Question

A hotel rents 240 240 rooms at a rate of $ 40 $40 per day. For each $ 2 $2 increase in the rate, four fewer rooms are rented. Find the room rate that maximizes daily revenue. The rate that maximizes revenue is $

Answer 1

The daily rate that maximizes the revenue is $40 and the maximum revenue is $9600

Explanation:

Given

`Rooms = 240`

`Price= \$40`

`Increment = \$2`

First, we calculate the revenue.

`Revenue (r) = Price (p) * Number\ of\ Rooms(n)`

`r = p*n`

Let x be the change in rate.

i.e.

For every 2x increment in price (p) i.e. p + 2x

There will be 4 fewer rooms (n) i.e. n - 4x

So, the revenue is updated as:

`r = (p + 2x) * (n - 4x)`

`Price = 40`

So:

`r = (40 + 2x) * (n - 4x)`

`Rooms = 240`

So:

`r = (40 + 2x) * (240 - 4x)`

Expand

`r = 40 * 240 - 40 * 4x + 2x * 240 - 2x*4x`

`r = 9600 - 160x + 480x - 8x^2`

Reorder

`r = -8x^2 - 160x + 480x + 9600`

`r = -8x^2 +320x + 9600`

The maximum of a function is calculated as:

`Max = -(b)/(2a)`

In:
`r = -8x^2 +320x + 9600`

`a = -8`
`b = 320`
`c = 9600`

So:

`Max = -(b)/(2a)`

`Max = -(320)/(-8)`

`Max = (320)/(8)`

`Max = 40`

To get the maximum revenue;

Substitute 40 for x in
`r = -8x^2 +320x + 9600`

`r = -8(40)^2 + 320 * 40 + 9600`

`r = 9600`

The daily rate that maximizes the revenue is $40 and the maximum revenue is $9600