# Calculus

with the equation
q=-p^2+33p+9 (18 less than or equal to p less than or equal to 28)
how do you determine the price to obtain the largest revenue?

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1. It's not factorable, so use a graphic calculator. Observe the graph. (Where is it highest?) Then, type that x-coordinate into the table of your graph to get the largest revenue (y).

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2. i don't do anything with take a derivative?

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3. Oh yea. You have to find the derivate and then follow those steps. Good eye. :)

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4. derivative*

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5. what about the inequality?

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6. i substituted the q with the equation with r=p*q but i don't kno what to do after that

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7. Don't worry about the inequality. To begin, simply find the derivative of:

q=-p^2+33p+9

(Use the power rule.)

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8. ok
q'=-2p+33

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9. Are you sure that q=-p^2+33p+9 isn't our derivative? That makes much more sense.

(With our current derivative, it's not profitable at all. We're losing money.)

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10. nope the paper says the demand for the latest best selling magazine is given by q=-p^2+33p+9 (18 less than or equal to p less than or equal to 28) copies sold per week when the price is p dollars. What is the price the company should charge to obtain the largest revenue? Taking into account storage and shipping cost the company C=9q+100 dollars to sell q copies of the magazine. What price should the company charge to get the largest weekly profit? What is the maximum possible weekly proft and how can you be certain the profit is maximized?

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11. To get the largest revenue, you don't need the derivative then. (Sorry about that!) Type the original equation into the graphing calc to see where it is highest. Set your window to Xmin=18, Xmax=28. Then, type that x-coordinate into your Table to get the y-coordinate.

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12. what about the cost function?

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13. I'm not 100% sure on this, but try graphing both equations and looking at where they intersect.

The x-coordinate is the price, and the y-coordinate is the max profit.

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