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
(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
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
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
The only possible solutions seem to be (cyclic variants of):
Case 1 -
Then
Case 2 -
Then
f(...) = 18+18=36
Case 3 -
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