Question

The Hotel Bellville has 700 rooms. Currently the hotel is filled . The daily rental is $ 500 per room.
For every $ 8 increase in rent the demand for rooms decreases by 7 rooms.
Let x = the number of $ 8 increases that can be made.
What should x be so as to maximize the revenue of the hotel ?
What is the rent per room when the revenue is maximized? $__
What is the maximum revenue? $__

Answer 1

if \(x\) is the number of $8 increases, then the revenue will be \((500+8x)(700-7x)\)

Answer 2

hard so imagine anyone who spends $500 a night would care about an $8 increase

Answer 3

in any case, multiply that out and find the vertex. that is what you are looking for

Answer 4

will that equal my max revenue?

Answer 5

the first coordinate will be the number of $8 increases you should have
the second coordinate will be the max revenue

Answer 6

i don't quite understand..

Answer 7

WAIT i got it lol. then what would be the rate per room when the revenue is max?

Answer 8

satellite73...?

Answer 9

hat would be the rate per room when the revenue is max?

Answer 10

Okay, so you multiply that out: (500+8x)(700−7x) = 350000 + 2100x - 56x^2

And you know the vertex is at -b/2a = -2100/-112 = 18.75

So, after 18.75 $8 increases, the revenue is maximized. 500 + 18.75*8 = $650.

Answer 11

\(x\) is the number of $8 increases

Answer 12

the cost will be \(500+8x\) and the revenue will be \((500+8x)(700-7x)\)

Answer 13

thank youuuuuuuu

Answer 14

yw, from both of us