# maximizing profit

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• November 28th 2008, 03:19 PM
xSMUxCHEERLEADINGx
maximizing profit
Use calculus to find the production level x that will maximize profit if the price (also known as demand) is p(x)=12-(x/500) and the cost is C(x)=680+4x+0.01x^2

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• November 29th 2008, 11:33 AM
CaptainBlack
Quote:

Originally Posted by xSMUxCHEERLEADINGx
Use calculus to find the production level x that will maximize profit if the price (also known as demand) is p(x)=12-(x/500) and the cost is C(x)=680+4x+0.01x^2

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${\rm{Profit}=Revenue-Costs}=xp(x)-C(x)=12x-\frac{x^2}{500}-680-4x-0.01x^2$ $=-0.012x^2+8x-640$

So to find the production that maximises profit we differentiate the quadratic on the right above and set the derivative to zero and solve for x.

CB