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 =
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.)
$_________________
6 answers
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.
Oh okay, thank you!
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
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
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 :)
A baseball team plays in a stadium that holds 58000 spectators. With the ticket price at $11 the average attendance has been 22000. When the price dropped to $10, the average attendance rose to 29000. Assume that attendance is linearly related to ticket price.