Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (anonymous):

A baseball team plays in the stadium that holds 68000 spectators. With the ticket price at 9 the average attendance has been 28000. When the price dropped to 7, the average attendance rose to 34000. a) Find the demand function p(x), where x is the number of the spectators. (assume p(x) is linear) p(x)= (-2/5000)x+18.3333333 b) How should be set a ticket price to maximize revenue? <<<<<<

OpenStudy (amistre64):

they give you 2 data points to construct a line with (9,28000) (7,34000)

OpenStudy (amistre64):

subtract the points to find your slope: (9,28000) -(7,34000) ----------- -2,-6000; slope = y/x = 6000/2 = 3000 now calibrate the line equation with one of the points to get: p(x) = 3000x -3000(9) + 28000 p(x) = 3000x + 1000

OpenStudy (amistre64):

my -2 is wrong :) 9-7 = 2 ; so the slope is gonna be -3000

OpenStudy (amistre64):

p(x) = -3000x +3000(9) + 28000 p(x) = -3000x +55000 maybe?

OpenStudy (anonymous):

I have my p(x) though. It's the second part which I don't quite understand how to get using my p(x)

OpenStudy (anonymous):

I think it is just P(x)* x which will give us parabola

OpenStudy (amistre64):

revenue = r(x) = number of people attending * price per ticket r(x) = p(x)*x r(x) = (-3000x +55000)x r(x) = -3000x^2 +55000x; take derivative or find vertex

OpenStudy (amistre64):

r(x) = -3x^2 + 55 ; for simplicity

OpenStudy (amistre64):

dropped an x :) r(x) = -3x^2 + 55x; r'(x) = -6x + 55; r(x)=0 when x=55/6

OpenStudy (amistre64):

1000* 9.1666666....... 9166.6666.... maybe?

OpenStudy (amistre64):

9.16 lol forget the 1000

OpenStudy (amistre64):

9.17 works better in the r(x) equation

OpenStudy (anonymous):

!!! I've been trying to get an answer for that for the past 2 days. Thanks, amistre :]

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!