Question

John has x children by first wife.mary has( x +1) children by first husband..they marry and have children of?
their own .the whole family has 24 children.assume that children of same parents don't fight.prove that maximum possible fight are​

Answer 1

Answer:

40 percentage of fighting.

Answer 2

the absolute maximum fights for an integer number of children from each marriage will be 191.

Explanation:

Total number of children that John & Marry have together is $24 - 2x - 1 = 23 - 2x$

Let the number N of pairs who fight is $(x)(x+1) + (x)(23 - 2x) + (x+1)(23 - 2x)$

$x^{2} + x + (2x+1)(23 - 2x)$

$N = x^{2} + x - 4x^{2} + 44x + 23 = 23 + 45x - 3x^{2}$

Differentiating with respect to $x$, $\frac{\mathrm{d}N}{\mathrm{d}x}= 45 - 3\times2x$

This can be zero for maximum fights, then  

$15 = 2x$ ⇒  $x = \frac{15}{2}$

x for max N is between 7 and 8.

and note that $\frac{\mathrm{d}^{2}N}{\mathrm{d}x^{2}} = - 6$

which is negative.

Hence, the whole numbers closest to the max are 7 and 8.

The number N of fights for$x = 7$ and $x = 8$ are

$N_{7} = 23 + 45\times7 - 3\times49 = 191$

And $N_{8} = 23 + 45\times8 - 3\times64 = 191$

so this is the absolute maximum fights for an integer number of children from each marriage will be 191.