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A baseball 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?

Answer :

opudodennis

Answer:

$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

[tex]m=\frac {31000-27000}{8-11}= -1333.33[/tex]

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)

[tex]R=-1333.33p^{2}+41666.67p[/tex]

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

0=2*-1333.33p+41666.67

[tex]p=-\frac {41666.67}{2(-1333.33)}= 15.625[/tex]

Therefore, the price that maximizes revenue is $15.625

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