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A baseball team plays in a stadium that holds 55,000 spectators. With ticket prices at $ \$ 10 $, the average attendance had been 27,000. When ticket prices were lowered to $ \$ 8 $, the average attendance rose to 33,000.(a) Find the demand function, assuming that it is linear.(b) How should ticket prices be set to maximize revenue?

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Calculus 1 / AB

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Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

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This problem asks to find a linear demand function and the number of tickets sold to maximize revenue. Given that At price $10 number of tickets sold is about 27,000. And at the price $8. The number of tickets sold is 33,000 say well let XB. The number of tickets and it's a p be the price for a ticket. So for the first statement let's go $10.00. And the number of tickets sold. 27,000, it's called this except one and then $8 it's called this P sub two and 33,000 is X. Up to. Because the demand function is linear. Then we need to find a slope for this function which is just the difference between the output or the price. So we have P sub two minus piece of one over the difference between the X values. So we have except to minus except one. So if peace up to is eight and piece of wand is 10, then we have 8 -10. This all over You have 33,000 -27000. This gives us negative two over 6000 Or that's just negative one over 3000. And so using point slope formula we have the linear equation P minus Piece of one. This is equal to the slope times X- Except one. So we have p minus piece of Languages 10. This is equal to negative one over 3000 times X -X. Sub one which is 27,000 simplifying this, we get p that's equal to negative one over 3000 X plus we have nine plus 10. So we have the linear demand function P which is equal to -1 over 3000 X plus 19. Now, revenue is just the price per unit times the number of units sold. So we have revenue. That's just p times X. Or that's just negative one over 3000 X 19. This times X will give us negative one over 3000 X squared plus 19. X. Not to find the maximum revenue. We take the derivative of the revenue function. We have our prime that's equal to -1 over 3000 Times two x plus 19. And then from here we want to set the Derivative to zero and Sulfur X. So we have zero. This is equal to negative X over 1500 plus 19. And so solving for X we have X over 1500. This is equal to 19 or X is just 28,500. From here we apply the concept of absolute extremist. Since We only have a limited value for X. That is from 0 to 55,000. So if access in between zero and 55,000 and we have a critical point 28,500, then evaluating the revenue using the endpoints of this interval and this critical point we have R of zero. This is equal to negative one over 3000 times zero squared Plus 19 times zero. This gives us zero and are of 28,500. This gives us negative one over 3000 Times 28,500 squared plus 19 times 28,500. This gives us 270,750. And if Or is 55,000, we get a revenue which is equal to negative one over 3000 times 55,000 squared Plus 19 times 55,000. That's equal to 36,600 66.67. And so, maximum revenue, Which is 270007 and 50 is attained when X is 28,500. Therefore, they need to sell 28,500 tickets to maximize revenue.

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