If composite function $f_1\left(f_2\right(f_3(....(\left(f_n\right(x\left)\right))n$ times is an increasing function and if r of $f_1’$ s are decreasing
function while rest are increasing, then the maximum value of $r(n–r)$ is

_{Let g(x)=}$f_1\left(f_2\right(f_3(....(\left(f_n\right(x\left)\right))n$
_{g(x) is a decreasing function.
Hence its derivative will be negative. 
Now, 
g′(x)=}$g\prime \left(x\right)=f_1\prime (f_2(f_3(...f_n(x\left)\right)...)\times f_2\prime (f_3(...f_n(x\left)\right)...)\times ...f{\text{'}}_{n-1}(f\left(x\right))\times f{\text{'}}_n(x)$ 
<sub>Since g′(x)<0 there must be odd number of negative terms out of the n terms in g'(x). 
It is given that r functions are decreasing. Hence, r must be odd. 

Now, we wish to maximise r(n−r)
Let p(r)=</sub> $r n-r^2$<sub>, for 0<r≤n
⇒p′(r)=n−2r  
⇒p′′(r)=−2
Hence maximum value of p(r) occurs at r=2n​
In our case r is an odd integer. Hence, let us consider different cases. 

Case I - n is an even integer. 
For p(r) to be maximum , r=2n​
For r to be an odd integer, we will have to consider sub cases.
 
Case I - (a)  n=4m+2 , where m is an integer
In this case, r=2n​ will be an odd integer. 
Hence, p(r)=</sub>$\frac{n^2}{4}$ 

_{Case I - (b)  n=4m , where m is an integer
In this case, 2n​ will not be an odd integer. 
Hence let us consider the integer before 2n​ and after it. 
i.e.} $r=\frac{n}{2}-1 or r=\frac{n}{2}+1$
_{In both the cases, maximum value of p(r)=}$\frac{n^2-4}{4}$_​. 

Case II - n is an odd number 
In this case let us consider the integers between which 2n​ lies. 
i.e. $\frac{n-1}{2}$ and $\frac{n+1}{2}$
In both the cases, the maximum value of p(r)=$\frac{n^2-1}{4}$