Thread: Optimization

Originally Posted by Truthbetold

You operate a tour service that offers the following rates:

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

For each addiational 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 the profit?




I know profit= revenue- cos

I also know cost= 6000 - 32x.

But I don't know the revenue in an equation. From there, I think I can take t what is the revenue?

You know you start out with 200 per person
f(x)=200x

When x > 50, the price will decrease $2 per person over the total of 50 people. So if you have 52, then you will decrease $2 * 2 people. So to find out how many people over 50 you have, you must subtract 50 from the total number of people. Then that number will be multiplied by negative 2
f(x)=200x - 2(x-50)

And since that is the difference for 1 person, we must multiply it by x to find the difference for all the people
f(x)=200x - 2(x-50)x

Since it also costs $32 per person to run the tour, you must subtract that amount of money as well
f(x)=200x-2(x-50)x-32x

simplify
f(x)=200x+(-2x+100)x-32x

simplify
f(x)=200x-2x^2+100x-32x

Simplify
f(x)=268x-2x^2

Now, that should be the total income based on how many people are coming (the fixed cost should not affect it, as the profit is the income - the expenses, the maximum of the income will be the same as the maximum of the profit)

So we just have to find where our formula is maximized. To do this, consider how it looks on a graph, that the maximum income will be the point where the graph is highest, which is also the point where it will turn around and head back down (unless the maximum profit is at x=50 or 80). So at this point, it's slope will be exactly horizontal. Consider what the derivative tells you, and use that to find the point where profit is maximized. (then also check your endpoints to make sure they aren't greater)

For a reference, I got x=67 for maximum.