Wikipedia:Reference desk/Archives/Mathematics/2007 December 13

= December 13 =

How do I solve this optimization problem with calculus?
For groups of 80 or more, a charter-bus company determines the rate per person according to the formula rate = $8 -$0.05(n - 80), n ≥ 80. What number of passengers will give the bus company maximum revenue? Thanks, anon. —Preceding unsigned comment added by 70.19.24.152 (talk) 02:13, 13 December 2007 (UTC)


 * You want to know when the discount given, meets the price of the ticket. In other words when: 8 - 0.05(n-80) = 0.05(n-80) 8 = 0.10(n-80) 80 = (n-80) 80 + 80 = n = 160. lol oops forgot the with claculus bit. You need to find when the derivate of the function is zero--Dacium (talk) 02:36, 13 December 2007 (UTC)


 * Firstly, you need your revenue function. If I understand your problem correctly, this is $$rate(n) \times n$$. This turns to be a quadratic function; hence, it has only one point where the derivative is zero (a critical point). This is a candidate for the revenue maximizer (the number of passengers that make the profits maximum). This is indeed the case. You may confirm this with, for instance, the second derivative test, that renders here a negative value (maximum). Maxima and minima is a good reference for you to take a look at. Pallida  Mors  05:17, 13 December 2007 (UTC)


 * Dacium's calculation is incorrect. You need to compare the price of a ticket with the marginal discount received from the extra passenger. Each passenger decreases the price per ticket by 0.05$, therefore the marginal discount is $0.05n. You can solve $$8-0.05(n-80)=0.05n$$. -- Meni Rosenfeld (talk) 09:48, 13 December 2007 (UTC)


 * Yes Meni is correcet, it comes out to n=120 at exactaly $6 per person. The full working if solving via derivative is: n*rate = cost n*(8-0.05(n-80)) = cost 12n - 0.5n^2 = cost derivative is: 12 - 0.1n = 0 n=120.--Dacium (talk) 00:34, 14 December 2007 (UTC)


 * Anyway, any bus company which uses this formula will quickly go bankrupt, since 240 passengers will travel for free, and even less passengers will pay a significantly lower sum than the cost to the company. -- Meni Rosenfeld (talk) 09:55, 13 December 2007 (UTC)


 * This is of course a stylized version of an economic problem. Rate function here is the inverse demand function. Such functions should not be linear, although linearity makes for very easy optimization problems. This is an issue of marginal decisions (number of passengers), albeit non-marginal questions like overall profit (as Meni points out) are also to be taken into account (information on costs is essential to such analysis, though). Pallida  Mors  15:01, 13 December 2007 (UTC)
 * Yes, it is clear to me that the problem was oversimplified to be suitable as a homework question. I am only somewhat annoyed with the tendency to give a problem in a "real world" setting on one hand, while making it totally unrealistic on the other hand. -- Meni Rosenfeld (talk) 21:18, 13 December 2007 (UTC)
 * A smarter discount formula is y=b&middot;xa where a is a constant, 0<a<1, and b=rate(1), 0<b. This formula gives positive rates y=rate(x) for all positive amounts x, and the ratio between the marginal price dy/dx and the average price y/x is the constant a: (dy/dx)/(y/x)=(dy/y)/(dx/x)=(d log y)/(d log x)=a. Bo Jacoby (talk) 08:57, 15 December 2007 (UTC).