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.)
$_________________

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.

To find the demand function (price p as a function of attendance x), we can use the information given in the problem. We are given two points on the demand function: (49,000, $10) and (51,000, $8).

Using the point-slope formula, we can calculate the slope of the demand function, which represents the rate of change of price with respect to attendance. The formula is:

m = (y2 - y1) / (x2 - x1)

m = (10 - 8) / (49,000 - 51,000)

m = -2 / -2,000

m = 0.001

Now that we have the slope, we can write the equation of the demand function in the slope-intercept form:

y = mx + b

where y represents the price (p) and x represents the attendance (x). Solving for b, the intercept:

10 = 0.001 * 49,000 + b

10 = 49 + b

b = 10 - 49

b = -39

Therefore, the demand function (price p as a function of attendance x) is:

P(x) = 0.001x - 39

To find the ticket prices that maximize revenue, we need to determine the attendance level that corresponds to the maximum revenue. Revenue is calculated by multiplying the attendance by the ticket price.

Revenue = Attendance * Ticket Price

R(x) = x * P(x)

Substituting P(x) from the demand function:

R(x) = x * (0.001x - 39)

R(x) = 0.001x^2 - 39x

To find the maximum of this quadratic function, we can take the derivative and set it equal to zero:

dR(x)/dx = 0.001(2x) - 39 = 0

0.002x - 39 = 0

0.002x = 39

x = 39/0.002

x = 19,500

Therefore, the attendance level that maximizes revenue is 19,500. To find the corresponding ticket price, we can substitute this attendance level into the demand function:

P(19,500) = 0.001 * 19,500 - 39

P(19,500) = 19.5 - 39

P(19,500) = -19.5

Note that the negative value doesn't make sense for a ticket price. Thus, we need to consider the practical range of attendance levels. Since the maximum capacity of the stadium is 53,000, the attendance level cannot exceed this value. Therefore, the maximum attainable attendance level is 53,000.

Substituting 53,000 into the demand function:

P(53,000) = 0.001 * 53,000 - 39

P(53,000) = 53 - 39

P(53,000) = 14

Therefore, to maximize revenue, the ticket prices should be set at $14.

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 =

Oh okay, thank you!

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.

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 :)

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