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anonymous
 5 years ago
A baseball team plays in a stadium that holds 52000 spectators. With the ticket price at $12 the average attendance has been 19000. When the price dropped to $9, the average attendance rose to 26000. Assume that attendance is linearly related to ticket price.
What ticket price would maximize revenue?
anonymous
 5 years ago
A baseball team plays in a stadium that holds 52000 spectators. With the ticket price at $12 the average attendance has been 19000. When the price dropped to $9, the average attendance rose to 26000. Assume that attendance is linearly related to ticket price. What ticket price would maximize revenue?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0P= price x= attendance

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x = kP +c for constants k and c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now you can form two eqns, solve for the constants c and k

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.019000 =12k +c 26000= 9k +c subtract the first from the second 7000= 3k > k = 7000/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sub that into one of the other eqns say the first , so c= 1900012k = 47000

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats what ive done so far but nothing from there is making sense. I also have to describe what im doing in words so i prove i understand it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so x= (7000/3) P +47000

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0revenue = R = attendance times cost price

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so R = P [ (7000/3)P +47000 ] differentiate with respect to P , we want to maximise R while varying P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0R = (7000/3) P^2 + 47000P dR/dP = (14000/3)P +47000 =0 for min/max

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now check, second derivative is d^2R/dP^2 = (14000/3) which is less than zero , so we do get a maximum . so ticket price = $3 maximises revenue

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0should be P = (47000) / ( 14000/3 ) = $10.07 , that maximises revenue

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0$10.07 is the right answer. I still however have no idea how you got it. We haven't went over any problems by differentiating. I have no idea what that is.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well , this can be done by other methods , I am assuming you have done quadratic equations before , ie , you know how to find the vertex of a parabola?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and you know that the vertex ( which is a maximum or a minimum ) occurs on the axis of symmetry , and axis of symmetry of y= ax^2 +bx +c is x=b / (2a)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes that's what we're doing now

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if you go back to this eqn R = (7000/3) P^2 + 47000P and find the axis of symmetry , which is P = b/ (2a) , this will give you the x coordinate of the maximum point

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which is 10.07, the answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok i know how you got this answer but why is there no c in the ax^2+bx+c equation? and how did you get 2 P's in the equation R= (7000/3)P^2 +47000P ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"why is there no c? " , there just isnt "where did the two Ps come from" , look at how revenue was calculated, revenue = price x number of people
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