(B^2+F^2-lambda)`

`0 = (del L)/(del B) = 2B(C^2+A^2-lambda)`

`0 = (del L)/(del C) = 2C(D^2+B^2-lambda)`

`0 = (del L)/(del D) = 2D(E^2+C^2-lambda)`

`0 = (del L)/(del E) = 2E(F^2+D^2-lambda)`

`0 = (del L)/(del F) = 2F(A^2+E^2-lambda)`

`0 = (del L)/(del lambda) = -(A^2+B^2+C^2+D^2+E^2+F^2 - 12)`

Note that in order for `f(...) > 0` we must have at least two adjacent non-zero values in `A, B, C, D, E, F`.

The only possible solutions seem to be (cyclic variants of):

`color(white)()`
Case 1 - `A, B != 0` and `C=D=E=F=0`

Then `A^2=B^2=lambda=12/2 = 6` and `f(...) = 36`

`color(white)()`
Case 2 - `A, B, C != 0` and `D=E=F=0`

Then `A^2+C^2 = B^2 = lambda`. Hence `lambda = 6`, and

`f(...) = 18+18=36`

`color(white)()`
Case 3 - `A, B, C, D, E, F != 0`

Then:

`lambda = B^2+F^2 = C^2+A^2=D^2+B^2=E^2+C^2=F^2+D^2=A^2+E^2`

Hence:

`A^2=B^2=C^2=D^2=E^2=F^2=2`

and `f(...) = 4+4+4+4+4+4=24`