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Hi i am confused how i would do these problems.

 Jun 13, 2020
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a) Let $x be the price of a ticket.

Price increase = $(x - 20).

 

For $1 increase in price, 15 fewer tickets are sold.

For $1(x - 20) increase in price, 15(x - 20) fewer tickets are sold.

 

Therefore if the price is $x, then the number of tickets sold would be 750 - 15(x - 20), which is 1050 - 15x upon simplification.

 

The revenue is equal to (number of tickets sold) * (price of each ticket), which is $x(1050 - 15x) = $15x(70 - x).

 

You can find the maximum revenue by completing the square.

 

b) is just similar.

 Jun 13, 2020

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