I need help with some calculus?

Question

You operate a tour service that offers the following rates:

$200 per person if 50 people (the min number to book the tour) go on the tour

For each additional person, up to a maximum of 80 people total, the rate per person is reduced by $2.

It costs $6000 (a fixed cost) plus $32 per person to conduct the tour. How many people does it take to maximize your profit?

Answer 1

We're working on the domain [50,80].

For that domain, the rate per person is 200 - 2(x-50) = 300 - 2x

Your total revenue is the rate per person times the number of people: (200 - 2x)x = 300x - 2x^2.

Your profit is the revenue minus the total cost. Here, the cost has two components, a fixed cost and a variable cost.

The total cost is 6000 + 32x.

Therefore, your profit function is (300x - 2x^2) - (6000 + 32x) =

f(x) = -6000 + 268x - 2x^2.

To find the extrema of this function, take the first derivative and set it equal to zero. This isn't rigorous, but since f(x) is a parabola that points down, it will have a single extremum that must be a maximum. (You'd really have to use the second derivative test to verify this.)

f'(x) = 268 - 4x = 0

268 = 4x

x = 67

With 67 people, the rate is $166. The revenue is $11122.

The costs are $6000 + 32*67 = $8144.

The profit is $2978.

Note that the profit at x = 50 is $10000 - $7600 = $2400.

The profit at x = 80 is $11200 - $8560 = $2640.

This doesn't prove that we've found the maximum. But given what we know about f(x) - namely, that it's a parabola that points downward, we should expect that it's rising slowly enough for its vertex to be at an x-coordinate greater than x = 80, or that it peaks at a point between x = 50 and x = 80 that is closer x = 50 than it is to x = 80.

Just as a second informal check, when x = 65, the revenue is $11050 and the costs are $8080, giving a profit of $2970. Clearly, we're approaching the maximum that occurs at x = 67.

Answer 2

Lets say you have x people (50 <= x <= 80).
Now, we need to know the cost per person. If we have x=50, we get 200. Every extra person decreases that by 2. Thus, we have 200 - 2*(number of people less than 50), ie 200-2(x-50) = 300-2x.
So the total revenue is number of people * cost per person = x(300-2x) = 300x - 2x^2.

So to get the total *profit*, subtract the cost: (300x - 2x^2) - (6000+32x) = -2x^2 +268x - 6000.

Differentiate and make equal to 0:
-4x + 268 = 0
x = 67. Since that was a negative quadratic, we know this must be a maximum (in other cases, you may have had to check the endpoints x=50 or 80, since maximums can also occur there).

So 67 people for a profit of $2978.

Answer 3

Let x = the number of people on the tour in excess of 50
Revenue is 50(200) +198 x [0 =< x =<30}
Costs are 6000 + (50+x)32
Profit= 10000+198 x-6000 - 50*32 - 32x
= 2400 + 136x
Thus profit is maximized when 80 people are on board.

Perhaps I misunderstood the question because no calculus was required to solve this problem.

Perhaps you meant that the price for the 51st person was $198, $196 for the 52nd person, $194 for the 53rd person ........ etc?

Answer 4

I'll have to agree with Stephen M's response and not bore you with an almost exact response.