Michael J. answered 12/24/15
Tutor
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Mastery of Limits, Derivatives, and Integration Techniques
Lets say you want to sell a number of products. A certain number of products will get you a maximum profit. If you know your maximum profit, you can use that information to create a derivative equation that equals zero. Then you would find the anti-derivative of that to find your profit for each amount of products sold.
For example:
Maximum profit is $500 at 20 chocolate bars are sold.
Create a derivative equation in terms of x that equal zero. When you solve for x, you should get 20.
-200(x - 20)(x + 10) = 0
Expand the left side of the equation.
-200(x2 - 10x - 200) = 0
-200x2 + 2000x + 40000 = 0
Next, find the anti-derivative of the left side.
P(x) = -66.67x3 + 1000x2 + 40000x + C
where C is any constant.
The value of P(x) is the profit earned for each number of chocolate bars sold.
Finally, we solve for C by plugging in x=20 and P(x)=500
500 = -66.67(20)3 + 1000(20)2 + 40000(20) + C
500 = 666640 + C
- 666140 = C
P(x) = -66.67x3 + 1000x2 + 40000x - 666140