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19 April, 05:20

A 210 210-room hotel is filled when the room rate is $ 50 $50 per day. for each $ 1 $1 increase in the rate, three fewer rooms are rented. find the room rate that maximizes daily revenue.

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  1. 19 April, 08:06
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    With 210 rooms filled the daily revenue is 10,557 $. The maximum amount of profit possible is with 180 rooms filled which means that the price is 60$ for each room. The revenue for this price is 10,800 $ daily. The closest daily revenues to this are: 183 rooms for the price of 59$=10,797$ 177 rooms for the price of 61$=10,797$
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