The Hotel Ventor has 400 rooms. Currently the hotel is filled. The daily rental is $ 600 per room. For every $6 increase in rent the demand for rooms decreases by 7 rooms. Let x = the number of $ 6 increases that can be made.
What should x be so as to maximize the revenue of the hotel?
I did p - 600 = (-6/7)(x-400) and got p=(-6x/7) + 942.8571429.
Then I multiplied that by x and got (-6x^2/7) + 942.8571429x.
Then I did -b/2a and got x=550.0000001, but that isn't the answer.
Can someone explain to me the right way to do this? thanks
Let me start fresh:
let the number of $6 increases be x
number of rooms rented = 400-7x
cost per room = 600 + 6x
Revenue = R = number of rooms x cost per room
= (400 - 7x)(600+6x)
=240000 -1800x - 42x^2
so the x of the vertex is -b/2a
= 1800/-84
= -150/7 or appr -21.4
since I defined x to be number of increases and x turned out negative, there will have to be DECREASES.
Assume that there will be multiples of $6 changes
if x = -21
number of rooms rented = 400 - (-21) = 421
BUT, they only have 400 rooms
looks like a trick question!
The cost should stay the same
Look at the graph
http://www.wolframalpha.com/input/?i=y+%3D+%28400+-+7x%29%28600%2B6x%29
Since the graph only makes sense for x≥0
the max value of the function is at x=0
namely when there is no increase or decrease in price.
My actual answer of -150/7 makes sense, notice that is where the vertex is located.
(BTW, $600 for a room ??
I will never pay that much for a room in my lifetime.)
Wow, thank you for such a thorough answer! =)
To maximize the revenue of the hotel, we need to find the value of x that maximizes the revenue function.
The revenue function is given by the product of the number of rooms rented and the rental price per room. Let's break it down step by step:
1. Let's start with the demand equation, which states that for every $6 increase in rent, the demand for rooms decreases by 7 rooms. We can write this as:
Demand = 400 - (7/6)x
Here, x represents the number of $6 increases in rent.
2. To find the revenue, we multiply the demand by the rental price per room, which is $600:
Revenue = (400 - (7/6)x) * 600
3. To maximize the revenue, we need to find the value of x that maximizes the revenue function. To do this, we can use calculus by taking the derivative of the revenue function with respect to x, and setting it equal to zero:
d(Revenue)/dx = 0
Let's find the derivative and solve for x:
d(Revenue)/dx = [d(400 - (7/6)x)/dx] * 600
= (-7/6) * 600
Now set the derivative equal to zero:
(-7/6) * 600 = 0
Solving this equation gives us x = 514.2857143
4. However, we need to check if this value of x maximizes the revenue function. To do this, we can calculate the second derivative and check its sign:
d²(Revenue)/dx² = [d((-7/6) * 600)/dx]
= 0
Since the second derivative is equal to zero, we cannot determine the sign. So, we need to test a value on each side of x = 514.2857143 to determine if it is a maximum or minimum.
5. Let's test two values – one slightly less than 514.2857143 and one slightly greater than it:
Test x = 514:
Evaluate the revenue at x = 514:
Revenue(x = 514) = (400 - (7/6) * 514) * 600 = $546,171.43
Test x = 515:
Evaluate the revenue at x = 515:
Revenue(x = 515) = (400 - (7/6) * 515) * 600 = $546,428.57
From these results, we can see that the revenue is higher at x = 515 than at x = 514, so the maximum revenue is achieved when x = 515.
Therefore, to maximize the revenue of the hotel, x should be 515.