Course: 23/SU Calculus 1 (MATH X VA MATH180: HW21- Sec 4.7 (F202 x M Inbox (111) - myassine3@hawkr x Open orders - EssayPro.com * *Course Hero x + X - C webassign.net/web/Student/Assignment-Responses/last?dep=32016420#Q14 5th GradeAlpha.pdf VitalSource Booksh.. bank mobile FA Order 3570395 - Es. b Success Confirmati V Z-Library Psychology in Ever. = Precalculus 2e- Op.. XYZ Homework - H. PB Technical and Prof. My Course- Macroe. MindTap - Cengage. 14. [6/9 Points] DETAILS PREVIOUS ANSWERS SCALCET9 4.7.AE.006. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Example 6 Video Example () A store has been selling 300 TV monitors a week at $450 each. A market survey indicates that for each $10 rebate offered to buyers, the number of monitors sold will increase by 20 a week. Find the demand function and the revenue function. How large a rebate should the store offer to maximize its revenue? Solution If x is the number of monitors sold per week, then the weekly increase in sales is 300 - x For each increase of 20 units sold, the price is decreased by $10. So for each additional unit sold, the decrease in price will be - 20 - x 10 and the demand function is X p(x) = 450 - 10/20 (x- 300) = 600 - 300x x The revenue function is R(x) = xp(x) = 600x - X Since R'(x) = 600 - x , we see that R'(x) = 0 when x = 600 . This value of x gives an absolute maximum by the First Derivative Test (or simply by observing that the graph of R is a parabola that opens downward). The corresponding price is p(600) = 300 and the rebate is 450 - 300 = 150 . Therefore, to maximize revenue, the store should offer a rebate of $ 150 Need Help? Read It Submit Answer Home My Assignments + Request Extension Copyright @ 1998 - 2023 Cengage Learning, Inc. All Rights Reserved | TERMS OF USE PRIVACY US _ Aug 4 10:29 +