drsparnum wrote:
I was hoping for some help with this problem from the quantitative practice set. Here is the problem...
The Go-Go Bus Line provides rides between New York and Boston. The bus line carries 18,000 passengers per month, each paying $50 for round trip. The company wants to raise the fare. However, the market research department estimates that for each $1 increase in fare, the bus line will lose 200 passengers per month. The total monthly cost of the bus line is $950,000.
What is the profit function f(x), where x is the dollar amount of increase in fare?
Since this is in the practice set the answer is also available just below. The answer is....
f(x) = (50 + x)(18,000 - 200x) - 950,000 = - 200x^2 + 8,000x - 50,000
So how is that function derived? Is that a standard formula? Yes, after testing it at a few points I could see that it works. However, I could not come up with that formula before looking at the answer. While working through it, I was very tempted to make a "revenue" term and "cost" term, which would have necessitated the introduction of a new variable (p, for passangers). Long story short, I was obviously on the wrong track there.
Not a standard formula - it's just derived from the question... profit is the revenue minus the cost of the bus... so you multiply the ticket price times number of passengers, and subtract cost. Think of doing the calculation without a variable fare... then you have to modify to allow for x... I'm sure this doesn't answer your question fully... but a quick half-response