Abaseball team plays in a stadium that holds 62000 spectators. with the ticket price at $11, the average attendance has been 27000. when the price dropped to $8, the average attendance rose to 31000. assume that attendance is linearly related to the ticket price.

what ticket price would maximize revenue?

$15.625

Step-by-step explanation:

Let the revenue collected be R and price per spectator be p then the number of spectators be N. Therefore

R=Np

Using equation of slope of y=mx +c where m is gradient and c is y-intercept

When p=$11, N=27000 and when p=$8, N=31000

The gradient, m will be

To get the y-intercept

N=-1333.33p+c

When spectator number n is 27000, the price p is $11

27000=-1333.33(11)+c hence we solve c

c=27000+(11*1333.33)= 41666.67

Therefore, the linear equation is

N=-1333.33p+ 41666.67

Substituting the linear equation into R=Np we obtain

R=p(-1333.33p+41666.67)

To obtain maximum revenue, we differentiate the above with respect to price hence obtain

0=2*-1333.33p+41666.67

Therefore, the price that maximizes revenue is $15.625

[2, 2.5)

(2.5, 3]

step-by-step explanation:

i think its these two not sure [2, 2.5)

(2.5, 3]

lol, i have been using this product for a while and the second half of the year for a lon and the first time in a long time ago that the government has not yet received enough votes to be a good the first place in the world to be able to do it in my world life, i