Question

The Hotel Ventor has 400 rooms. Currently the hotel is filled. The daily rental is $ 600 per room. For every $6 increase in rent the demand for rooms decreases by 7 rooms. Let x = the number of $ 6 increases that can be made.

What should x be so as to maximize the revenue of the hotel?



I did p - 600 = (-6/7)(x-400) and got p=(-6x/7) + 942.8571429.
Then I multiplied that by x and got (-6x^2/7) + 942.8571429x.
Then I did -b/2a and got x=550.0000001, but that isn't the answer.

Can someone explain to me the right way to do this? thanks

Answer 1

Let me start fresh:
let the number of $6 increases be x

number of rooms rented = 400-7x
cost per room = 600 + 6x

Revenue = R = number of rooms x cost per room
= (400 - 7x)(600+6x)
=240000 -1800x - 42x^2

so the x of the vertex is -b/2a
= 1800/-84
= -150/7 or appr -21.4

since I defined x to be number of increases and x turned out negative, there will have to be DECREASES.

Assume that there will be multiples of $6 changes
if x = -21
number of rooms rented = 400 - (-21) = 421
BUT, they only have 400 rooms

looks like a trick question!
The cost should stay the same

Look at the graph
http://www.wolframalpha.com/input/?i=y+%3D+%28400+-+7x%29%28600%2B6x%29

Since the graph only makes sense for x≥0
the max value of the function is at x=0
namely when there is no increase or decrease in price.
My actual answer of -150/7 makes sense, notice that is where the vertex is located.

(BTW, $600 for a room ??
I will never pay that much for a room in my lifetime.)

Answer 2

Wow, thank you for such a thorough answer! =)