Respuesta :
A theater wants to raise the ticket by $20.
For each $20 increase they lose 10 subscribers.
At what price can they maximize the revenue?
:
Let x = no. $20 raises in price, and no. of 10 subscriber losses
:
Revenue = no. of subscribers * ticket price
r = (300-10x) * (400+20x)
FOIL
r = 120000 + 6000x - 4000x - 200x^2
A quadratic equation
r(x) = -200x^2 + 2000x + 120000
maximum occurs at the axis of symmetry, find that x = -b/(2a)
in this equation a = -200, b=2000
x = 
x = +5
:
Max revenue occurs when ticket price raise is $20 * 5 or $100 to $500
You will lose 5*10 = 50 subscribers or 250
so the revenue will be:
500 * 250 = $125,000
:
The required $125,000 price would maximize the revenue from the season tickets.
Given,
Company has total no. of season ticket subscribes = 300
Current price of the ticket = $400
A survey of the subscribers has determined that, for every $20 increase in price, 10 subscribes would not renew their season tickets.
Let x = no. $20 raises in price, and no. of 10 subscriber losses.
Revenue = ( no. of subscriber ) × ( ticket price)
r = ( 300 - 10x ) ( 400 + 20x )
r = 120000 - 6000x - 4000x - 200[tex]x^{2}[/tex]
The quadratic equation is ,
r = -[tex]200x^{2} - 10000x + 120000[/tex]
Maximum occurs at the axis of symmetry,
To find that,
x = [tex]\frac{-b}{2a}[/tex]
In this equation, a = -200, b=2000
[tex]x = \frac{-2000}{2 (-200)} \\\\[/tex]
x = 5
Max revenue occurs when ticket price raise is = $20 × 5 or $100 to $500
Company will lose 5×10 = 50 subscribers or 250
So the revenue will be,
500 × 250 = $125,000
Hence, The required $125,000 price would maximize the revenue from the season tickets.
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