In a certain theatre, there are 300 seats. When the price of a

We can let n = the number of $3 increases in ticket price (or the number of 10-seat decreases in audience). Recall that the total revenue, R, is the product of the number of seats and ticket price. Thus, R, in terms of n, is:

R = (300 - 10n)(60 + 3n)

R = 18000 + 900n - 600n - 30n^2

R = -30n^2 + 300n + 18000

Notice that R is a quadratic function in terms of n, and its graph is a downward-opening parabola with maximum value at the vertex. To find the value of n that maximizes the revenue, we use
the formula n = -b/(2a). Thus, when n = -300/(2(-30)) = -300/-60 = 5, R will be maximized. Finally, the maximum revenue (when n = 5) is:

R = (300 - 10(5))(60 + 3(5))

R = 250 x 75

R = 18750 dollars

Answer: B

Answer must be (B)

The question is similar to the one below, with slight change in wordings -

https://gmatclub.com/forum/in-an-opera- ... 07959.html

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