As the number of rooms that will be occupied, based on the price being charged, is u(p) = p^2 - 12p + 45u(p)=p2−12p+45 with u(p)<=58u(p)≤58 and u(p)inIu(p)∈I.
As such revenue rr will be given by p(p^2-12p+45)p(p2−12p+45) and this will be maximized when d/(dp)r(p)=0ddpr(p)=0, where r(p)=p^3-12p^2+45pr(p)=p3−12p2+45p and second derivative d^2/(dp)^2r(p)<0d2(dp)2r(p)<0 for maxima and d^2/(dp)^2r(p)>0d2(dp)2r(p)>0 for minima.
As d/(dp)r(p)=3p^2-24p+45ddpr(p)=3p2−24p+45 and 3p^2-24p+45=03p2−24p+45=0 and dividing each term by 33, we get
p^2-8p+15=0p2−8p+15=0 i.e. (p-5)(p-3)=0(p−5)(p−3)=0
and as d^2/(dp)^2r(p)=6p-24=6(p-4)d2(dp)2r(p)=6p−24=6(p−4)
while for p=5p=5, d^2/(dp)^2r(p)=6d2(dp)2r(p)=6
for p=3p=3, d^2/(dp)^2r(p)=-6d2(dp)2r(p)=−6
Hence, we have a local maxima at p=3p=3 and a local minima at x=5x=5
At p=3p=3, we have r(3)=3^3-12xx3^2+45xx3=54r(3)=33−12×32+45×3=54 and at p=5p=5, we have u(5)=5^3-12xx5^2+45xx5=50u(5)=53−12×52+45×5=50, but the latter is a local maxima and as subsequently r(p)r(p) continues to rise and is limited only by u(p)<=58u(p)≤58.
And at u(p)=58u(p)=58 and p^2-12p+45=58p2−12p+45=58 is p^2-12p-13=0p2−12p−13=0 i.e. (p-13)(p+1)=0(p−13)(p+1)=0
Revenue is maximized at p=13p=13, where it is
13^3-12xx13^2+45xx13=2197-2028+585=754133−12×132+45×13=2197−2028+585=754, when occupancy is 5858.
However, as even beyond p=13p=13 i.e. for p>13p>13, demand for rooms continues to increase. Hence, answer is that there is no limit post the price p>13p>13,
graph{x^3-12x^2+45x [-5, 15, -50, 800]}
