Posted by **Anonymous** on Sunday, July 15, 2012 at 1:51pm.

A baseball team plays in a stadium that holds 53,000 spectators. With ticket prices at $10, the average attendance had been 49,000. When ticket prices were lowered to $8, the average attendance rose to 51,000.

(a) Find the demand function (price p as a function of attendance x), assuming it to be linear.

P(x)=_____

How should ticket prices be set to maximize revenue? (Round your answer to the nearest cent.)

$_________________

- Calculus -
**Damon**, Sunday, July 15, 2012 at 4:03pm
k is for thousands

(49 k, 10) and (51 k, 8)

slope = (8-10)/(51 k - 49 k) = -1/1k

p = -(1/k)x+ b

10 = -(1/k)49 k + b

59 = b

so

p = -(1/k)x + 59 or p = -x/1000 + 59

r = x p

r = -x^2/1000 + 59 x

x^2 -59000 x = 1000 r

x^2 - 59000 x + 29,500^2 = 1000 r + 29,500^2

(x-29,500)^2 = 1000(r + 870,250)

so max r at x = 29,500 people and revenue of 870,250

then price = 870,250/29,500 = $29.50

then p =

- Calculus -
**Damon**, Sunday, July 15, 2012 at 4:05pm
By the way there is no need to use calculus for these since you can complete the square to find the vertex of a parabola with algebra 2.

- Calculus -
**Anonymous**, Sunday, July 15, 2012 at 6:40pm
Oh okay, thank you!

- Calculus - sign error (does not matter) -
**Damon**, Sunday, July 15, 2012 at 6:49pm
r = x p

r = -x^2/1000 + 59 x

x^2 -59000 x = -1000 r

x^2 - 59000 x + 29,500^2 = -1000 r + 29,500^2

(x-29,500)^2 = 1000(r - 870,250)

so max r at x = 29,500 people and revenue of 870,250

then price = 870,250/29,500 = $29.50

- Calculus -
**bob**, Wednesday, October 17, 2012 at 11:54pm
i feel like you solved for when the stadium holds 59000 not 53000 like in his question hahaha WHICH IS PERFECT BECAUSE MY PROBLEM IS WITH 59000! yay :)

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