help me please

Question

Answer 1

The rent for each room is 350 + 20x

Where 20x is rental incerment for x rooms is $20

The number of occupied rooms will be 550 - 4x

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

R = (550 - 4x) * (350 + 20x)

R = 192500 - 350*4x + 550*20x - 80x²

R = 192500 - 64 x² +50*8x

R =  - 80x² + 9600x + 192500

To maximize the revenue we must find its first derivative and solve for zero.

R' = -80*2 x + 9600

R' =-160 x + 9600

Now solve for x

-160 x + 9600 = 0

160x = 9600

x = 9600/160

x = 60.

(a)

x should be 60 to get maximum revenue of the hotel 

x = 60

(b)

 The rent per room when the revenue is maximized is 350 + 20x = 350 + 20*60

 Rent per room  = $1550.

(c)     

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

= 1550* (550 - 4*60)

=1550*310

= 480,500

The maximum revenue is  $ 480,500.