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A 300 room hotel is filled to capacity at \$80 a night. If the charge is increased by \$3 it rents 9 less rooms. If it costs \$10 to clean a rented room the next day, how much should the inn keeper charge in order to maximize its profit?

I thought the question was really straight forward and that I'd be able to do the following to get my answer:

Revenue = (# Of Rooms * Room Charge) - (# Of Rooms * Clean Charge)

Full Inn

Revenue = (300 * 80) - (300 * 10) which is 24,000 - 3,000 so Revenue = \$21,000

Not Full Inn

Revenue = (291 * 83) - (291 * 10) which is 24,153 - 2,910 so Revenue = \$21,243

Therefore the inn keeper should charge \$83 a room. I got the question wrong, so can someone explain what I should have done?

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Is that (homework)? – Rob Dec 5 '12 at 19:48
@Rob No, it was on an exam and I got the question wrong so I just want to know why. – StrugglingWithMath Dec 5 '12 at 20:01
You are right in being a bit confused. The intent of the question is that if you increase charge by $3y$, you decrease occupancy by $9y$, for any reasonable $y$. That means you need to explore charging more than $83$, it might lead to even greater profit. – André Nicolas Dec 5 '12 at 20:04
@AndréNicolas Ahh, I wasn't aware the question was asking for an overall maximized profit. I thought it was asking between the two. It would have been nice if the question was worded better so that there was no question about what it was asking. I'll check your response now and see what I can make of it. – StrugglingWithMath Dec 5 '12 at 20:06
Instead of my $x$, you may wish to replace it everywhere by $3x$. So $80+3x$, rent $300-9x$, and so on. I used my version because it makes for marginally simpler-looking numbers. Kind of an automatic reflex! – André Nicolas Dec 5 '12 at 20:15
up vote 1 down vote accepted

Let us charge $80+x$. Then we rent $300-3x$ rooms.

Net Income, after cleaning: $(80+x)(300-3x) -(10)(300-3x)$.

Do the usual stuff to maximize, not forgetting about endpoints. There is also the complication that the number that maximizes our function will not necessarily lead to an integer number of rooms rented, so we may have to make a mild adjustment.

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So, I gave it a go: $R = (80+3x)(300-9x)-(10)(80+3x)$ so that is: $-720x^2 + 843x -23,200$ which means, $R' = 1440x -843$ set to $0$ leaves us with $x = 843/1440$. Our Revenue would be 23,278.56 with that x-value. Is that correct? – StrugglingWithMath Dec 5 '12 at 20:29
Redid it and came out with: $R = 27x^2 - 150x + 23,200$. So, $R' = 54x - 150$ set it to $0$ and $x = 150/54$. Plugging that in, max revenue gets to 23,408.33? – StrugglingWithMath Dec 5 '12 at 20:52
I messed up! Got confused between rate and rooms. Sorry! Changed answer. With your notation, we want $(80+3x)(300-9x)-(10)(300-9x)$. I will dothe calculation. – André Nicolas Dec 5 '12 at 21:09
The numbers turn out real simple. The correct net income is $-27x^2+270x+21000$. Max at $x=5$. Max revenue is $21675$. – André Nicolas Dec 5 '12 at 21:14
$(80+3x)(300-9x)=-27x^2 +180x+24000$. Subtract $10(300-9x)=3000-90x$. So we get $(-27x^2+180x+24000)-(3000-90x)$. Rest is easy, remember to change signs, for $-(-90x)=90x$. – André Nicolas Dec 5 '12 at 21:23

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