1. Solve this problem by writing a quadratic equation and maximizing it using algebra.

Motel managers advertise that they will provide dinner, dancing, and drinks for $50 per couple for a New Year's Eve party. They must have a guarantee of 30 couples. Furthermore, they agree that for each couple in excess of 30, they will reduce the price per couple for all attending by $0.50.

A) How many couple will it take to maximize the morel's revenue?
B)What is the maximum revenue for the motel?

For each additional couple after 30, the price will go down by 0.50x each time, where 'x' is the number of additional couples, resulting in a final price of $ (50 - 0.5x)

The total revenue will be equal to the number of couples times the ticket price.
The number of couples is given by (30+x), where x is the number of additional couples, as stated earlier. The revenue is therefore (50 - 0.5x)(30+x).

Expand and find the derivative of that expression. Set for zero to find the maximum value of x.

The rest should come easily. Post your answer! :)

In my opinion, you need more cost information. You need to know the cost to the hotel for each added couple in excess of 30.
The cost of the band remains constant, but the hotel's cost of dinner and drinks should decrease as the number of couples increases beyond 30. Does that make sense?

In my opinion, you need more cost information. You need to know the cost to the hotel for each added couple in excess of 30.
The cost of the band remains constant, but the hotel's cost of dinner and drinks should decrease as the number of couples increases beyond 30. Does that make sense?

I can't see your point... :confused:

For 30 couples, the cost will be $ 50.00 each.
For 31 couples, the cost will be $ 49.50 each.
For 32 couples, the cost will be $ 49.00 each.
For 33 couples, the cost will be $ 48.50 each.

The cost decreases for all the present couples, so, the cost does not remain for a certain group of people. See in bold and underlined.


Furthermore, they agree that for each couple in excess of 30, they will reduce the price per couple for all attending by $0.50.

The cost is not specified, I agree, but it really doesn't have to. The overall cost per couple is given, and that is ample. The problem is really simple and should not be made more complicated. I hope you get what I mean, without taking it offensively. This is a public forum, and I want that people understand things well as they should be.

Thanks for posting :)

Unk8 is on-point (as usual ;))... you'd need to know the motel's cost function if you were maximizing their profit function. But in this case it's asking to optimize the revenues, and so a knowledge of the cost situation isn't needed.

Employing the first derivative, as Unknown008 has suggested, is a great 'all-purpose' method that handles the optimization of a wide variety of functions. But Matt, I'm wondering from your first post if you need an algebraic-based approach.

If so, put Unk8's revenue formula into "vertex" form, and, well... there you go. The answer to (A) is the x-coordinate of the vertex point, and the answer to (B) is the y-coordinate.

Nevertheless, it's good to keep in mind that the derivatives approach isn't limited to quadratics.

If I may. One may also solve this without calcarooney.

Note that the x coordinate for the vertex of a parabola is \underbrace{x=\frac{-b}{2a}}_{\text{no. of couples}}

\underbrace{y=c-\frac{b^{2}}{4a}}_{\text{max revenue}}

Since this involves a quadratic, it can be solved by this method as well. Unless, of course, you are required to use calc.

How's this look Unknown008?

-0.5x^2 + 35x + 1500 ------> x^2 + 70 + 3000
= -1(x^2 - 70x) + 3000
= -1(x^2 - 70x + (-35)^2) + 3000 + (-35)^2
= -1(x^2 - 70x + 1225) + 3000 + 1225
y= -(x-35)^2 + 4225

Max=4225, x=35

y=2112.5

Pardon my previous incorrect assumption.

Here is a spreadsheet print out of the
Answer to the original request

60 35.00 2,100.00 5.50
61 34.50 2,104.50 4.50
62 34.00 2,108.00 3.50
63 33.50 2,110.50 2.50
64 33.00 2,112.00 1.50
65 32.50 2,112.50 0.50 MAX. REVENUE[/B]
66 32.00 2,112.00 -0.50
67 31.50 2,110.50 -1.50
68 31.00 2,108.00 -2.50
69 30.50 2,104.50 -3.50
70 30.00 2,100.00 -4.50

Yes, when you multiplied by 2, this is what happened:

y = -0.5x^2 + 35x + 1500

2y = - x^2 + 70x + 3000

And you proceeded, giving 2y = -(x+35)^2 + 4225

You should have then looked for y, giving

y = -\frac{1}{2}(x+35)^2 + \frac{4225}{2}

Apart form that mistake, all was good, well done! :)