In an opera theater, there are 300 seats available.

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In an opera theater, there are 300 seats available. When all the seats in the theater are sold out, the price of each ticket is $60. For every $1 increase in the price of the ticket, the number of seats sold decreases by 2.

How much should the theater owner charge for each seat to make the maximum profit?

A)87 $

b)95 $

c)105 $

d)120 $

e)127 $

oa

c
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by GMATGuruNY » Fri Apr 03, 2015 6:07 pm
gmat_winter wrote:In an opera theater, there are 300 seats available. When all the seats in the theater are sold out, the price of each ticket is $60. For every $1 increase in the price of the ticket, the number of seats sold decreases by 2.

How much should the theater owner charge for each seat to make the maximum profit?

A)87 $

b)95 $

c)105 $

d)120 $

e)127 $
We can PLUG IN THE ANSWERS, which represent the price required to maximize the total revenue.
Since 2 fewer seats are sold for every $1 increase in the ticket price, the seat decrease is twice the price increase.
Test the three middle answer choices and LOOK FOR A PATTERN.

Answer choice D: 120
Price increase = 120-60 = 60.
Seat decrease = 2*60 = 120.
Sold seats = 300-120 = 180.
Total revenue = (sold seats)I(ticket price) = 180*120 = 21,600.

Answer choice C: 105
Price increase = 105-60 = 45.
Seat decrease = 2*45 = 90.
Sold seats = 300-90 = 210.
Total revenue = (sold seats)(ticket price) = 210*105 = 22,050.

Answer choice B: 95
Price increase = 95-60 = 35.
Seat decrease = 2*35 = 70.
Sold seats = 300-70 = 230.
Total revenue = (sold seats)(ticket price) = 230*95 = 21,850.

The resulting totals -- 21600, 22050, 21850 -- imply that the total revenue is maximized by answer choice C.

The correct answer is C.
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by Brent@GMATPrepNow » Sat Apr 04, 2015 7:02 am
gmat_winter wrote:In an opera theater, there are 300 seats available. When all the seats in the theater are sold out, the price of each ticket is $60. For every $1 increase in the price of the ticket, the number of seats sold decreases by 2.

How much should the theater owner charge for each seat to make the maximum profit?

A)87 $
b)95 $
c)105 $
d)120 $
e)127 $
I'm not a big fan of this question. The only approach (without using techniques that are out of scope for the GMAT) is TEST THE ANSWERS, and the great thing about most GMAT math problems is that they can be solved using more than 1 approach.

To see what I mean, consider this algebraic approach:

Let x = the number of dollars added to the price of the ticket.
So, the new ticket price = 60+x dollars.

For every $1 increase in the price of the ticket, the number of seats sold decreases by 2.
So, 300-2x = the number of seats sold.

Revenue = (ticket price)(# seats sold)
So, revenue = (60+x)(300-2x)
Expand to get: revenue = 18,000 + 300x - 120n - 2x²
Simplify and rearrange to get: revenue = -2x² + 180x + 18,000
So our goal is to find the value of x that MAXIMIZES the value of -2x² + 180x + 18,000

1) One approach is to use calculus and find the derivative of -2x² + 180x + 18,000 and blah blah blah...(OUT OF SCOPE!)

2) Another approach is to recognize that y = -2x² + 180x + 18,000 represents a downward parabola AND the coordinates of the vertex (at the TOP of the parabola) represent the information that MAXIMIZES profits.
So, let's use approach #2 and try to determine the vertex of the parabola described by y = -2x² + 180x + 18,000
We'll use a process called COMPLETING THE SQUARE (which is also beyond the scope of the GMAT) [more on completing the square here https://www.mathsisfun.com/algebra/compl ... quare.html]

Start with: y = -2x² + 180x + 18,00
Factor -2 from first two terms to get: y = -2(x² - 90x) + 18,000
Complete the square to get: y = -2(x² - 90x + 2025 - 2025) + 18,000
Expand and remove -2025 from brackets to get: y = -2(x² - 90x + 2025) + 4050 +18,000

NOTE: Through our handwork, x² - 90x + 2025 is a square that can be factored!
Simplify and rewrite to get: y = -2(x - 45)² + 22,050
Now that we've rewritten our equation in this nice form, we can see that the vertex of the parabola has coordinates (45, 22050)

22050 is the HIGHEST POINT of the parabola, which means the MAXIMUM revenue is $22,050
45 is the value of x that achieves this maximum revenue
So, to maximize revenue, we must add $45 to the ticket price to get $60 + $45 = [spoiler]$105[/spoiler]
Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by gmat_winter » Sun Apr 05, 2015 5:42 am
Thanks Brent

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by Jeff@TargetTestPrep » Fri Jan 05, 2018 7:25 am
gmat_winter wrote:In an opera theater, there are 300 seats available. When all the seats in the theater are sold out, the price of each ticket is $60. For every $1 increase in the price of the ticket, the number of seats sold decreases by 2.

How much should the theater owner charge for each seat to make the maximum profit?

A)87 $

b)95 $

c)105 $

d)120 $

e)127 $
We can let x = the number of $1 increments in price (or the number of 2-seat decrements) and y = total revenue. Thus, the expression (60 + x) is the increased ticket price and (300 - 2x) represents the decreased number of tickets sold at that increased price. Recall that revenue is the product of the ticket price and the number of tickets sold. Thus, we have:

y = (60 + x)(300 - 2x)

Expanding the right-hand side, we have:

y = 18000 + 180x - 2x^2

Notice that the above is a quadratic equation. Let's rearrange the terms so that it's in standard ax^2 + bx + c form:

y = -2x^2 + 180x + 18000

We see that the graph of the quadratic equation is a downward parabola, so the maximum revenue is the y-value of the vertex. To get the y-value of the vertex, we must get the x-value of the vertex first, using the formula x = -b/(2a):

x = -180/(2(-2)) = -180/-4 = 45

Thus, to maximize the revenue, the ticket price should be 60 + 45 = $105.

(Note: We are using the term "revenue" instead of "profit" because the amount of money received from sales is technically called "revenue," not "profit." As we know, the term "profit" is the difference between revenue and cost. The problem uses the term "profit" instead of "revenue" perhaps because cost is either zero or negligible.)

Answer: C

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