Demand for rooms, of a hotel which has 58 rooms, is a function of price charged given by u(p)=p^2-12p+45. Find out at what price the revenue is maximized and what is the revenue?

1 Answer
Jan 26, 2017

There is no limit to revenue for p>13, but demand cannot be fulfilled beyond 58 rooms, which is for p=13 and maximum revenue at this level is 754.

Explanation:

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]}