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.?

1 Answer
Sep 5, 2017

(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#