A large hotel finds that it can rent 95 rooms when it charge $100 per room. For each $1 increase in room cost it rents one room less. (Likewise, for each $1 decrease in room cost it rents one room more) What price should it charge in order to maximize the revenue?

Well, revenue = total money taken in which = (price)(per room)

Let x = price
Revenue = x(190 - x) = -x^2 + 190x
x = -190 / 1(-1) = -190 / -1 = $190

See this is the answer Im coming up with but It doesnt look right, please help!

You defined x as the price, but did not use it as such

I will re-define x as the number of $1 increases

new price = (100+x)
number of rooms rented = 95-x

Revenue = R(x) = (100+x)(95-x)
= 9500 - 5x - x^2
R ' (x) = -5 - 2x
= 0 for a max of R(x)
2x=-5
x = -2.5
I will assume you want x to be a whole number, since you couldn't rent 92.5 rooms
so x = -3 or x = -2

if x = -3, rent = 97, number rented = 98
revenue = 97x98 = $9506
if x = -2, rent = 98, number rented = 97
revenue = 98x97 = $9506

So they should charge either $97$ or $98

To find the price that maximizes revenue, we need to find the vertex of the quadratic equation. The equation you have is correct: Revenue = -x^2 + 190x.

The vertex of a quadratic equation in the form of ax^2 + bx + c is given by the x-coordinate of the vertex: x = -b / (2a).

In our case, a = -1, and b = 190. Substituting these values, we get:

x = -190 / (2*(-1)) = -190 / (-2) = 95

So the price that maximizes revenue is $95.

To verify this, we can substitute x = 95 into the revenue equation:

Revenue = -x^2 + 190x
Revenue = -(95^2) + 190(95)
Revenue = -9025 + 18050
Revenue = 9025

Therefore, the maximum revenue is $9025 when the price is $95 per room.

It is important to note that while the price per room is $95, the hotel will not actually be able to rent 95 rooms. The maximum number of rooms that can be rented is given by the equation 190 - x, which in this case would be 190 - 95 = 95 rooms.