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theater has 300 suscribers who each pay $400 for a season ticket.
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)
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 = +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
You can also check this by finding the vertex of this equation
Substitute 5 for x and solve for r(x)