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

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Retired Moderator
Status: Preparing for GMAT
Joined: 25 Nov 2015
Posts: 1041
Location: India
GPA: 3.64
In a certain theatre, there are 300 seats. When the price of a  [#permalink]

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09 Jan 2018, 10:34
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Difficulty:

45% (medium)

Question Stats:

76% (03:37) correct 24% (03:48) wrong based on 64 sessions

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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
Intern
Joined: 24 Nov 2016
Posts: 30
Re: In a certain theatre, there are 300 seats. When the price of a  [#permalink]

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10 Jan 2018, 04:42
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
##### General Discussion
Retired Moderator
Joined: 25 Feb 2013
Posts: 1124
Location: India
GPA: 3.82
In a certain theatre, there are 300 seats. When the price of a  [#permalink]

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09 Jan 2018, 11:31
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
VP
Joined: 07 Dec 2014
Posts: 1252
Re: In a certain theatre, there are 300 seats. When the price of a  [#permalink]

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09 Jan 2018, 19:43
2
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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10 Jan 2018, 07:34
1
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 [ 7.85 KiB | Viewed 967 times ]

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

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16 Jan 2018, 16:45
1
1
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

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Re: In a certain theatre, there are 300 seats. When the price of a   [#permalink] 16 Jan 2018, 16:45