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

Question

In a certain theater, there are 300 seats. When the price of a ticket was $60, the theater ran to a full house. For every $3 increase in the price of the ticket, the strength of the audience dropped by 10. The maximum possible revenue that the theater owner will earn from the sale of the tickets is
A) $18600
B) $18750
C) $19050
D) $19350
E) $19600

JouviMat · Accepted Answer

Price = 60 + 3x
Seats = 300 - 10x
Revenue = Price * Seats = (60+3x)(300-10x)

= 18000 - 600x + 900x -30x^2
= -30x^2 + 300x + 18000

This is a down parabola equation so the maxium = -b/2a
= -300/-60 = 5

(60+3(5)) * (300 - 10(5)) = 75 * 350 = 18750

niks18 · Answer

The question is similar to the one below, with slight change in wordings - https://gmatclub.com/forum/in-an-opera- ... 07959.html

gracie · Answer

let x=number of tickets sold
(60+3x)(300-10x)>  
60x>270
x>4.5
x=5
(60+3*5)(300-10*5)=$18750
B

Abhishek009 · Answer

Revenue.PNG Answer must be (B)

ScottTargetTestPrep · Answer

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

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

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