# Thread: maximizing profit

1. ## 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

2. 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

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

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