# Calculus

A baseball team plays in a stadium that holds 68000 spectators. With the ticket price at \$11 the average attendence has been 27000. When the price dropped to \$10, the average attendence rose to 34000. Assume that attendence is linearly related to ticket price.
What ticket price would maximize revenue? \$
I have no idea how to go about doing this optimization question, I've been stuck on trying to find a formula for this, I've been around the internet trying to find out how to do this question. If someone can break this down for me, that would be greatly appreciated...GREATLY!

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1. Treat it as two ordered pairs,
(11,27000) and (10,34000) and then find the equation of the line like you did in grade 9

slope = (34000-27000)/(10-11)
= -7000

y = mx + b
y = -7000x + b
using (10,34000)
34000 = 10(-7000) + b
b = 104000

y = -7000x + 104000 , y ≤68000

so at a maximum of y, y = 68000
68000 = -7000x + 104000
7000x = 36000
x = 36/7 = 5.14..

They will fill the stadium when the price of a ticket is \$5.14

At that price, the revenue
= 68000(5.14) = \$349,714.29

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2. I tried this way several times and it is not correct. Are there any other ways to doing this problem?

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3. y = -7000x + 104000

the person above got you to here, which is the line function of the amount of people vs. cost of ticket

we are looking for revenue, so (number of people) x (cost of ticket)
that will get us revenue, the cost of ticket is x,
so
(-7000x + 104000)(x)
R = -7000x^2 + 104000x = Revenue

now take the derivative,
dR/dx = -14000x + 104000

then find the critical number, (what makes this equal 0)

and that should be the best answer but it is possible you would want to find out the revenue is everyone showed up and what the cost would be then, but that one is unlikely.

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