Question #189957

A 40-room hotel is fully occupied if Php 3000 is charged per day per room. For every 𝑥

hundred-peso increase in the daily rate, there are 𝑥 units vacant. What rate will maximize the

revenue of the hotel operation? 


1
Expert's answer
2022-01-11T15:41:16-0500

for every hundred-peso increase in price, the number of rooms goes down by 1

So the income can be represented by:

R(x)=(40x)(3000+100x)R(x)=(40-x)(3000+100x)

then:

R(x)=200x+1000=0R'(x)=-200x+1000=0

x=1000/200=5x=1000/200=5

rate that maximize the revenue:

5100=5005\cdot100=500 peso/day


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Comments

Assignment Expert
12.06.21, 09:27

Dear Hampikiti, thank you for leaving a feedback.


Hampikiti
19.05.21, 18:07

The revenue function must be R(x) = (3000 + 100x) (40 - x) this will give you a rate of Php 3500 to maximize revenue :)

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