OHaganBooks.com is now offering a wide range of online books. Demand for The Secret Loves
of John Q, a romance novel by Margo Dufon that flopped after two weeks on the market, is
given by
q = −2 p2 + 5 p + 6 (0 ≤ p ≤ 3.3)
copies sold per week when the price is p dollars. Taking into account storage and shipping, it
costs OHaganBooks.com
C = 3q
dollars to sell q copies of The Secret Loves of John Q in a week. Please help me determine the
price OHaganBooks should charge to obtain the largest weekly revenue and determine the
largest weekly revenue. At this price, is the cost increasing or decreasing? What does this tell
you about the profit?
I am also interested in determining the price to charge for the maximum weekly profit. What is
the maximum weekly profit? Please explain any difference in the price that gives the maximum
revenue and the price that gives the maximum profit.
I would like your typed report, 2 – 3 pages in length, in two weeks. Include in the report all of
the necessary details so I fully understand your solutions. In addition to answering my questions,
you must show all of your calculations and include an explanation of how you arrived at your
answers. You must use calculus techniques and convince me that you have found the maximum
revenue and maximum profit. Your solution needs to be clear, variables need to be defined and
any equations used should be clearly explained.
Categorized | Affiliate Marketing 101
Using Calculus Techniques Can Anyone Help Me With The Problem And Show Me How?
Posted on April 17, 2011.
q = – 2 p^2 + 5 p + 6 … where (0 ≤ p ≤ 3.3) <== copies sold revenue = copies sold * price each ... (q * p) Let revenue = R R = - 2 p^3 + 5p^2 + 6p . . . R is maximized when R ' = 0 R ' = - 6 p^2 + 10p + 6 - 6 p^2 + 10p + 6 = 0 p = - 0.46 <=== invalid p = 2.13504 ... assume you round to 2.14 R (2.14) = -2 * 2.14^3 + 5*2.14^2 + 6*2.14 Revenue (max) = 16.14 per week when p = 2.14 ******* Cost = 3q = 3 (- 2 p^2 + 5 p + 6) C = - 6 p^2 + 15p + 18 C(2.14) = 22.62 <== ouch, costs are more than revenue C ' = -12 p + 15 C ' (2.14) = - 10.68 <=== negative value indicates costs are decreasing. This shows It may be more profitable to raise the price and sell fewer copies, but with a greater margin. ********* Profit P = revenue - costs P = R - C P = - 2 p^3 + 5p^2 + 6p - (- 6 p^2 + 15p + 18) P = -2 p^3 + 11 p^2 - 9 p - 18 . . . P is maximized when P ' = 0 P ' - 6p^2 + 22p - 9 - 6p^2 + 22p - 9 = 0 p = 0.47 p = 3.20 <== both within limits. One is a maximum, the other is a minimum . . . second derivative test to determine max or min P ' ' = -12p + 22 P ' ' (0.47) = -12 * 0.47 + 22 = 16.46 . . . positive value indicates p=0.47 is a minimum P ' ' (3.20) = - 12*3.2 + 22 = - 16.4 . . . negative value indicates p=3.20 is a maximum You should price the book at 3.20 for maximum profit. P (3.20) = 0.30 per week ... (flop is an understatement)