Answer:
The daily rate that maximizes the revenue is $40 and the maximum revenue is $9600
Step-by-step explanation:
Given
[tex]Rooms = 240[/tex]
[tex]Price= \$40[/tex]
[tex]Increment = \$2[/tex]
First, we calculate the revenue.
[tex]Revenue (r) = Price (p) * Number\ of\ Rooms(n)[/tex]
[tex]r = p*n[/tex]
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:
[tex]r = (p + 2x) * (n - 4x)[/tex]
[tex]Price = 40[/tex]
So:
[tex]r = (40 + 2x) * (n - 4x)[/tex]
[tex]Rooms = 240[/tex]
So:
[tex]r = (40 + 2x) * (240 - 4x)[/tex]
Expand
[tex]r = 40 * 240 - 40 * 4x + 2x * 240 - 2x*4x[/tex]
[tex]r = 9600 - 160x + 480x - 8x^2[/tex]
Reorder
[tex]r = -8x^2 - 160x + 480x + 9600[/tex]
[tex]r = -8x^2 +320x + 9600[/tex]
The maximum of a function is calculated as:
[tex]Max = -\frac{b}{2a}[/tex]
In: [tex]r = -8x^2 +320x + 9600[/tex]
[tex]a = -8[/tex] [tex]b = 320[/tex] [tex]c = 9600[/tex]
So:
[tex]Max = -\frac{b}{2a}[/tex]
[tex]Max = -\frac{320}{-8}[/tex]
[tex]Max = \frac{320}{8}[/tex]
[tex]Max = 40[/tex]
To get the maximum revenue;
Substitute 40 for x in [tex]r = -8x^2 +320x + 9600[/tex]
[tex]r = -8(40)^2 + 320 * 40 + 9600[/tex]
[tex]r = 9600[/tex]
The daily rate that maximizes the revenue is $40 and the maximum revenue is $9600
