As the number of rooms that will be occupied, based on the price being charged, is u(p) = p^2 - 12p + 45 with u(p)<=58 and u(p)inI.
As such revenue r will be given by p(p^2-12p+45) and this will be maximized when d/(dp)r(p)=0, where r(p)=p^3-12p^2+45p and second derivative d^2/(dp)^2r(p)<0 for maxima and d^2/(dp)^2r(p)>0 for minima.
As d/(dp)r(p)=3p^2-24p+45 and 3p^2-24p+45=0 and dividing each term by 3, we get
p^2-8p+15=0 i.e. (p-5)(p-3)=0
and as d^2/(dp)^2r(p)=6p-24=6(p-4)
while for p=5, d^2/(dp)^2r(p)=6
for p=3, d^2/(dp)^2r(p)=-6
Hence, we have a local maxima at p=3 and a local minima at x=5
At p=3, we have r(3)=3^3-12xx3^2+45xx3=54 and at p=5, we have u(5)=5^3-12xx5^2+45xx5=50, but the latter is a local maxima and as subsequently r(p) continues to rise and is limited only by u(p)<=58.
And at u(p)=58 and p^2-12p+45=58 is p^2-12p-13=0 i.e. (p-13)(p+1)=0
Revenue is maximized at p=13, where it is
13^3-12xx13^2+45xx13=2197-2028+585=754, when occupancy is 58.
However, as even beyond p=13 i.e. for p>13, demand for rooms continues to increase. Hence, answer is that there is no limit post the price p>13,
graph{x^3-12x^2+45x [-5, 15, -50, 800]}
