As the number of rooms that will be occupied, based on the price being charged, is u(p)=p2−12p+45 with u(p)≤58 and u(p)∈I.
As such revenue r will be given by p(p2−12p+45) and this will be maximized when ddpr(p)=0, where r(p)=p3−12p2+45p and second derivative d2(dp)2r(p)<0 for maxima and d2(dp)2r(p)>0 for minima.
As ddpr(p)=3p2−24p+45 and 3p2−24p+45=0 and dividing each term by 3, we get
p2−8p+15=0 i.e. (p−5)(p−3)=0
and as d2(dp)2r(p)=6p−24=6(p−4)
while for p=5, d2(dp)2r(p)=6
for p=3, d2(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)=33−12×32+45×3=54 and at p=5, we have u(5)=53−12×52+45×5=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 p2−12p+45=58 is p2−12p−13=0 i.e. (p−13)(p+1)=0
Revenue is maximized at p=13, where it is
133−12×132+45×13=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]}
