Question

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? 

Answer 1

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

So the income can be represented by:

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

then:

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

$x=1000/200=5$

rate that maximize the revenue:

$5\cdot100=500$ peso/day