word problem help

Question

The Hotel Plenrich has 650 rooms. Currently the hotel is filled . The daily rental is $ 600 per room. 
For every $ 8 increase in rent the demand for rooms decreases by 8 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? $  
(Hint: see problem 32 on page 127. This problem has a similar setup.)

Answer 1

The daily rental for a room is 600 + 8x

Where 8x is $ 8 increase in rent for x rooms

The number of occupied rooms will be 650 - 8x

Total revenue (R) = number of rooms occupied * rent for each room

R = (600 + 8x)(650 - 8x)

R = 422500 + 650*8x -600*8x -64 x²

R = 422500 -64 x² +50*8x

R =  -64 x² + 400x + 422500

To maximize the revenue we must find its first derivative 

R' = -64*2 x + 400

R' =-128 x + 400

Now solve for x

-128 x + 400 = 0

128x = 400

x = 400/128

x = 3.125

Since x must be an integer we take the greatest value x = 4

(a)

x should  be 4 so as to maximize the revenue of the hotel 

x = 4

(b)

 The rent per room when the revenue is maximized is 600 + 8x

                                                                                              = 600 + 8*4

 Rent per room  = 632 $

(c)     

The maximum revenue is   number of rooms occupied * rent for each room

= 632* (650-32)

=632*618

=390576

The maximum revenue is  390576 $