Q.140. Maximum Value

Question 140.
What is the maximum value of ab + bc + ca, if a, b, and c satisfy the relations
a^2 + ab + b^2 = 3 and b^2 + bc + c^2 = 16?
 

[ This question was asked in Yahoo Answers! by Zo Maar whose monumental high percentage of the best answers to the challenging questions asked in Yahoo Answers! is a proof of his expertise in the subject. He also posts very challenging questions which I find very difficult to answer.

 

Dragan K, whose contribution in Yahoo Answers! is praiseworthy, posted an elegant solution of the above question which was obviously adjudged as the Best Answer. I liked this solution so much that I sought his permission to let me post it here on my blog where I post only my best answers given in YA! I felt that the readers of my blog will enjoy reading this unique solution of this challenging question. 

For the benefit of the average students of mathematics, I have added some intermediate steps for clarity of understanding. ]

Answer 140.

Refer to the  figure as above. O is a point in ΔABC to which each side of the triangle subtends 120°.

Let OA = a, OB = b and OC = c.

Using Cosine Rule for Δ OAB,

cos120° = - 1/2 = (a^2 + b^2 - AB^2) / 2ab

=> AB^2 = a^2 + ab + b^2

=> AB = √p = √(a^2 + ab + b^2)

Similarly,

BC = √q = √(b^2 + bc + c^2).

Area of

(ΔOAB + ΔOBC + ΔOCA) = area of ΔABC

=> (1/2) √(pq) sinB = (1/2) (ab + bc +ca) sin120°

=> ab + bc + ca = 2√(pq/3) sinB

ab +bc + ca will have the maximum value when sinB = 1, i.e., when B = π/2 and

the maximum value of ab + bc + ca = 2√(pq/3).

For the given problem, p = 3 and q = 16

=> maximum value of ab + bc +ca = 2√(3*16)/3 = 8.

[ It can be shown that a = 7/√31, b = 4/√31 and c = 20/√31. ]

LINK to YA!