The price at which the seats should be priced has to be determined to maximize the revenue earned. If the price of each seat is kept at P, the revenue from the stadium is given by:

R = P*Q

Substituting P = 22 - 0.2Q, gives

R = P*((22 - P)/0.2)

= 22*P/0.2-P^2/0.2

= 110P - P^2/0.2

Differentiating R with respect to P gives R' = 110-2*P/0.2 = 110-10P

The value of R is maximized at the point where R' is equal to 0.

Solving R' = 0 for P gives (110 - 10P) = 0

110 = 10P

P = 11

When P = 11, the demand is Q = (22 - 11)/0.2 = 55 (in thousands).

As this is lower than the maximum capacity of the stadium, it is possible to accommodate 55000 people in the stadium.

The revenue in this case is 11*55000 = 605000

As the owner offers 10% of the revenue, the earning per game is $60,500.