How do you find the price for a ticket that will maximize revenue using quadratic equations?
The question is:
The city bus company usually transports 12 000 rides per day at a ticket price of $1. The company wants to raise the ticket price and knows that for every $0.10 increase the number of riders decreases by 400.
I figured it up to: y = 12000 + 800x - 40^2, but how do I find the the ticket price?
riders = 12,000 - 400 * increase / 0.1
revenue = riders(1 + increase)
substitue riders into bottom equation:
revenue = (12,000 - 4000 * increase )(1 +increase)
= 12000 + 8000 i-4000 i ^2
take the first derivative which is f ‘ (i) = 8000-8000i, solve for zero, that is either a max or min, here is a max, i = 1 dollar, check to see if it makes sense in the equation and if it is a local max. good luck
riders = 12,000 - 400 * increase / 0.1
revenue = riders(1 + increase)
substitue riders into bottom equation:
revenue = (12,000 - 4000 * increase )(1 +increase)
= 12000 + 8000 i-4000 i ^2
take the first derivative which is f ‘ (i) = 8000-8000i, solve for zero, that is either a max or min, here is a max, i = 1 dollar, check to see if it makes sense in the equation and if it is a local max. good luck
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