A professional basketball team plays in a stadium that holds 23 000 spectators. WIth ticket prices at $60, the average attendance had been 18 000. When ticket prices were lowered to $55, the average attendance rose to 20 000. Based on this pattern, how should ticket prices be set to maximize revenue?
2007-01-08 14:37:12 · 4 answers · asked by Mickey mouse 1 in Science & Mathematics ➔ Mathematics
(60, 18000)
(55, 20000)
Find the line that represents these two points.
m=(20000-18000)/(55-60)
=2000/-5 = -400
y-18000=-400(x-60) This is the demand equation so solve for y
y-18000=-400x+24000
y=-400x+42000
Revenue R=x(-400x+42000)
=-400x^2+42000x
The maximum is x=-b/(2a)=-42000/2(-400)=52.5 so the price should be $52.50 to maximize the revenue.
2007-01-08 14:44:02 · answer #1 · answered by Professor Maddie 4 · 0⤊ 0⤋
If every price drop of $5 gets an extra 2000 "butts on seats" then dropping another $5 should get the average to 22,000. Trouble is, at bot $50 AND $55 your total income is $1,100,000
So split it, make the price $52.50. Crowds should average 21,000 and revenue becomes $1,102,500
2007-01-08 22:59:11 · answer #2 · answered by Alan 6 · 0⤊ 0⤋
47.50
Why? because for each 5 dollars you take of the price you get 2000 spectators more so if from 55 you lower it to 50 you get 2000 spectators more= 22000, and know since you only want 1000 (Half) more to get 23000 spectators you take half of 5 dollars= 2.50 so 50-2.50=47.50. Hope you understand what I try yo say.
2007-01-08 22:42:59 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Wouldn't $47.50 fill the stadium????
2007-01-08 22:44:54 · answer #4 · answered by stimpy36 2 · 0⤊ 0⤋