DLS
  • DLS
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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DLS
  • DLS
@dan815 @experimentX
experimentX
  • experimentX
1, 5, 4
anonymous
  • anonymous
hmph... I was going to guess 3,4,3

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More answers

DLS
  • DLS
what are u guys saying ? :|
experimentX
  • experimentX
children
anonymous
  • anonymous
@DLS 3 children together, 4 children of Mrs. B's first marriage, 3 children of Mr. A's
DLS
  • DLS
is all this a guess? 1,5,3 3,4,3?
dan815
  • dan815
|dw:1371204653776:dw|
experimentX
  • experimentX
more children from different parent, more fight .. minimize the number of childrent hey have.
DLS
  • DLS
so 1 5 4?
dan815
  • dan815
|dw:1371204764329:dw|
DLS
  • DLS
well children cant be 0 :|
dan815
  • dan815
well the statment can say she has 0 children cant it
DLS
  • DLS
@experimentX @oldrin.bataku 154 or 901?
experimentX
  • experimentX
they can't have even number of children
DLS
  • DLS
why?
anonymous
  • anonymous
I'm going with 29 fights with Mr. A having 3 kids.
experimentX
  • experimentX
any if they get 1 more, then they have a football team with coach and referee.
DLS
  • DLS
@oldrin.bataku definitely not the right answer :P
dan815
  • dan815
i say 3 4 3 9+12+12 fights?
DLS
  • DLS
winner!
dan815
  • dan815
yay :D
DLS
  • DLS
why did we go with 343?
experimentX
  • experimentX
woops!! looks like i missed the fact that .. A's first children can fight with his latter children.
dan815
  • dan815
because we gotta maximisee fights as we are trying to raise rocky
anonymous
  • 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
  • DLS
why was 154 incorrect?
experimentX
  • experimentX
I told you i thought ... the wouldn't fight with the younger children.
dan815
  • 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
  • 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
  • experimentX
x is close to 3
experimentX
  • experimentX
i thought the older kids with different parent would find each other ... lol should have thought new born have different parentage.
dan815
  • dan815
you can maximize setting constaint as integers only too on wolfram i think
anonymous
  • anonymous
That's actually exactly what I did :-p it was just an integer optimiation problem.
anonymous
  • anonymous
@dan815 I think with Mathematica but I don't know about vanilla WolframAlpha.
anonymous
  • anonymous
I optimizes the wrong polynomial, though; I ended up with \(-3x^2+19x-1\).
DLS
  • DLS
|dw:1371205372374:dw|
DLS
  • DLS
is it so?
DLS
  • DLS
235 should work too..:/
anonymous
  • 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
  • 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
  • dan815
^
DLS
  • DLS
okay :| TY :)
anonymous
  • 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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