Mr. A has x children by his first wife and
Mrs.B has (x + 1) children by her first
husband. They marry & have children of their
own. Now there are total ten children.
Assuming that children of same parents don’t
fight, maximum no. of fights among children?

- DLS

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- DLS

@dan815 @experimentX

- experimentX

1, 5, 4

- anonymous

hmph... I was going to guess 3,4,3

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## More answers

- DLS

what are u guys saying ? :|

- experimentX

children

- anonymous

@DLS 3 children together, 4 children of Mrs. B's first marriage, 3 children of Mr. A's

- DLS

is all this a guess?
1,5,3
3,4,3?

- dan815

|dw:1371204653776:dw|

- experimentX

more children from different parent, more fight .. minimize the number of childrent hey have.

- DLS

so 1 5 4?

- dan815

|dw:1371204764329:dw|

- DLS

well children cant be 0 :|

- dan815

well the statment can say she has 0 children cant it

- DLS

@experimentX @oldrin.bataku
154 or 901?

- experimentX

they can't have even number of children

- DLS

why?

- anonymous

I'm going with 29 fights with Mr. A having 3 kids.

- experimentX

any if they get 1 more, then they have a football team with coach and referee.

- DLS

@oldrin.bataku definitely not the right answer :P

- dan815

i say
3 4 3
9+12+12 fights?

- DLS

winner!

- dan815

yay :D

- DLS

why did we go with 343?

- experimentX

woops!! looks like i missed the fact that .. A's first children can fight with his latter children.

- dan815

because we gotta maximisee fights as we are trying to raise rocky

- anonymous

3,4,3 just seemed logical... you want to minimize the number of kids in each family while also striving to maximize the number of kids in each. 3,4,3 just balanced.

- DLS

why was 154 incorrect?

- experimentX

I told you i thought ... the wouldn't fight with the younger children.

- dan815

man can u imagine being that 1 guy in 1 , 5,4 lol that one kid is gonna be some fighter when he grows up

- experimentX

http://www.wolframalpha.com/input/?i=Maximize+x%28x%2B1%29+%2B+x%2810-2x+-+1%29+%2B+%28x%2B1%29%2810+-+2x+-+1%29

- experimentX

x is close to 3

- experimentX

i thought the older kids with different parent would find each other ... lol should have thought new born have different parentage.

- dan815

you can maximize setting constaint as integers only too on wolfram i think

- anonymous

That's actually exactly what I did :-p it was just an integer optimiation problem.

- anonymous

@dan815 I think with Mathematica but I don't know about vanilla WolframAlpha.

- anonymous

I optimizes the wrong polynomial, though; I ended up with \(-3x^2+19x-1\).

- DLS

|dw:1371205372374:dw|

- DLS

is it so?

- DLS

235 should work too..:/

- anonymous

Ah! Drats. I didn't see the \(-2x,10\) terms when simplifying hence the \(19x-1\) rather than \(17x+9\)!
$$F(x)=x(x+1)+x(10-(2x+1))+(x+1)(10-(2x+1))\\\ \ \ \ \ \ \ =x^2+x+10x-2x^2-x+10x-2x^2-x+10-2x-1\\\ \ \ \ \ \ \ =-3x^2+17x+9$$From there's it's merely an integer optimization problem; note the maximum occurs nearest to \(x=3\) giving us \(-3(3)^2+17(3)+9=-27+51+9=60-27=33\)

- anonymous

3,4,3 was obvious though since you want each to be as near in size as the others so that you're getting the biggest bang for your buck essentially... it just seemed intuitively optimal

- dan815

^

- DLS

okay :|
TY :)

- anonymous

That polynomial we derived pretty easily, too. We have three groups of kids of sizes x, x+1, and 10-(x+x+1)=10-(2x+1) respectively. We know that kids of different groups are all capable of fighting, and since we're interested in maximizing the number of possible fights, we consider specifically that case:
if all x kids from the first group fight with the x+1 kids in the second, we have x(x+1) fights
if all x kids from the first group fight with the 10-(2x+1) kids in the third, we have x(10-(2x+1)) fights
if all x+1 kids in the second group fight with the 10-(2x+1) kids in the third, we have (x+1)(10-(2x+1)) fights
so the total number of fights is just their sum. Simplifying it we can that nice polynomial in x above, whose maximum is nearest to the integer x=3

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