# Problem 3 of MIDTERM Page 288-289 #9 Group 6 Ritchie Roi Chua Ramon Jose Miguel Naguiat Mary Jane L. Paglumotan Mark Anthony Salazar Mark Anthony Salazar.

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Problem 3 of MIDTERM Page 288-289 #9 Group 6 Ritchie Roi Chua Ramon Jose Miguel Naguiat Mary Jane L. Paglumotan Mark Anthony Salazar Mark Anthony Salazar Charlotte Tang Tong Ya(Nicole)

Problem 3 (Page 289 #9)  A research organization made a survey of the listening habits of the residents in a certain population between 4pm and midnight. The survey reveals that the fraction of the adult population that listens to a particular radio station X hours after 4pm is given by   F(X)= -2X 3 + 27X 2 -108X + 300 A. A.At what time between 4pm and midnight are the most people listening to the station? B. B.At what time between 4pm and midnight are the least people listening to the station?

Solution  F(X)=-2X 3 +27X 2 -108X+300 to get the critical points: F’(X)= -6X 2 +54X-108 F’(X)= 0 -6X 2 +54X-108=0 6X 2- 54X+108=0 6(X 2- 9X+18)=0 X 2- 9X+18=0 (X-3)(X-6)=0 X=3 and X=6 at X=3,Y=192 at X=6, Y=165

Solution  To Check whether X=3 and X=6 are min/max points:  F”(X)=-12X+54 At X=6: F”(X)=-12(6)+54=-18 F”(X)<0 therefore F(X) is a MAXIMUM at X=6 At X=3: F”(X)=-12(3)+54=18 F”(X)>0 therefore F(X) is a MINIMUM at X=3

Conclusion A. The time wherein the MOST people are listening to the station is 10 pm =4pm+X hours =4+6=10pm B. The time wherein the LEAST people are listening to the station is 7 pm =4pm+X hours =4+3=7pm

Graph X=-2X 3 +27X 2 -108X+300 X = 3 ; 7pm X = 6 ; 10 pm

Web Links  http://www.math.hawaii.edu/~lee/calculus/ max-min.pdf http://www.math.hawaii.edu/~lee/calculus/ max-min.pdf http://www.math.hawaii.edu/~lee/calculus/ max-min.pdf  http://www.math.com/tables/derivatives/ext rema.htm http://www.math.com/tables/derivatives/ext rema.htm http://www.math.com/tables/derivatives/ext rema.htm  http://astro.temple.edu/~dhill001/maxmin/ maxmin.html http://astro.temple.edu/~dhill001/maxmin/ maxmin.html http://astro.temple.edu/~dhill001/maxmin/ maxmin.html

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