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Mathematics 7 Online
OpenStudy (precal):

Maximum Profit (Optimization - Calculus)

OpenStudy (anonymous):

how 'bout maximum time to write out a problem?

OpenStudy (precal):

City Cyles Inc finds that it cost $70 to manufacture each bicycle, and fixed costs are $100 per day. The price function is p(x)=270-10x, where p is the price in dollars at which x bicycles will be sold. Find the quantity City Cyles should produce and the price it should charge to maximinze profit. Also, find the maximum profit.

OpenStudy (anonymous):

wouldn't C(x) = 70x + 100 because fixed costs are 100 and it costs $70 for each bicycle?

OpenStudy (precal):

yes you are correct

OpenStudy (precal):

I do have the solution

OpenStudy (anonymous):

Profit = Total Revenue - Total Cost Profit = P(x) - C(x) Profit = 270 - 10x - (70x + 100) Profit = f = 170 - 80x Profit is maximized when x = 0, because if x is greater than 0, then profit is less than 170. Maximum profit is 170. Using calculus, derivative of profit is -80, so you have to check the endpoints for profit, 0 and 170/80 = 2.125. And 0 is the maximum out of those two endpoints.

OpenStudy (precal):

I have 10 bikes per day at $170 per bike, max profit is 900

OpenStudy (anonymous):

is your price function right? P(x) = 270 - 10x ?

OpenStudy (precal):

yes it was given in the problem

OpenStudy (precal):

Revenue is price times quantity, R=p times x

OpenStudy (anonymous):

oh

OpenStudy (precal):

R(x)=(270-10x)x

OpenStudy (anonymous):

ok so Profit = R(x) - C(x)

OpenStudy (precal):

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