Abaseball team plays in a stadium that holds 58,000 spectators. with ticket prices at $12, the average attendance had been 23,000. when ticket prices were lowered to $10, the average attendance rose to 29,000.
(a) find the demand function (price p as a function of attendance x), assuming it to be linear.
(b) how should ticket prices be set to maximize revenue? (round your answer to the nearest cent.)

  a)  p = -x/3000 + 19 2/3

  b)  $9.83

Step-by-step explanation:

a) The 2-point form of the equation for a line can be used with the two given points. The attendance is said to be the independent variable.

  y = (y2 -y1)/(x2 -x1)(x -x1) + y1

Here "y" is the price (p), and "x" is the attendance, so we have ...

  p = (10 -12)/(29000 -23000)(x -23000) +12

  p = -1/3000(x -23000) +12

  p = -x/3000 + 19 2/3 . . . . price as a function of attendance

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b) Revenue is the product of attendance and price. We can find the attendance associated with maximum revenue, then find the corresponding price.

  R(x) = x·p = x(-x/3000 +19 2/3)

This has zeros at x=0 and x=3000(19 2/3) = 59,000. The maximum revenue corresponds to attendance halfway between these values, at x = 29,500. The demand function tells us the ticket price should be ...

  p = -29500/3000 +19 2/3 = 9.83 . . . . dollars

Comment on the problem working

I might have written the "demand function" to use price as the independent variable. Then the price is what you know when you find the maximum revenue; you don't have to do an extra step to find it.

  x = 3000(19 2/3 -p)

  xp = 3000p(19 2/3 -p) is maximized at p = (19 2/3)/2 = 9 5/6 ≈ 9.83

answer is d-29

step-by-step explanation:

since you didn't include the picture, i can give you the description so you know how to pick the correct answer. the third quartile of a data is always shown in a box plot. in the middle of your graph should be a box. the third quartile is the value that goes with the far right edge of the box.