Question

How should ticket prices be set to maximize revenue? (Round your answer to the nearest cent.) The equation is p(x)=(58000-x)/(2000).

Answer 1

I times the whole thing by x, get derivative, and set equal to zero. But it's wrong. what am i doing wrong?

Answer 2

is p(x) function of revenue

Answer 3

or p(x) is the price?

Answer 4

while x is the quantity?

Answer 5

Well , no need to guess; need respond if he/she needs help

Answer 6

Usually, p(x) is the price function, which is the price of some item where x is the quantity sold


The revenue function r(x) you want to maximize is then


r(x) = x*p(x)

Answer 7

For finding the answer, set x*p(x) = 0
the substitute that x value to the equation:
(58000-x)/(2000).

Answer 8

yes you guys are right sorry this thing lags for me p(x) is price function.

Answer 9

x is not your final answer, you need to substitue x back to the original equation

Answer 10

oh wait...i get 58000 then substitute it back in and 0/2000?

Answer 11

that's wrong you should get 29000 as x

Answer 12

oh...hmm