Ahotel finds that its revenue is given by r=8000 +760x-30x2 when it charges 80 +10x dollars for a room. what is the maximum?

I think what you might be asking is the maximum amount of money that can be made when you have that revenue existing when that amount is charged for a room.  Mathematically, what that means is that you need to see where the 2 functions intersect on a graph. That will be the x value of the number of people in a room and the corresponding y value, which is the money to be made at that number of rented rooms with that number of people per room. When you set them equal to each other, you have to then do the math on them to combine like terms and set the whole thing equal to 0 in order to complete the square on it. Completing the square is what will give you the max value, which is the vertex of the quadratic. The equation, after you complete the square, looks like this: r(x)= -30(x - 12.5)^2 + 12,607.50

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.

Hello!
your answer should be c