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(C) Given equations are:$q = p(4-p) \dots (i)$$r = q(4-q) \dots (ii)$$p = r(4-r) \dots (iii)$Adding equations $(i)$,$(ii)$,and $(iii)$,we get:$p+q+r =

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$9$

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(C) Given equations are:$q = p(4-p) \dots (i)$$r = q(4-q) \dots (ii)$$p = r(4-r) \dots (iii)$Adding equations $(i)$,$(ii)$,and $(iii)$,we get:$p+q+r = 4(p+q+r) - (p^2+q^2+r^2)$$p^2+q^2+r^2 = 3(p+q+r)$Consider the case where $p=q=r$. Then $p = p(4-p) \Rightarrow p = 4p - p^2 \Rightarrow p^2 - 3p = 0$,which gives $p=0$ or $p=3$.If $p=q=r=0$,then $p+q+r = 0$.If $p=q=r=3$,then $p+q+r = 3+3+3 = 9$.Since $p^2+q^2+r^2 = 3(p+q+r)$,and by Cauchy-Schwarz inequality $(p+q+r)^2 \le 3(p^2+q^2+r^2)$,we have $(p+q+r)^2 \le 3(3(p+q+r)) = 9(p+q+r)$.Thus,$p+q+r \le 9$.The maximum value is $9$.

Suppose $p, q, r$ are real numbers such that $q=p(4-p)$,$r=q(4-q)$,and $p=r(4-r)$. The maximum possible value of $p+q+r$ is

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