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

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In a certain theatre, there are 300 seats. When the price of a  [#permalink]

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New post 09 Jan 2018, 11:34
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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

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Re: In a certain theatre, there are 300 seats. When the price of a  [#permalink]

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New post 10 Jan 2018, 05:42
5
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
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In a certain theatre, there are 300 seats. When the price of a  [#permalink]

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New post 09 Jan 2018, 12:31
souvonik2k wrote:
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


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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Re: In a certain theatre, there are 300 seats. When the price of a  [#permalink]

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New post 09 Jan 2018, 20:43
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souvonik2k wrote:
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


let x=number of tickets sold
(60+3x)(300-10x)>[60+3(x+1)][300-10(x+1)]
60x>270
x>4.5
x=5
(60+3*5)(300-10*5)=$18750
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Re: In a certain theatre, there are 300 seats. When the price of a  [#permalink]

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New post 10 Jan 2018, 08:34
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souvonik2k wrote:
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
Attachment:
Revenue.PNG
Revenue.PNG [ 7.85 KiB | Viewed 599 times ]

Answer must be (B)
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Re: In a certain theatre, there are 300 seats. When the price of a  [#permalink]

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New post 16 Jan 2018, 17:45
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souvonik2k wrote:
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


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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