What ticket price results in the greatest revenue and what is the greatest revenue in the case below?
the cost of a ticket to a basketball gym seating 700 fans is $3. At this price all seats are filled. The owner estimates that if the price is increased, attendance with fall by 100 for every dollar increase.
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When the cost of a ticket to a basketball gym seating 700 fans is $3 all seats are filled. The owner estimates that if the price is increased, attendance with fall by 100 for every dollar increase. It is assumed that the price of tickets can fall in terms of fractions of a dollar.
There is no point of decreasing the ticket price as the capacity cannot exceed 700.
For a $1 increase, attendance falls by 100 for a x increase attendance falls by 100*x
The revenue is given by R = (3+x)*(700 - 100*x)
= 2100 + 700x - 300x - 100x^2
To maximize revenue solve R' = 0
=> 700 - 300 - 200x = 0
=> 200x = 400
=> x = 2
The revenue is maximized at x = 5 and is equal to $2500
Lets say the revenue is A. If we say our ticket price is n where n>3 and `n in Z` ;
Decrease in attendance = `(n-3)*100`
So total attendance = `700-(n-3)*100`
Revenue of owner (A) = `n*(700-(n-3)*100)`
A = `n*(700-(n-3)*100)`
A = `700n-100n^2+300n`
A = `1000n-100n^2`
For the maximum or minimum revenue `(dA)/dx = 0` .If the revenue is maximum then `(d^2A)/(dx^2) lt0`
`(dA)/dx` = `1000-2*100n`
0 = `1000-200n`
`200n` = `1000`
`n` = `5`
`(d^2A)/(dx^2)` = `1000-2*100n`
So `(d^2A)/(dx^2)lt0` . So revenue is a maximum.
Maximum revenue obtained when the ticket price is $5.
Maximum `A = 5*(700-(5-3)*100) = 2500`
So maximum revenue is $2500.
It will be:2500
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