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]}