# Please help with this revenue and profit maximization Calculus problem.

• Nov 20th 2009, 03:23 AM
pcapone23
Please help with this revenue and profit maximization Calculus problem.
Demand for the latest best selling book is given by q= -p^2 + 33p + 9 (18< or equal to p < or equal to 28) copies sold per week when the price is p dollars. Can you help me determine what price to charge to obtain the largest revenue? It costs C= 9q+100 dollars to sell q copies of this book in a week. What price should the company charge to get the largest weekly profit? What is the maximum weekly profit and how can you be sure that the profit is maximized?

Must use calculus.

Thank you
• Nov 20th 2009, 05:29 AM
pcapone23
• Nov 20th 2009, 05:39 AM
HallsofIvy
Quote:

Originally Posted by pcapone23
Demand for the latest best selling book is given by q= -p^2 + 33p + 9 (18< or equal to p < or equal to 28) copies sold per week when the price is p dollars. Can you help me determine what price to charge to obtain the largest revenue? It costs C= 9q+100 dollars to sell q copies of this book in a week. What price should the company charge to get the largest weekly profit? What is the maximum weekly profit and how can you be sure that the profit is maximized?

Must use calculus.

Thank you

You got upset because no one had responded within 2 hours? We do have lives, you know!
Revenue is, of course, price per book times number of books. Here that is pq= p(-p^2+ 33p+ 9)= -p^3+ 33p+ 9. Set the derivative of that equal to 0 and solve for p.

Profit is revenue minus costs: Profit= -p^3+ 33p+ 9- (9q+100)= -p^3+ 33p+ 9- 9(-p^2+ 33p+ 9)- 100= -p^3+ 9p^2- 265p- 172. Again, set the derivative of that equal to 0 and solve for p. To see if that really maximizes the profit, see what happens if you change the price by, say, 1 penny.
• Nov 20th 2009, 05:53 AM
pcapone23
Quote:

Originally Posted by HallsofIvy
You got upset because no one had responded within 2 hours? We do have lives, you know!
Revenue is, of course, price per book times number of books. Here that is pq= p(-p^2+ 33p+ 9)= -p^3+ 33p+ 9. Set the derivative of that equal to 0 and solve for p.

Profit is revenue minus costs: Profit= -p^3+ 33p+ 9- (9q+100)= -p^3+ 33p+ 9- 9(-p^2+ 33p+ 9)- 100= -p^3+ 9p^2- 265p- 172. Again, set the derivative of that equal to 0 and solve for p. To see if that really maximizes the profit, see what happens if you change the price by, say, 1 penny.

I wasnt getting upset I was just trying to bump the thread, but thanks alot. I was trying to solve the maximizing revenue equation exactly how you said but I keep getting screwy numbers. Are you sure that p(-p^2+ 33p+ 9)= -p^3+ 33p+ 9 because I get -p^3+33p^2+9p. I may be wrong though.
• Nov 20th 2009, 08:18 AM
tom@ballooncalculus
No, that is a typo - follow the advice but with your pq.