The owner of a baseball team and local stadium has commissioned a study that shows the demand by fans for stadium seats (per playing date) to be P = 22 - 0.2Q, where P is the average price of a ticket and Q represents the number of seats (expressed in thousands). The local stadium seats a maximum of 56,000 per game. Suppose the owner offers you 10% of the revenues. If you can only choose a uniform per-ticket price, what is the maximum amount you can earn per game? (Note: assume that all seats and all games are the same in this problem.)

To find the maximum amount you can earn per game, you can graph the formula as a line. The greatest revenue that can be earned is equivalent to the area of the largest rectangle that can be inscribed under the right triangle formed by the line as well as the x and y axis. That rectangle will have a length and width that are 1/2 the length of the triangle's legs.

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The relationship between the average ticket price and the number of seats is linear, and so one way to find the maximum revenue that can be earned is to graph the function. If you substitute y for P and x for q, you can write the equation as y =...

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The relationship between the average ticket price and the number of seats is linear, and so one way to find the maximum revenue that can be earned is to graph the function. If you substitute y for P and x for q, you can write the equation as y = -1/5x + 22. Plotting this line, you get a y-intercept of (0, 22) and an x intercept of (110, 0). In essence, this means that if the ticket price was $22, no one would come, and if the price was $0, 110,000 fans would attend. Since the stadium seats 56,000, though, the point with the greatest fan to price ratio is (56, 10.8). In other words, at a price of $10.80 per ticket, 56,000 fans would buy tickets. 56 x 10.8 = 604.8 or $604,800.

The maximum is slightly higher, however, if Q = 55, because then P = 11. 55 x 11 = 605: a price of $11 will attract 55,000 fans and earn a revenue of $605,000. This answer can, again, be understood in terms of the graph. The region under the line in the first quadrant is a right triangle with sides of 22 and 110. The product of P and Q is equivalent to the area of a rectangle inscribed in the triangle. The greatest area of such a rectangle is obtained when the length and width are 1/2 as long as the legs of the triangle: 11 and 55. You can confirm this by plugging in 54 and 56 for Q: in both cases, you get a P that is slightly less than 605.

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