Question 813771: A 180-room hotel is filled when the room rate is $50 per day. For each $1 increase in the rate, two fewer rooms are rented. Find the room rate that maximizes daily revenue.
The rate that maximizes revenue is
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! use the revenue equation:
r = p * n
where
p = price per item
n = number of items sold
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because we know that changing the price (p) also changes the number sold (n), we need to introduce a variable to the revenue equation to measure the changes in p and n.
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so just make up a variable, call it x, such that:
p + x means a one dollar price increase and
p - x means a one dollar price decrease and
n + x means a one item number-sold increase and
n - x means a one item number-sold decrease
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for each $1 price increase (p + x) there are 2 fewer items sold (n - 2x), so:
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r = (p + x) * (n - 2x)
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we know that at $50 per item, we sell 180 items, so:
r = (50 + x) * (180 - 2x)
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r = (50 + x) * (180 - 2x)
r = 9000 - 100x + 180x - 2xx
r = -2xx + 80x + 9000
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the above quadratic equation is in standard form, with a=-2, b=80, and c=9000
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to solve the quadratic equation, plug this:
-2 80 9000
into this: https://sooeet.com/math/quadratic-equation-solver.php
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Answer 1:
the quadratic equation has a maximum at: ( x=20, r=9800 ), so:
maximum revenue is $9800
when price per item is $20
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