Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $a, b, c, d, e$ and $f$ are non negative real numbers such that $a + b + c + d + e + f = 1$, then what is the maximum value of $ab + bc + cd + de + ef$?

share|improve this question
    
I'm pretty sure this should be $\frac{1}{4}$, similarly to simply maximizing $ab$ subject to the same constraint. –  gnometorule Jan 17 '13 at 4:46
    
@gnometorule agreed. –  Rustyn Jan 17 '13 at 4:48
    
Yes the Answer is 1/4..still me a bit confused –  user58452 Jan 17 '13 at 5:26
1  
My Approach : a + c + e + b + d + f = 1 So maximum value of (a + c + e) . (b + d + f) is (1/2).(1/2) i.e. (ab + bc + cd + de + ef) + (ad + af + cf + be) = (1/4) and now i am stuck...!! –  user58452 Jan 17 '13 at 5:28
    
As you said, $(a+c+e)(b+d+f) \le \frac{1}{4},$ from which $(ab + bc + cd + de + ef) + (ad + af + cf + be) \le \frac{1}{4}.$ Since all of our variables are non-negative, the second expression is also non-negative, which means that $(ab + bc + cd + de + ef) \le \frac{1}{4}$ as well. The maximum occurs when the second expression is $0$ and when $a+c+e = b+d+f,$ which is easy to achieve. –  lyj Jan 17 '13 at 5:34

3 Answers 3

I'm pretty sure the answer is $1/4$, but I can't prove it rigorously...

We want to multiply the two largest numbers we can together, so simply letting $a=1/2$ and $b=1/2$ will do the trick.

In fact if we let any of $b, c, d, e$ equal $1/2$, and let the number before and after it add to $1/2$, we will also get $1/4$.

So for example we can let $e=1/2$, and also let $d=1/8$ and $f=3/8$, we will get a maximum value of $1/4$.

share|improve this answer

Note that lyj's comment pretty much answers this question. On top of that, we can achieve this maximum by taking $(a, b, c, d, e, f) = (0, 0, 11/32, 1/2, 5/32, 0)$.

I.e., We finish showing the following.

  1. The quantity in question has an upper bound $1/4$ (by lyj's argument).

  2. The upper bound $1/4$ can be achieved.

share|improve this answer

Here's the full solution in case the comments weren't enough.

Note that $(a+c+e)(b+d+f) = (a+c+e)(1-(a+c+e)) \le \frac{1}{4},$ with equality iff $a + c + e = \frac{1}{2}.$ Expanding the first expression above gives $(ab+bc+cd+de+ef)+(ad+af+be+cf) \le \frac{1}{4}.$ Since all of our variables are non-negative, $ad + af + be + cf \ge 0,$ or $-(ad+af+be+cf) \le 0,$ which gives $ab+bc+cd+de+ef \le \frac{1}{4} - (ad+af+be+cf) \le \frac{1}{4}.$ To achieve equality, we need the following:

1) $a + c + e = \frac{1}{2},$ our original condition.

2) $\frac{1}{4} - (ad + af + be + cf) = \frac{1}{4},\, \textrm{ i.e. } ad + af + be + cf = 0.$

So we have shown that the upper bound is $\frac{1}{4},$ and that we can achieve this upper bound. For example, let $a = \frac{1}{2} = b$ and $c = d = e = f = 0.$ Another example is $a = 0,\, b = \frac{1}{3},\, c = \frac{1}{2},\, d = \frac{1}{6}, e = 0,\, f = 0.$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.